Chapter 12 Introduction to Three Dimensional Geometry (Additional Questions)

Solution:

A point in three-dimensional space is represented by three coordinates, which specify its position relative to the origin along the three mutually perpendicular axes (usually denoted as the x, y, and z axes).

Let the coordinates along the x-axis, y-axis, and z-axis be $x$, $y$, and $z$ respectively.

The coordinates of a point in three-dimensional space are therefore represented as an ordered triplet $(x, y, z)$.

Looking at the options provided:

• (A) $(x, y)$ represents a point in two-dimensional space.
• (B) $(x, y, z)$ represents a point in three-dimensional space.
• (C) $(r, \theta)$ represents a point in two-dimensional polar coordinates.
• (D) $(x, y, 0)$ represents a point in three-dimensional space, but specifically one that lies on the xy-plane (where the z-coordinate is zero). While it is a point in 3D space, option (B) represents the general case for *any* point in 3D space.

Therefore, the correct option is (B).

Answer:

(B) $(x, y, z)$

Solution:

In a three-dimensional Cartesian coordinate system, points are represented by coordinates $(x, y, z)$.

The xy-plane is a specific plane in this system.

It consists of all points where the position along the z-axis is zero.

Geometrically, it is the plane formed by the x and y axes.

Any point $(x, y, z)$ lies in the xy-plane if and only if its z-coordinate is equal to 0.

So, a point in the xy-plane has the form $(x, y, 0)$, where $x$ and $y$ can be any real values.

Looking at the given options:

• (A) A positive value: Incorrect. Only points with $z > 0$ are above the xy-plane.
• (B) A negative value: Incorrect. Only points with $z < 0$ are below the xy-plane.
• (C) 0: Correct. Points in the xy-plane have $z=0$.
• (D) Any real value: Incorrect. The z-coordinate must be specifically 0 for a point to be on the xy-plane.

Therefore, the z-coordinate of a point in the xy-plane is always 0.

Answer:

(C) 0

Given:

Point 1: $P_1 = (1, -2, 3)$

Point 2: $P_2 = (4, 3, -1)$

To Find:

The distance between $P_1$ and $P_2$.

Solution:

Let the coordinates of $P_1$ be $(x_1, y_1, z_1)$ and the coordinates of $P_2$ be $(x_2, y_2, z_2)$.

So, $x_1 = 1$, $y_1 = -2$, $z_1 = 3$ and $x_2 = 4$, $y_2 = 3$, $z_2 = -1$.

The distance between two points in three-dimensional space is given by the formula:

Substitute the coordinates of $P_1$ and $P_2$ into the formula:

Distance $= \sqrt{(4 - 1)^2 + (3 - (-2))^2 + (-1 - 3)^2}$

Distance $= \sqrt{(3)^2 + (3 + 2)^2 + (-4)^2}$

Distance $= \sqrt{(3)^2 + (5)^2 + (-4)^2}$

Distance $= \sqrt{9 + 25 + 16}$

Distance $= \sqrt{50}$

We can simplify $\sqrt{50}$:

$\sqrt{50} = \sqrt{25 \times 2} = \sqrt{25} \times \sqrt{2} = 5\sqrt{2}$

So, the distance is $\sqrt{50}$ or $5\sqrt{2}$.

Now, let's compare our result with the given options:

• (A) $\sqrt{25} = 5$
• (B) $\sqrt{36} = 6$
• (C) $5\sqrt{2}$
• (D) $\sqrt{50}$

Our calculated distance is $\sqrt{50}$, which matches option (D).

Also, our calculated distance simplifies to $5\sqrt{2}$, which matches option (C).

Both options (C) and (D) represent the correct distance between the given points.

Answer:

Both (C) $5\sqrt{2}$ and (D) $\sqrt{50}$ are correct.

Given:

Point 1: $P_1 = (x, y, z)$

Point 2: The Origin $O = (0, 0, 0)$

To Find:

The distance between the point $(x, y, z)$ and the origin $(0, 0, 0)$.

Solution:

Let the coordinates of $P_1$ be $(x_1, y_1, z_1) = (x, y, z)$ and the coordinates of the Origin $O$ be $(x_2, y_2, z_2) = (0, 0, 0)$.

The distance formula between two points $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$ in three-dimensional space is:

Distance $= \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}$

Substitute the coordinates of the point $(x, y, z)$ and the origin $(0, 0, 0)$ into the formula:

Distance $= \sqrt{(0 - x)^2 + (0 - y)^2 + (0 - z)^2}$

Distance $= \sqrt{(-x)^2 + (-y)^2 + (-z)^2}$

Distance $= \sqrt{x^2 + y^2 + z^2}$

This is the formula for the distance of a point $(x, y, z)$ from the origin.

Compare this result with the given options:

• (A) $\sqrt{x^2 + y^2}$: This is the distance from the origin in 2D space (or the distance from the z-axis in 3D).
• (B) $\sqrt{x^2 + y^2 + z^2}$: This matches our calculated distance.
• (C) $|x| + |y| + |z|$: This is the Manhattan distance or L1 norm, not the standard Euclidean distance.
• (D) $\sqrt{(x+y+z)^2} = |x+y+z|$: This does not represent the distance from the origin.

The correct formula for the standard Euclidean distance from the origin in 3D space is $\sqrt{x^2 + y^2 + z^2}$.

Answer:

(B) $\sqrt{x^2 + y^2 + z^2}$

Given:

Point P: $P(1, 2, 3)$

Point Q: $Q(4, 5, 6)$

Ratio of internal division: $m:n = 1:2$

To Find:

The coordinates of the point that divides the line segment PQ internally in the ratio $1:2$.

Solution:

Let the coordinates of point P be $(x_1, y_1, z_1) = (1, 2, 3)$ and the coordinates of point Q be $(x_2, y_2, z_2) = (4, 5, 6)$.

Let the ratio of internal division be $m:n = 1:2$.

The coordinates of a point $(x, y, z)$ that divides the line segment joining $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$ internally in the ratio $m:n$ are given by the Section Formula in three dimensions:

Substitute the given values $x_1=1, y_1=2, z_1=3$, $x_2=4, y_2=5, z_2=6$, $m=1$, and $n=2$ into the formula:

For the x-coordinate:

$x = \frac{(1)(4) + (2)(1)}{1 + 2}$

$x = \frac{4 + 2}{3}$

$x = \frac{6}{3}$

$x = 2$

For the y-coordinate:

$y = \frac{(1)(5) + (2)(2)}{1 + 2}$

$y = \frac{5 + 4}{3}$

$y = \frac{9}{3}$

$y = 3$

For the z-coordinate:

$z = \frac{(1)(6) + (2)(3)}{1 + 2}$

$z = \frac{6 + 6}{3}$

$z = \frac{12}{3}$

$z = 4$

The coordinates of the point that divides the line segment PQ internally in the ratio $1:2$ are $(2, 3, 4)$.

Comparing this result with the given options:

• (A) $(2, 3, 4)$: This matches our calculated coordinates.
• (B) $(\frac{1 \cdot 4 + 2 \cdot 1}{1+2}, \frac{1 \cdot 5 + 2 \cdot 2}{1+2}, \frac{1 \cdot 6 + 2 \cdot 3}{1+2})$: This is the formula expression which evaluates to $(2, 3, 4)$. While technically representing the coordinates, option (A) provides the final simplified values.
• (C) $(3, 4, 5)$: Incorrect.
• (D) $(2, 4, 6)$: Incorrect.

The most direct answer representing the coordinates is $(2, 3, 4)$.

Answer:

(A) $(2, 3, 4)$

Given:

Point 1: $A(2, -1, 3)$

Point 2: $B(4, 5, -1)$

To Find:

The coordinates of the midpoint of the line segment AB.

Solution:

Let the coordinates of point A be $(x_1, y_1, z_1) = (2, -1, 3)$ and the coordinates of point B be $(x_2, y_2, z_2) = (4, 5, -1)$.

The coordinates of the midpoint $(x, y, z)$ of a line segment joining $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$ in three dimensions are given by the Midpoint Formula:

Substitute the given values $x_1=2, y_1=-1, z_1=3$ and $x_2=4, y_2=5, z_2=-1$ into the formula:

For the x-coordinate:

$x = \frac{2 + 4}{2}$

$x = \frac{6}{2}$

$x = 3$

For the y-coordinate:

$y = \frac{-1 + 5}{2}$

$y = \frac{4}{2}$

$y = 2$

For the z-coordinate:

$z = \frac{3 + (-1)}{2}$

$z = \frac{3 - 1}{2}$

$z = \frac{2}{2}$

$z = 1$

The coordinates of the midpoint are $(3, 2, 1)$.

Comparing this result with the given options:

• (A) $(3, 2, 1)$: This matches our calculated coordinates.
• (B) $(\frac{2+4}{2}, \frac{-1+5}{2}, \frac{3+(-1)}{2})$: This is the formula expression, which evaluates to $(3, 2, 1)$.
• (C) $(3, -2, 1)$: Incorrect y-coordinate.
• (D) $(3, -2, 2)$: Incorrect y and z coordinates.

The correct coordinates are $(3, 2, 1)$.

Answer:

(A) $(3, 2, 1)$

Evaluation of Assertion (A):

The formula for the distance between two points $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$ in three-dimensional space is indeed given by $\sqrt{(x_2-x_1)^2 + (y_2-y_1)^2 + (z_2-z_1)^2}$.

Thus, Assertion (A) is True.

Evaluation of Reason (R):

The distance formula in three dimensions is derived using the Pythagorean theorem. Consider the difference in coordinates along each axis: $\Delta x = x_2 - x_1$, $\Delta y = y_2 - y_1$, and $\Delta z = z_2 - z_1$.

The distance between the two points can be seen as the length of the hypotenuse of a right triangle in 3D space. If we project the line segment onto the xy-plane, the length of the projection is $\sqrt{(\Delta x)^2 + (\Delta y)^2}$ by the Pythagorean theorem in 2D.

Now, consider a right triangle where the base is this projection in the xy-plane and the height is the vertical difference $|\Delta z|$ along the z-axis. The distance between the points in 3D is the hypotenuse of this triangle. Applying the Pythagorean theorem again:

Distance$^2 = (\text{Projection Length})^2 + (\Delta z)^2$

Distance$^2 = (\sqrt{(\Delta x)^2 + (\Delta y)^2})^2 + (\Delta z)^2$

Distance$^2 = (\Delta x)^2 + (\Delta y)^2 + (\Delta z)^2$

Distance $= \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}$

Therefore, the distance formula is indeed an application of the Pythagorean theorem extended to three dimensions.

Thus, Reason (R) is True.

Relationship between A and R:

Both Assertion (A) and Reason (R) are true, and Reason (R) correctly explains the origin and derivation of the distance formula stated in Assertion (A).

Conclusion:

Based on the evaluation, both A and R are true, and R is the correct explanation of A.

Answer:

(A) Both A and R are true and R is the correct explanation of A.

Solution:

In a three-dimensional Cartesian coordinate system, the coordinate planes are formed by setting one of the coordinates to zero.

The yz-plane is the plane that contains the y-axis and the z-axis.

For any point to lie in the yz-plane, its distance from the yz-plane must be zero.

The distance of a point $(x, y, z)$ from the yz-plane is given by $|x|$.

Therefore, for a point to be in the yz-plane, the x-coordinate must be 0.

The y and z coordinates can take any real values.

So, the coordinates of a point in the yz-plane are of the form $(0, y, z)$.

Let's examine the options:

• (A) $(x, y, z)$: Represents a general point in 3D space.
• (B) $(0, y, z)$: Represents a point where the x-coordinate is 0, which is the definition of a point in the yz-plane.
• (C) $(x, 0, z)$: Represents a point in the xz-plane.
• (D) $(x, y, 0)$: Represents a point in the xy-plane.

The correct form of coordinates for a point in the yz-plane is $(0, y, z)$.

Answer:

(B) $(0, y, z)$

Solution:

We need to identify which plane or axis each given point lies on based on its coordinates.

Recall the definitions:

• A point $(x, y, z)$ lies on the xy-plane if $z=0$.
• A point $(x, y, z)$ lies on the yz-plane if $x=0$.
• A point $(x, y, z)$ lies on the xz-plane if $y=0$.
• A point $(x, y, z)$ lies on the x-axis if $y=0$ and $z=0$.
• A point $(x, y, z)$ lies on the y-axis if $x=0$ and $z=0$.
• A point $(x, y, z)$ lies on the z-axis if $x=0$ and $y=0$.

Let's examine each point:

(i) Point $(5, -2, 0)$: The z-coordinate is 0. This point lies on the xy-plane.

So, (i) matches with (c).

(ii) Point $(0, 3, -4)$: The x-coordinate is 0. This point lies on the yz-plane.

So, (ii) matches with (b).

(iii) Point $(-1, 0, 6)$: The y-coordinate is 0. This point lies on the xz-plane.

So, (iii) matches with (a).

(iv) Point $(0, 0, 7)$: The x-coordinate is 0 and the y-coordinate is 0. This point lies on the z-axis.

So, (iv) matches with (d).

The correct matching is:

• (i) - (c)
• (ii) - (b)
• (iii) - (a)
• (iv) - (d)

Comparing this with the given options, we find that option (B) corresponds to this matching.

Answer:

(B) (i)-(c), (ii)-(b), (iii)-(a), (iv)-(d)

Given:

The point is $P(3, 4, 5)$.

To Find:

The distance of the point $P(3, 4, 5)$ from the x-axis.

Solution:

The distance of a point $(x_1, y_1, z_1)$ from the x-axis is the distance from the point to its projection onto the x-axis.

The projection of the point $P(3, 4, 5)$ onto the x-axis is the point where the y and z coordinates are zero, while the x coordinate remains the same.

So, the projection of $P(3, 4, 5)$ onto the x-axis is the point $Q(3, 0, 0)$.

Now, we find the distance between $P(3, 4, 5)$ and $Q(3, 0, 0)$ using the distance formula:

Distance $= \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}$

Substitute the coordinates $(x_1, y_1, z_1) = (3, 4, 5)$ and $(x_2, y_2, z_2) = (3, 0, 0)$:

Distance $= \sqrt{(3 - 3)^2 + (0 - 4)^2 + (0 - 5)^2}$

Distance $= \sqrt{(0)^2 + (-4)^2 + (-5)^2}$

Distance $= \sqrt{0 + 16 + 25}$

Distance $= \sqrt{41}$

Alternatively, the formula for the distance of a point $(x_1, y_1, z_1)$ from the x-axis is simply $\sqrt{y_1^2 + z_1^2}$.

Using this formula with $(3, 4, 5)$, we get:

Distance $= \sqrt{4^2 + 5^2}$

Distance $= \sqrt{16 + 25}$

Distance $= \sqrt{41}$

Comparing our result with the given options:

• (A) 3
• (B) 4
• (C) 5
• (D) $\sqrt{4^2 + 5^2}$

Our calculated distance $\sqrt{41}$ matches the expression in option (D).

Answer:

(D) $\sqrt{4^2 + 5^2}$

Given:

Point A: $A(-2, 4, 7)$

Point B: $B(3, -5, 8)$

To Find:

The ratio in which the yz-plane divides the line segment AB.

Solution:

Let the yz-plane divide the line segment joining $A(x_1, y_1, z_1)$ and $B(x_2, y_2, z_2)$ in the ratio $\lambda:1$.

Here, $(x_1, y_1, z_1) = (-2, 4, 7)$ and $(x_2, y_2, z_2) = (3, -5, 8)$.

The coordinates of the point of division are given by the Section Formula:

A point lies on the yz-plane if its x-coordinate is 0.

So, the x-coordinate of the point of division must be 0.

From the section formula, the x-coordinate is $\frac{\lambda x_2 + 1 x_1}{\lambda + 1}$.

Set this equal to 0:

$\frac{\lambda x_2 + x_1}{\lambda + 1} = 0$

Substitute the values $x_1 = -2$ and $x_2 = 3$:

$\frac{\lambda(3) + (-2)}{\lambda + 1} = 0$

$\frac{3\lambda - 2}{\lambda + 1} = 0$

For this fraction to be zero, the numerator must be zero (assuming the denominator is not zero, which is true for $\lambda \neq -1$).

$3\lambda - 2 = 0$

$3\lambda = 2$

$\lambda = \frac{2}{3}$

The ratio of division is $\lambda:1 = \frac{2}{3}:1$. Multiplying by 3 to get integer ratio, we get $2:3$.

Since the value of $\lambda$ is positive ($\lambda = \frac{2}{3} > 0$), the division is internal.

The ratio is $2:3$ internally.

Comparing this result with the given options:

• (A) 2 : 3 internally: Matches our result.
• (B) 2 : 3 externally: Incorrect (ratio is positive).
• (C) 3 : 2 internally: Incorrect ratio.
• (D) 3 : 2 externally: Incorrect ratio and division type.

Answer:

(A) 2 : 3 internally

Solution:

In a three-dimensional Cartesian coordinate system, points are located using three coordinates $(x, y, z)$.

The z-axis is one of the three coordinate axes.

It is defined as the line where both the x-coordinate and the y-coordinate are zero.

Any point that lies on the z-axis has its x-coordinate equal to 0 and its y-coordinate equal to 0.

The z-coordinate can take any real value, representing the position along the z-axis.

Therefore, the coordinates of a point on the z-axis are of the form $(0, 0, z)$, where $z$ is a real number.

Let's analyze the given options:

• (A) $(x, 0, 0)$: These are coordinates of a point on the x-axis.
• (B) $(0, y, 0)$: These are coordinates of a point on the y-axis.
• (C) $(0, 0, z)$: These are coordinates of a point on the z-axis.
• (D) $(x, y, z)$: These are coordinates of a general point in three-dimensional space, not necessarily on any axis or plane.

The coordinates $(0, 0, z)$ correctly represent any point lying on the z-axis.

Answer:

(C) $(0, 0, z)$

Given:

The point is $P(6, -2, 3)$.

To Find:

The distance of the point $P(6, -2, 3)$ from the xy-plane.

Solution:

In a three-dimensional coordinate system, the xy-plane is defined by the equation $z = 0$.

The distance of any point $(x_1, y_1, z_1)$ from the xy-plane is the absolute value of its z-coordinate.

Distance from xy-plane $= |z_1|$

For the given point $(6, -2, 3)$, the coordinates are $x_1 = 6$, $y_1 = -2$, and $z_1 = 3$.

The distance of the point $(6, -2, 3)$ from the xy-plane is $|3|$.

Distance $= |3| = 3$

Let's consider why this is the case. Imagine the point $P(6, -2, 3)$. Its projection onto the xy-plane is the point $Q(6, -2, 0)$, which lies on the xy-plane. The distance between $P$ and the xy-plane is the distance between $P$ and its projection $Q$.

Using the distance formula between $P(6, -2, 3)$ and $Q(6, -2, 0)$:

Distance $= \sqrt{(6 - 6)^2 + (-2 - (-2))^2 + (3 - 0)^2}$

Distance $= \sqrt{(0)^2 + (0)^2 + (3)^2}$

Distance $= \sqrt{0 + 0 + 9}$

Distance $= \sqrt{9}$

Distance $= 3$

This confirms that the distance is simply the absolute value of the z-coordinate.

Comparing our result with the given options:

• (A) 6: This is the x-coordinate.
• (B) -2: This is the y-coordinate. Distance must be non-negative.
• (C) $|3|$: This is the absolute value of the z-coordinate, which is 3. This matches our result.
• (D) $\sqrt{6^2 + (-2)^2}$: This is the distance from the z-axis, not the xy-plane.

The distance of the point from the xy-plane is $|3|$.

Answer:

(C) $|3|$

Given:

Point A: $A(1, 2, 3)$

Point B: $B(4, 5, 6)$

Point D divides the line segment AB internally in the ratio $1:2$.

To Find:

The coordinates of point D.

Solution:

Let the coordinates of point A be $(x_1, y_1, z_1) = (1, 2, 3)$.

Let the coordinates of point B be $(x_2, y_2, z_2) = (4, 5, 6)$.

The ratio of internal division is $m:n = 1:2$.

The coordinates of a point $(x, y, z)$ that divides the line segment joining $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$ internally in the ratio $m:n$ are given by the Section Formula in three dimensions:

Substitute the given values $x_1=1, y_1=2, z_1=3$, $x_2=4, y_2=5, z_2=6$, $m=1$, and $n=2$ into the formula:

For the x-coordinate of D:

$x_D = \frac{(1)(4) + (2)(1)}{1 + 2}$

$x_D = \frac{4 + 2}{3}$

$x_D = \frac{6}{3}$

$x_D = 2$

For the y-coordinate of D:

$y_D = \frac{(1)(5) + (2)(2)}{1 + 2}$

$y_D = \frac{5 + 4}{3}$

$y_D = \frac{9}{3}$

$y_D = 3$

For the z-coordinate of D:

$z_D = \frac{(1)(6) + (2)(3)}{1 + 2}$

$z_D = \frac{6 + 6}{3}$

$z_D = \frac{12}{3}$

$z_D = 4$

The coordinates of point D are $(2, 3, 4)$.

Comparing this result with the given options:

• (A) $(2, 3, 4)$: This matches our calculated coordinates.
• (B) $(2.5, 3.5, 4.5)$: These would be the coordinates of the midpoint of AB.
• (C) $(3, 3, 3)$: Incorrect.
• (D) $(\frac{1 \cdot 4 + 2 \cdot 1}{1+2}, \frac{1 \cdot 5 + 2 \cdot 2}{1+2}, \frac{1 \cdot 6 + 2 \cdot 3}{1+2})$: This is the expression from the section formula before simplification, which evaluates to $(2, 3, 4)$. Option (A) gives the simplified coordinates.

Both (A) and (D) represent the correct coordinates of point D, with (A) being the simplified form.

Answer:

(A) $(2, 3, 4)$

Given:

Point A: $A(1, 2, 3)$

Point C: $C(-1, 0, 2)$

Point E is the midpoint of the line segment AC.

To Find:

The coordinates of point E.

Solution:

Let the coordinates of point A be $(x_1, y_1, z_1) = (1, 2, 3)$ and the coordinates of point C be $(x_2, y_2, z_2) = (-1, 0, 2)$.

The coordinates of the midpoint $(x, y, z)$ of a line segment joining $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$ in three dimensions are given by the Midpoint Formula:

$(x, y, z) = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}, \frac{z_1 + z_2}{2} \right)$

Substitute the given values $x_1=1, y_1=2, z_1=3$ and $x_2=-1, y_2=0, z_2=2$ into the formula:

For the x-coordinate of E:

$x_E = \frac{1 + (-1)}{2}$

$x_E = \frac{0}{2}$

$x_E = 0$

For the y-coordinate of E:

$y_E = \frac{2 + 0}{2}$

$y_E = \frac{2}{2}$

$y_E = 1$

For the z-coordinate of E:

$z_E = \frac{3 + 2}{2}$

$z_E = \frac{5}{2}$

$z_E = 2.5$

The coordinates of point E are $(0, 1, 2.5)$.

Comparing this result with the given options:

• (A) $(\frac{1+(-1)}{2}, \frac{2+0}{2}, \frac{3+2}{2})$: This is the expression from the midpoint formula before simplification, which evaluates to $(0, 1, 2.5)$.
• (B) $(0, 1, 2.5)$: This matches our calculated coordinates.
• (C) $(-1, 1, 2.5)$: Incorrect x-coordinate.
• (D) $(1, 1, 2.5)$: Incorrect x-coordinate.

Both (A) and (B) represent the correct coordinates of point E, with (B) being the simplified form.

Answer:

(B) $(0, 1, 2.5)$

Given:

Point D: $D(2, 3, 4)$ (from Question 14)

Point E: $E(0, 1, 2.5)$ (from Question 15)

To Find:

The distance between point D and point E.

Solution:

Let the coordinates of point D be $(x_1, y_1, z_1) = (2, 3, 4)$.

Let the coordinates of point E be $(x_2, y_2, z_2) = (0, 1, 2.5)$.

The distance between two points $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$ in three-dimensional space is given by the Distance Formula:

Substitute the coordinates of D and E into the formula:

Distance $= \sqrt{(0 - 2)^2 + (1 - 3)^2 + (2.5 - 4)^2}$

Distance $= \sqrt{(-2)^2 + (-2)^2 + (-1.5)^2}$

Calculate the squares:

$(-2)^2 = 4$

$(-2)^2 = 4$

$(-1.5)^2 = (-3/2)^2 = 9/4 = 2.25$

Substitute these values back into the distance formula:

Distance $= \sqrt{4 + 4 + 2.25}$

Distance $= \sqrt{8 + 2.25}$

Distance $= \sqrt{10.25}$

The distance between points D and E is $\sqrt{10.25}$.

Comparing this result with the given options:

• (A) $\sqrt{(2-0)^2 + (3-1)^2 + (4-2.5)^2}$: This is the correct setup of the distance calculation, which evaluates to $\sqrt{10.25}$.
• (B) $\sqrt{10}$: Incorrect. $\sqrt{10} = \sqrt{10.00}$
• (C) $\sqrt{10.25}$: This matches our calculated distance.
• (D) $\sqrt{10.5}$: Incorrect. $\sqrt{10.5} = \sqrt{10.50}$

Both (A) and (C) represent the correct distance, with (C) providing the evaluated numerical value.

Answer:

(C) $\sqrt{10.25}$

Solution:

In a three-dimensional Cartesian coordinate system, the three coordinate planes are the xy-plane ($z=0$), the yz-plane ($x=0$), and the xz-plane ($y=0$).

These three planes are mutually perpendicular and intersect at the origin $(0, 0, 0)$.

Just as two perpendicular lines (the x and y axes) divide a 2D plane into four regions called quadrants, the three perpendicular coordinate planes divide the 3D space into regions.

Each region is defined by the signs of the x, y, and z coordinates.

There are $2 \times 2 \times 2 = 8$ possible combinations of signs (e.g., (+, +, +), (+, +, -), etc., including boundary cases), defining eight distinct regions.

These eight regions are called octants.

Let's look at the options:

• (A) Quadrants: These are the four regions in a 2D plane divided by two axes.
• (B) Octants: These are the eight regions in 3D space divided by the three coordinate planes.
• (C) Hemispheres: These are halves of a sphere, typically divided by a plane passing through the center.
• (D) Sectors: These are regions of a circle or sphere bounded by radii or planes, typically less than the full space.

The correct term for the eight regions created by the coordinate planes in 3D space is octants.

Answer:

(B) Octants

Given:

Points in three-dimensional space.

To Find:

Which point lies in the first octant.

Solution:

In a three-dimensional Cartesian coordinate system, the space is divided into eight octants by the three coordinate planes (xy-plane, yz-plane, and xz-plane).

The first octant is the region where all three coordinates (x, y, and z) are positive.

That is, a point $(x, y, z)$ lies in the first octant if $x > 0$, $y > 0$, and $z > 0$.

Let's examine the coordinates of each given point:

• (A) $(2, 3, 4)$: Here, $x = 2$, $y = 3$, $z = 4$. All three coordinates are positive ($2>0$, $3>0$, $4>0$).
• (B) $(-1, 2, 3)$: Here, $x = -1$. Since $x$ is negative, this point is not in the first octant.
• (C) $(1, -2, 3)$: Here, $y = -2$. Since $y$ is negative, this point is not in the first octant.
• (D) $(1, 2, -3)$: Here, $z = -3$. Since $z$ is negative, this point is not in the first octant.

Only the point $(2, 3, 4)$ has all positive coordinates, so it lies in the first octant.

Answer:

(A) $(2, 3, 4)$

Solution:

The formula for the coordinates of a point that divides the line segment joining two points $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$ in three-dimensional space depends on whether the division is internal or external.

The Section Formula for internal division in the ratio $m:n$ is:

$(x, y, z) = \left( \frac{m x_2 + n x_1}{m + n}, \frac{m y_2 + n y_1}{m + n}, \frac{m z_2 + n z_1}{m + n} \right)$

The Section Formula for external division in the ratio $m:n$ is obtained by replacing the '+' sign in the internal division formula with a '-' sign in both the numerator and the denominator.

This formula is valid provided $m \neq n$.

Comparing this formula with the given options:

• (A) $(\frac{mx_2 - nx_1}{m-n}, \frac{my_2 - ny_1}{m-n}, \frac{mz_2 - nz_1}{m-n})$: This exactly matches the formula for external division.
• (B) $(\frac{mx_2 + nx_1}{m+n}, \frac{my_2 + ny_1}{m+n}, \frac{mz_2 + nz_1}{m+n})$: This is the formula for internal division.
• (C) $(\frac{nx_1 - mx_2}{m-n}, \frac{ny_1 - my_2}{m-n}, \frac{nz_1 - mz_2}{m-n})$: This is not the standard form and is generally incorrect. For example, the x-coordinate is $\frac{nx_1 - mx_2}{m-n} = \frac{-(mx_2 - nx_1)}{-(n-m)} = \frac{mx_2 - nx_1}{n-m}$, which is different from $\frac{mx_2 - nx_1}{m-n}$ unless $n-m = -(m-n)$.
• (D) $(\frac{nx_1 + mx_2}{m+n}, \frac{ny_1 + my_2}{m+n}, \frac{nz_1 + mz_2}{m+n})$: This is also the internal division formula, but the order of terms in the numerator is swapped (which is permissible for addition, but it's written in a way that doesn't match the standard form).

The correct formula for external division is given in option (A).

Answer:

(A) $(\frac{mx_2 - nx_1}{m-n}, \frac{my_2 - ny_1}{m-n}, \frac{mz_2 - nz_1}{m-n})$

Given:

Point A: $A(2, -3, 5)$

Point B: $B(1, 4, -6)$

Ratio of external division: $m:n = 2:3$

To Find:

The coordinates of the point that divides the line segment AB externally in the ratio $2:3$.

Solution:

Let the coordinates of point A be $(x_1, y_1, z_1) = (2, -3, 5)$ and the coordinates of point B be $(x_2, y_2, z_2) = (1, 4, -6)$.

Let the ratio of external division be $m:n = 2:3$.

The coordinates of a point $(x, y, z)$ that divides the line segment joining $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$ externally in the ratio $m:n$ are given by the Section Formula for External Division:

Substitute the given values $x_1=2, y_1=-3, z_1=5$, $x_2=1, y_2=4, z_2=-6$, $m=2$, and $n=3$ into the formula:

For the x-coordinate:

$x = \frac{(2)(1) - (3)(2)}{2 - 3}$

$x = \frac{2 - 6}{-1}$

$x = \frac{-4}{-1}$

$x = 4$

For the y-coordinate:

$y = \frac{(2)(4) - (3)(-3)}{2 - 3}$

$y = \frac{8 - (-9)}{-1}$

$y = \frac{8 + 9}{-1}$

$y = \frac{17}{-1}$

$y = -17$

For the z-coordinate:

$z = \frac{(2)(-6) - (3)(5)}{2 - 3}$

$z = \frac{-12 - 15}{-1}$

$z = \frac{-27}{-1}$

$z = 27$

The coordinates of the point that divides the line segment AB externally in the ratio $2:3$ are $(4, -17, 27)$.

Comparing this result with the given options:

• (A) $(4, -17, 27)$: This matches our calculated coordinates.
• (B) $(\frac{2 \cdot 1 - 3 \cdot 2}{2-3}, \frac{2 \cdot 4 - 3 \cdot (-3)}{2-3}, \frac{2 \cdot (-6) - 3 \cdot 5}{2-3})$: This is the expression from the external section formula before simplification, which evaluates to $(4, -17, 27)$.
• (C) $(-4, 17, -27)$: Incorrect signs.
• (D) $(4, 17, 27)$: Incorrect y-coordinate.

Both (A) and (B) represent the correct coordinates, with (A) being the simplified form.

Answer:

(A) $(4, -17, 27)$

Solution:

In a three-dimensional Cartesian coordinate system, the yz-plane is defined by the equation $x = 0$. This plane contains all points whose x-coordinate is zero.

The distance of a point $(x_1, y_1, z_1)$ from a plane $Ax + By + Cz + D = 0$ is given by the formula:

Distance $= \frac{|Ax_1 + By_1 + Cz_1 + D|}{\sqrt{A^2 + B^2 + C^2}}$

For the yz-plane, the equation is $1 \cdot x + 0 \cdot y + 0 \cdot z + 0 = 0$. So, $A=1$, $B=0$, $C=0$, and $D=0$.

For the point $(x, y, z)$, we have $x_1=x$, $y_1=y$, and $z_1=z$.

Substitute these values into the distance formula:

Distance $= \frac{|(1)(x) + (0)(y) + (0)(z) + 0|}{\sqrt{(1)^2 + (0)^2 + (0)^2}}$

Distance $= \frac{|x + 0 + 0 + 0|}{\sqrt{1 + 0 + 0}}$

Distance $= \frac{|x|}{\sqrt{1}}$

Distance $= |x|$

The distance of the point $(x, y, z)$ from the yz-plane is the absolute value of its x-coordinate.

Alternatively, consider the projection of the point $(x, y, z)$ onto the yz-plane. This projection is the point $(0, y, z)$. The distance between the point $(x, y, z)$ and its projection $(0, y, z)$ on the yz-plane is the distance from the point to the plane.

Using the distance formula for points $(x_1, y_1, z_1) = (x, y, z)$ and $(x_2, y_2, z_2) = (0, y, z)$:

Distance $= \sqrt{(0 - x)^2 + (y - y)^2 + (z - z)^2}$

Distance $= \sqrt{(-x)^2 + (0)^2 + (0)^2}$

Distance $= \sqrt{x^2}$

Distance $= |x|$

This confirms that the distance from the yz-plane is the absolute value of the x-coordinate.

Comparing this result with the given options:

• (A) $|x|$: This matches our calculated distance.
• (B) $|y|$: This is the distance from the xz-plane.
• (C) $|z|$: This is the distance from the xy-plane.
• (D) $\sqrt{y^2 + z^2}$: This is the distance from the x-axis.

The correct distance is $|x|$.

Answer:

(A) $|x|$

Solution:

In a three-dimensional Cartesian coordinate system, the coordinate planes are defined by setting one of the coordinates to zero.

The xy-plane is the plane that contains the x-axis and the y-axis.

Any point that lies on the xy-plane has its z-coordinate equal to zero.

The coordinates of any point on the xy-plane are of the form $(x, y, 0)$, where $x$ and $y$ can be any real numbers.

Therefore, the equation that describes all points on the xy-plane is $z = 0$.

Let's examine the options:

• (A) $x = 0$: This is the equation of the yz-plane.
• (B) $y = 0$: This is the equation of the xz-plane.
• (C) $z = 0$: This is the equation where the z-coordinate is zero, which defines the xy-plane.
• (D) $x + y = 0$: This represents a plane that is perpendicular to the xy-plane and passes through the z-axis. It is not the xy-plane itself.

The equation of the xy-plane is $z = 0$.

Answer:

(C) $z = 0$

Solution:

In a three-dimensional Cartesian coordinate system, the origin is the point where the three coordinate axes (the x-axis, the y-axis, and the z-axis) intersect.

This point serves as the reference point from which all other points in space are located.

At the origin, the displacement along the x-axis is 0, the displacement along the y-axis is 0, and the displacement along the z-axis is 0.

Therefore, the coordinates of the origin are $(0, 0, 0)$.

Let's look at the given options:

• (A) $(1, 1, 1)$: This represents a point located 1 unit away from the origin along the positive x, y, and z directions.
• (B) $(0, 0, 0)$: This represents the point where all coordinates are zero, which is the definition of the origin.
• (C) $(x, y, z)$: This represents a general point in 3D space, not a specific point like the origin.
• (D) $(0, 0, 1)$: This represents a point on the z-axis, 1 unit away from the origin along the positive z direction.

The coordinates of the origin in 3D space are $(0, 0, 0)$.

Answer:

(B) $(0, 0, 0)$

Evaluation of Assertion (A):

The given point is $(-1, 2, -3)$.

The coordinates have the following signs:

• x-coordinate is $-1$ (Negative, -)
• y-coordinate is $2$ (Positive, +)
• z-coordinate is $-3$ (Negative, -)

The point $(-1, 2, -3)$ has signs (-, +, -).

Based on the standard definition of octants, the octants are defined by the signs of the coordinates $(x, y, z)$:

• 1st Octant: (+, +, +)
• 2nd Octant: (-, +, +)
• 3rd Octant: (-, -, +)
• 4th Octant: (+, -, +)
• 5th Octant: (+, +, -)
• 6th Octant: (-, +, -)
• 7th Octant: (-, -, -)
• 8th Octant: (+, -, -)

The signs (-, +, -) correspond to the sixth octant.

Thus, Assertion (A) states that the point $(-1, 2, -3)$ lies in the sixth octant, which is True.

Evaluation of Reason (R):

Reason (R) states that in the sixth octant, the x-coordinate is negative, and the y and z coordinates are positive.

According to the standard definition of the sixth octant, the signs are (-, +, -).

This means:

• x-coordinate is Negative
• y-coordinate is Positive
• z-coordinate is Negative

Reason (R) claims that the z-coordinate is positive in the sixth octant, which is incorrect according to the standard definition (-, +, -).

Thus, Reason (R) is False.

Relationship between A and R:

Assertion (A) is true, but Reason (R) is false as it provides an incorrect definition for the sixth octant.

Conclusion:

Assertion (A) is true, and Reason (R) is false.

Answer:

(C) A is true but R is false.

Given:

The point is $(2, 3, 5)$.

To Find:

The distance of the point $(2, 3, 5)$ from the yz-plane.

Solution:

In a three-dimensional coordinate system, the yz-plane is the plane where the x-coordinate is always zero. Its equation is $x = 0$.

The distance of a point $(x_1, y_1, z_1)$ from the yz-plane is the absolute value of its x-coordinate, $|x_1|$.

For the given point $(2, 3, 5)$, the coordinates are $x_1 = 2$, $y_1 = 3$, and $z_1 = 5$.

The distance from the yz-plane is $|x_1| = |2|$.

Distance $= 2$

The distance of the point $(2, 3, 5)$ from the yz-plane is 2.

Comparing our result with the given options:

• (A) 2: This matches our calculated distance.
• (B) 3: This is the y-coordinate.
• (C) 5: This is the z-coordinate.
• (D) $\sqrt{3^2 + 5^2}$: This is the distance from the x-axis.

The correct distance is 2.

Answer:

(A) 2

Given:

The point is $(1, 2, 3)$.

To Find:

The distance of the point $(1, 2, 3)$ from the xz-plane.

Solution:

In a three-dimensional Cartesian coordinate system, the xz-plane is defined by the equation $y = 0$. This plane contains all points whose y-coordinate is zero.

The distance of a point $(x_1, y_1, z_1)$ from the xz-plane is the absolute value of its y-coordinate, $|y_1|$.

For the given point $(1, 2, 3)$, the coordinates are $x_1 = 1$, $y_1 = 2$, and $z_1 = 3$.

The distance from the xz-plane is $|y_1| = |2|$.

Distance $= |2| = 2$

The distance of the point $(1, 2, 3)$ from the xz-plane is 2.

Comparing our result with the given options:

• (A) 1: This is the x-coordinate.
• (B) $|2|$: This is the absolute value of the y-coordinate, which is 2. This matches our result.
• (C) 3: This is the z-coordinate.
• (D) $\sqrt{1^2 + 3^2}$: This is the distance from the y-axis.

The correct distance is $|2|$, which equals 2.

Answer:

(B) $|2|$

Given:

The point is $(a, b, c)$.

The point is equidistant from the yz-plane and the xz-plane.

To Find:

The condition that must be true based on the given information.

Solution:

The distance of a point $(x_1, y_1, z_1)$ from the yz-plane (equation $x=0$) is $|x_1|$.

For the point $(a, b, c)$, the distance from the yz-plane is $|a|$.

The distance of a point $(x_1, y_1, z_1)$ from the xz-plane (equation $y=0$) is $|y_1|$.

For the point $(a, b, c)$, the distance from the xz-plane is $|b|$.

We are given that the point $(a, b, c)$ is equidistant from the yz-plane and the xz-plane.

So, we must have:

This condition means that the absolute value of the x-coordinate is equal to the absolute value of the y-coordinate.

Comparing this condition with the given options:

• (A) $|a| = |b|$: This matches our derived condition.
• (B) $|a| = |c|$: This would be the condition if the point were equidistant from the yz-plane and the xy-plane ($z=0$).
• (C) $|b| = |c|$: This would be the condition if the point were equidistant from the xz-plane and the xy-plane ($z=0$).
• (D) $a = b = c$: This is a stronger condition than necessary. For example, the point $(-2, 2, 5)$ satisfies $|a|=|b|$ but not $a=b=c$.

The condition that must be true for the point $(a, b, c)$ to be equidistant from the yz-plane and the xz-plane is $|a| = |b|$.

Answer:

(A) $|a| = |b|$

Given:

Point 1: $P_1(0, 0, 0)$ (The Origin)

Point 2: $P_2(6, -3, 9)$

Ratio of internal division: $m:n = 1:2$

To Find:

The coordinates of the point that divides the line segment joining $P_1$ and $P_2$ internally in the ratio $1:2$.

Solution:

Let the coordinates of $P_1$ be $(x_1, y_1, z_1) = (0, 0, 0)$ and the coordinates of $P_2$ be $(x_2, y_2, z_2) = (6, -3, 9)$.

Let the ratio of internal division be $m:n = 1:2$.

The coordinates of a point $(x, y, z)$ that divides the line segment joining $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$ internally in the ratio $m:n$ are given by the Section Formula in three dimensions:

Substitute the given values $x_1=0, y_1=0, z_1=0$, $x_2=6, y_2=-3, z_2=9$, $m=1$, and $n=2$ into the formula:

For the x-coordinate:

$x = \frac{(1)(6) + (2)(0)}{1 + 2}$

$x = \frac{6 + 0}{3}$

$x = \frac{6}{3}$

$x = 2$

For the y-coordinate:

$y = \frac{(1)(-3) + (2)(0)}{1 + 2}$

$y = \frac{-3 + 0}{3}$

$y = \frac{-3}{3}$

$y = -1$

For the z-coordinate:

$z = \frac{(1)(9) + (2)(0)}{1 + 2}$

$z = \frac{9 + 0}{3}$

$z = \frac{9}{3}$

$z = 3$

The coordinates of the point that divides the line segment joining the origin and $(6, -3, 9)$ internally in the ratio $1:2$ are $(2, -1, 3)$.

Comparing this result with the given options:

• (A) $(2, -1, 3)$: This matches our calculated coordinates.
• (B) $(\frac{1 \cdot 6 + 2 \cdot 0}{1+2}, \frac{1 \cdot (-3) + 2 \cdot 0}{1+2}, \frac{1 \cdot 9 + 2 \cdot 0}{1+2})$: This is the expression from the section formula before simplification, which evaluates to $(2, -1, 3)$.
• (C) $(1, -0.5, 1.5)$: Incorrect coordinates.
• (D) $(2, -2, 6)$: Incorrect y and z coordinates.

Both (A) and (B) represent the correct coordinates of the point, with (A) being the simplified form.

Answer:

(A) $(2, -1, 3)$

Given:

The point is $(a, b, c)$.

To Find:

The distance of the point $(a, b, c)$ from the z-axis.

Solution:

The distance of a point from an axis is the perpendicular distance from the point to the axis.

Consider the point $P(a, b, c)$ and the z-axis.

The z-axis consists of all points of the form $(0, 0, z)$.

The point on the z-axis that is closest to $P(a, b, c)$ is the projection of $P$ onto the z-axis. This projection point has the same z-coordinate as $P$, but its x and y coordinates are zero. The projection is $Q(0, 0, c)$.

The distance of $P$ from the z-axis is the distance between $P(a, b, c)$ and $Q(0, 0, c)$.

Using the Distance Formula between two points $(x_1, y_1, z_1) = (a, b, c)$ and $(x_2, y_2, z_2) = (0, 0, c)$:

Distance $= \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}$

Substitute the coordinates:

Distance $= \sqrt{(0 - a)^2 + (0 - b)^2 + (c - c)^2}$

Distance $= \sqrt{(-a)^2 + (-b)^2 + (0)^2}$

Distance $= \sqrt{a^2 + b^2 + 0}$

Distance $= \sqrt{a^2 + b^2}$

The distance of the point $(a, b, c)$ from the z-axis is $\sqrt{a^2 + b^2}$.

Comparing this result with the given options:

• (A) $|c|$: This is the distance from the xy-plane.
• (B) $\sqrt{a^2 + b^2}$: This matches our calculated distance.
• (C) $\sqrt{a^2 + c^2}$: This would be the distance from the y-axis.
• (D) $\sqrt{b^2 + c^2}$: This would be the distance from the x-axis.

The correct distance from the z-axis is $\sqrt{a^2 + b^2}$.

Answer:

(B) $\sqrt{a^2 + b^2}$

Given:

A list of points in three-dimensional space.

To Find:

The point that does NOT lie on the xz-plane.

Solution:

The xz-plane in a three-dimensional Cartesian coordinate system is defined by the equation $y = 0$.

A point $(x, y, z)$ lies on the xz-plane if and only if its y-coordinate is equal to 0.

We examine the y-coordinate of each given point:

• (A) Point $(5, 0, 2)$: The y-coordinate is $0$. Since $y=0$, this point lies on the xz-plane.
• (B) Point $(0, 0, -3)$: The y-coordinate is $0$. Since $y=0$, this point lies on the xz-plane (it is on the z-axis, which is part of the xz-plane).
• (C) Point $(-1, 0, 4)$: The y-coordinate is $0$. Since $y=0$, this point lies on the xz-plane.
• (D) Point $(2, 1, 5)$: The y-coordinate is $1$. Since $y \neq 0$, this point does NOT lie on the xz-plane.

The point that does not satisfy the condition $y=0$ is $(2, 1, 5)$.

Answer:

(D) $(2, 1, 5)$

Given:

Point 1: $P(p, q, r)$

Point 2: $Q(2p, 2q, 2r)$

Origin: $O(0, 0, 0)$

The distance of point $P$ from the origin is equal to the distance of point $Q$ from the origin.

To Find:

The relation between $p, q, r$.

Solution:

The distance between two points $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$ in three-dimensional space is given by the formula:

Distance $= \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}$

The distance of a point $(x, y, z)$ from the origin $(0, 0, 0)$ is $\sqrt{x^2 + y^2 + z^2}$.

Distance of point $P(p, q, r)$ from the origin $O(0, 0, 0)$ is:

$OP = \sqrt{(p - 0)^2 + (q - 0)^2 + (r - 0)^2} = \sqrt{p^2 + q^2 + r^2}$

Distance of point $Q(2p, 2q, 2r)$ from the origin $O(0, 0, 0)$ is:

$OQ = \sqrt{(2p - 0)^2 + (2q - 0)^2 + (2r - 0)^2}$

$OQ = \sqrt{(2p)^2 + (2q)^2 + (2r)^2}$

$OQ = \sqrt{4p^2 + 4q^2 + 4r^2}$

$OQ = \sqrt{4(p^2 + q^2 + r^2)}$

$OQ = 2\sqrt{p^2 + q^2 + r^2}$

According to the given condition, the distance of point $P$ from the origin is equal to the distance of point $Q$ from the origin.

Square both sides of the equation (i):

$(\sqrt{p^2 + q^2 + r^2})^2 = (2\sqrt{p^2 + q^2 + r^2})^2$

$p^2 + q^2 + r^2 = 4 (p^2 + q^2 + r^2)$

$p^2 + q^2 + r^2 = 4p^2 + 4q^2 + 4r^2$

Rearrange the terms:

$0 = 4p^2 - p^2 + 4q^2 - q^2 + 4r^2 - r^2$

$0 = 3p^2 + 3q^2 + 3r^2$

Divide by 3:

For real numbers $p$, $q$, and $r$, the square of each number is non-negative ($p^2 \geq 0$, $q^2 \geq 0$, $r^2 \geq 0$).

The sum of non-negative numbers can only be zero if each number is zero.

Therefore, $p^2 = 0$, $q^2 = 0$, and $r^2 = 0$, which implies $p = 0$, $q = 0$, and $r = 0$.

This means the point $(p, q, r)$ must be the origin itself.

The condition that must be true is $p^2 + q^2 + r^2 = 0$.

Comparing this result with the given options:

• (A) $p^2 + q^2 + r^2 = 0$: This matches our derived relation.
• (B) $p+q+r=0$: This allows for non-zero values (e.g., $(1, -1, 0)$).
• (C) $p^2 + q^2 + r^2 \neq 0$: This is the opposite of our derived relation.
• (D) $p=q=r$: This allows for non-zero equal values (e.g., $(1, 1, 1)$).

The correct relation is $p^2 + q^2 + r^2 = 0$.

Answer:

(A) $p^2 + q^2 + r^2 = 0$

Given:

The vertices of the triangle are $A(1, -1, 2)$, $B(2, 3, -1)$, and $C(-3, 1, 4)$.

To Find:

The coordinates of the centroid of the triangle ABC.

Solution:

Let the vertices of the triangle be $(x_1, y_1, z_1) = (1, -1, 2)$, $(x_2, y_2, z_2) = (2, 3, -1)$, and $(x_3, y_3, z_3) = (-3, 1, 4)$.

The coordinates of the centroid $(x, y, z)$ of a triangle with vertices $(x_1, y_1, z_1)$, $(x_2, y_2, z_2)$, and $(x_3, y_3, z_3)$ are given by the Centroid Formula:

$$(x, y, z) = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3}, \frac{z_1 + z_2 + z_3}{3} \right)$$

Substitute the coordinates of the given vertices into the formula:

For the x-coordinate of the centroid:

$x = \frac{1 + 2 + (-3)}{3}$

$x = \frac{1 + 2 - 3}{3}$

$x = \frac{0}{3}$

$x = 0$

For the y-coordinate of the centroid:

$y = \frac{-1 + 3 + 1}{3}$

$y = \frac{3}{3}$

$y = 1$

For the z-coordinate of the centroid:

$z = \frac{2 + (-1) + 4}{3}$

$z = \frac{2 - 1 + 4}{3}$

$z = \frac{5}{3}$

The coordinates of the centroid are $(0, 1, \frac{5}{3})$.

Comparing this result with the given options:

• (A) $(0, 1, 5/3)$: This matches our calculated coordinates.
• (B) $(\frac{1+2+(-3)}{3}, \frac{-1+3+1}{3}, \frac{2+(-1)+4}{3})$: This is the expression from the centroid formula before simplification, which evaluates to $(0, 1, 5/3)$.
• (C) $(0, 1/3, 5/3)$: Incorrect y-coordinate.
• (D) $(1, 1, 5/3)$: Incorrect x-coordinate.

Both (A) and (B) represent the correct coordinates, with (A) being the simplified form.

Answer:

(A) $(0, 1, 5/3)$

Solution:

In a three-dimensional Cartesian coordinate system, the xz-plane is the plane where the y-coordinate is always zero. Its equation is $y = 0$.

The distance of a point $(x_1, y_1, z_1)$ from the xz-plane is the absolute value of its y-coordinate, $|y_1|$.

For the point $(x, y, z)$, the coordinates are $x_1 = x$, $y_1 = y$, and $z_1 = z$.

The distance from the xz-plane is $|y_1| = |y|$.

Distance $= |y|$

The distance of the point $(x, y, z)$ from the xz-plane is the absolute value of the y-coordinate.

Let's verify this. The projection of the point $(x, y, z)$ onto the xz-plane is the point $(x, 0, z)$. The distance between $(x, y, z)$ and $(x, 0, z)$ is:

Distance $= \sqrt{(x - x)^2 + (0 - y)^2 + (z - z)^2}$

Distance $= \sqrt{(0)^2 + (-y)^2 + (0)^2}$

Distance $= \sqrt{y^2}$

Distance $= |y|$

This confirms that the distance from the xz-plane is $|y|$.

Comparing this result with the given options:

• (A) $|x|$: This is the distance from the yz-plane.
• (B) $|y|$: This matches our calculated distance.
• (C) $|z|$: This is the distance from the xy-plane.
• (D) $\sqrt{x^2 + z^2}$: This is the distance from the y-axis.

The correct distance is $|y|$.

Answer:

(B) $|y|$

To Determine:

The correct option based on the truthfulness of Assertion (A) and Reason (R) regarding the location of the point $(2, -3, 1)$ in the three-dimensional coordinate system.

Solution:

In three-dimensional space, the coordinate axes divide the space into eight regions known as octants. The location of a point $(x, y, z)$ in an octant is determined by the signs of its coordinates.

The signs of the coordinates $(x, y, z)$ for each of the eight octants are as follows:

• Octant I: $(+, +, +)$
• Octant II: $(-, +, +)$
• Octant III: $(-, -, +)$
• Octant IV: $(+, -, +)$
• Octant V: $(+, +, -)$
• Octant VI: $(-, +, -)$
• Octant VII: $(-, -, -)$
• Octant VIII: $(+, -, -)$

Now, let's analyze the given Assertion and Reason.

Assertion (A): The point $(2, -3, 1)$ lies in the fourth octant.

For the point $(2, -3, 1)$, the coordinates are $x=2$, $y=-3$, and $z=1$.

The sign of the x-coordinate is positive ($2 > 0$).

The sign of the y-coordinate is negative ($-3 < 0$).

The sign of the z-coordinate is positive ($1 > 0$).

Thus, the signs of the coordinates for the point $(2, -3, 1)$ are $(+, -, +)$.

Comparing these signs with the octant definitions, we find that the signs $(+, -, +)$ correspond to the fourth octant.

Therefore, Assertion (A) is True.

Reason (R): In the fourth octant, the x and z coordinates are positive, and the y-coordinate is negative.

According to the definition of the fourth octant, the signs of the coordinates $(x, y, z)$ are $(+, -, +)$. This means the x-coordinate is positive, the y-coordinate is negative, and the z-coordinate is positive.

This statement matches the definition of the fourth octant.

Therefore, Reason (R) is also True.

We have determined that both Assertion (A) and Reason (R) are true.

Now, let's check if Reason (R) is the correct explanation for Assertion (A).

Assertion (A) states that the point $(2, -3, 1)$ is in the fourth octant. Reason (R) describes the characteristic sign pattern of the fourth octant, which is $(+, -, +)$. Since the point $(2, -3, 1)$ has exactly this sign pattern, Reason (R) correctly explains why the point is located in the fourth octant.

Thus, Reason (R) is the correct explanation for Assertion (A).

Based on our analysis, both Assertion (A) and Reason (R) are true, and Reason (R) provides the correct explanation for Assertion (A).

Final Answer:

Both A and R are true and R is the correct explanation of A.

The correct option is (A).

Let the two points be $P_1 = (x_1, y_1, z_1) = (a \cos \theta, a \sin \theta, 0)$ and $P_2 = (x_2, y_2, z_2) = (a \cos \phi, a \sin \phi, 0)$.

The distance between two points in 3D space is given by the formula:

$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}$

Substitute the coordinates of the given points into the distance formula:

$d = \sqrt{(a \cos \phi - a \cos \theta)^2 + (a \sin \phi - a \sin \theta)^2 + (0 - 0)^2}$

Simplify the expression:

$d = \sqrt{a^2 (\cos \phi - \cos \theta)^2 + a^2 (\sin \phi - \sin \theta)^2 + 0^2}$

$d = \sqrt{a^2 [(\cos \phi - \cos \theta)^2 + (\sin \phi - \sin \theta)^2]}$

$d = a \sqrt{(\cos^2 \phi - 2 \cos \phi \cos \theta + \cos^2 \theta) + (\sin^2 \phi - 2 \sin \phi \sin \theta + \sin^2 \theta)}$

Rearrange the terms and use the identity $\cos^2 x + \sin^2 x = 1$:

$d = a \sqrt{(\cos^2 \phi + \sin^2 \phi) + (\cos^2 \theta + \sin^2 \theta) - 2 (\cos \phi \cos \theta + \sin \phi \sin \theta)}$

$d = a \sqrt{1 + 1 - 2 (\cos \phi \cos \theta + \sin \phi \sin \theta)}$

$d = a \sqrt{2 - 2 (\cos \phi \cos \theta + \sin \phi \sin \theta)}$

Use the trigonometric identity $\cos A \cos B + \sin A \sin B = \cos (A - B)$:

$d = a \sqrt{2 - 2 \cos (\phi - \theta)}$

Since $\cos (\phi - \theta) = \cos (\theta - \phi)$, we have:

$d = a \sqrt{2 - 2 \cos (\theta - \phi)}$

Factor out 2:

$d = a \sqrt{2 (1 - \cos (\theta - \phi))}$

Use the half-angle identity $1 - \cos x = 2 \sin^2 \left(\frac{x}{2}\right)$:

$d = a \sqrt{2 \left(2 \sin^2 \left(\frac{\theta - \phi}{2}\right)\right)}$

$d = a \sqrt{4 \sin^2 \left(\frac{\theta - \phi}{2}\right)}$

Simplify the square root:

$d = a \sqrt{\left(2 \sin \left(\frac{\theta - \phi}{2}\right)\right)^2}$

$d = a \left|2 \sin \left(\frac{\theta - \phi}{2}\right)\right|$

$d = 2a \left|\sin \left(\frac{\theta - \phi}{2}\right)\right|$

Comparing this result with the given options, we see that the steps and the final simplified form match option (D).

The final answer is $2a \left|\sin \left(\frac{\theta - \phi}{2}\right)\right|$.

Thus, the correct option is (D).

Let the points be $P(x_1, y_1, z_1) = (a \cos \theta, a \sin \theta, 0)$ and $Q(x_2, y_2, z_2) = (a \cos \phi, a \sin \phi, 0)$.

The distance $d$ between $P$ and $Q$ is given by the 3D distance formula:

$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}$

Substituting the coordinates:

$d = \sqrt{(a \cos \phi - a \cos \theta)^2 + (a \sin \phi - a \sin \theta)^2 + (0 - 0)^2}$

$d = \sqrt{a^2 (\cos \phi - \cos \theta)^2 + a^2 (\sin \phi - \sin \theta)^2}$

$d = a \sqrt{(\cos \phi - \cos \theta)^2 + (\sin \phi - \sin \theta)^2}$

Expand the squares:

$d = a \sqrt{(\cos^2 \phi - 2 \cos \phi \cos \theta + \cos^2 \theta) + (\sin^2 \phi - 2 \sin \phi \sin \theta + \sin^2 \theta)}$

$d = a \sqrt{(\cos^2 \phi + \sin^2 \phi) + (\cos^2 \theta + \sin^2 \theta) - 2 (\cos \phi \cos \theta + \sin \phi \sin \theta)}$

Using the identities $\cos^2 x + \sin^2 x = 1$ and $\cos A \cos B + \sin A \sin B = \cos (A - B)$:

$d = a \sqrt{1 + 1 - 2 \cos (\phi - \theta)}$

$d = a \sqrt{2 - 2 \cos (\theta - \phi)}$

$d = a \sqrt{2 (1 - \cos (\theta - \phi))}$

Using the half-angle identity $1 - \cos x = 2 \sin^2 \left(\frac{x}{2}\right)$:

$d = a \sqrt{2 \left(2 \sin^2 \left(\frac{\theta - \phi}{2}\right)\right)}$

$d = a \sqrt{4 \sin^2 \left(\frac{\theta - \phi}{2}\right)}$

Simplifying the square root:

$d = a \left|2 \sin \left(\frac{\theta - \phi}{2}\right)\right|$

$d = 2a \left|\sin \left(\frac{\theta - \phi}{2}\right)\right|$

This matches the final form presented in option (D).

The correct option is (D) $2a \left|\sin\left(\frac{\theta-\phi}{2}\right)\right|$.

Let the point be $P(x, y, z) = (1, 2, -3)$.

To find the distance of a point from the y-axis in 3D space, we consider the projection of the point onto the y-axis.

The projection of the point $(x, y, z)$ onto the y-axis is $(0, y, 0)$.

For the given point $(1, 2, -3)$, its projection onto the y-axis is $(0, 2, 0)$.

The distance of the point $(x, y, z)$ from the y-axis is the distance between $(x, y, z)$ and $(0, y, 0)$.

Using the distance formula between two points $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$, which is $\sqrt{(x_2-x_1)^2 + (y_2-y_1)^2 + (z_2-z_1)^2}$:

Distance from y-axis $= \sqrt{(1-0)^2 + (2-2)^2 + (-3-0)^2}$

Simplify the expression:

Distance $= \sqrt{(1)^2 + (0)^2 + (-3)^2}$

Distance $= \sqrt{1^2 + (-3)^2}

Distance $= \sqrt{1 + 9}

Distance $= \sqrt{10}$

Compare this result with the given options:

(A) $|2| = 2$

(B) $\sqrt{1^2 + (-3)^2} = \sqrt{1+9} = \sqrt{10}$

(C) $\sqrt{1^2 + 2^2 + (-3)^2}$ (This is the distance from the origin)

(D) $\sqrt{2^2} = |2| = 2$

The calculated distance is $\sqrt{10}$, which matches option (B).

Thus, the correct option is (B) $\sqrt{1^2 + (-3)^2} = \sqrt{10}$.

The xz-plane is the plane that contains the x-axis and the z-axis.

Any point that lies on the xz-plane has its y-coordinate equal to zero.

For example, points on the x-axis have coordinates of the form $(x, 0, 0)$. Points on the z-axis have coordinates of the form $(0, 0, z)$. Any point in the xz-plane can be represented as $(x, 0, z)$.

Since the y-coordinate is always 0 for any point on the xz-plane, the equation of the xz-plane is $y=0$.

Let's examine the given options:

(A) $x = 0$: This is the equation of the yz-plane (where the x-coordinate is always zero).

(B) $y = 0$: This is the equation of the xz-plane (where the y-coordinate is always zero).

(C) $z = 0$: This is the equation of the xy-plane (where the z-coordinate is always zero).

(D) $x + z = 0$: This is an equation of a plane, but not one of the coordinate planes. For instance, the point $(1, 0, -1)$ lies on this plane, but also on the xz-plane ($y=0$). However, the point $(1, 1, -1)$ also satisfies $x+z=0$ but is not on the xz-plane.

Therefore, the equation of the xz-plane is $y=0$.

Thus, the correct option is (B) $y = 0$.

Let the point be $R(x, y, z)$ that divides the line segment joining $P(x_1, y_1, z_1)$ and $Q(x_2, y_2, z_2)$ externally in the ratio $m:n$.

The section formula for external division in 3D is given by:

$x = \frac{m x_2 - n x_1}{m-n}$

$y = \frac{m y_2 - n y_1}{m-n}$

$z = \frac{m z_2 - n z_1}{m-n}$

In this question, the ratio of external division is given as $\lambda:1$. This means $m=\lambda$ and $n=1$.

Substitute these values into the external section formula:

$x = \frac{\lambda \cdot x_2 - 1 \cdot x_1}{\lambda-1} = \frac{\lambda x_2 - x_1}{\lambda-1}$

$y = \frac{\lambda \cdot y_2 - 1 \cdot y_1}{\lambda-1} = \frac{\lambda y_2 - y_1}{\lambda-1}$

$z = \frac{\lambda \cdot z_2 - 1 \cdot z_1}{\lambda-1} = \frac{\lambda z_2 - z_1}{\lambda-1}$

So, the coordinates of the point are $\left(\frac{\lambda x_2 - x_1}{\lambda-1}, \frac{\lambda y_2 - y_1}{\lambda-1}, \frac{\lambda z_2 - z_1}{\lambda-1}\right)$.

Compare this result with the given options:

(A) This is the formula for internal division.

(B) This matches the derived formula for external division in the ratio $\lambda:1$.

(C) This is the formula for internal division in the ratio $1:\lambda$.

(D) This is the formula for external division in the ratio $1:\lambda$ or $\lambda:1$ if the points are swapped.

The coordinates of the point that divides the line segment joining $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$ externally in the ratio $\lambda:1$ are indeed $\left(\frac{\lambda x_2 - x_1}{\lambda-1}, \frac{\lambda y_2 - y_1}{\lambda-1}, \frac{\lambda z_2 - z_1}{\lambda-1}\right)$.

Thus, the correct option is (B) $(\frac{\lambda x_2 - x_1}{\lambda-1}, \frac{\lambda y_2 - y_1}{\lambda-1}, \frac{\lambda z_2 - z_1}{\lambda-1})$.

Let the given point be $P(x, y, z) = (-1, -2, -3)$.

The yz-plane is the plane defined by the equation $x=0$.

The distance of a point $(x_0, y_0, z_0)$ from the plane $Ax+By+Cz+D=0$ is given by the formula $\frac{|Ax_0+By_0+Cz_0+D|}{\sqrt{A^2+B^2+C^2}}$.

For the yz-plane, the equation is $x=0$, which can be written as $1 \cdot x + 0 \cdot y + 0 \cdot z + 0 = 0$.

So, $A=1$, $B=0$, $C=0$, and $D=0$.

The point is $(x_0, y_0, z_0) = (-1, -2, -3)$.

The distance of the point $(-1, -2, -3)$ from the yz-plane is:

$d = \frac{|1 \cdot (-1) + 0 \cdot (-2) + 0 \cdot (-3) + 0|}{\sqrt{1^2 + 0^2 + 0^2}}$

$d = \frac{|-1 + 0 + 0 + 0|}{\sqrt{1}}$

$d = \frac{|-1|}{1}

$d = |-1|$

$d = 1$

Alternatively, the distance of a point $(x, y, z)$ from the yz-plane is simply the absolute value of its x-coordinate, which is $|x|$.

For the point $(-1, -2, -3)$, the distance from the yz-plane is $|-1| = 1$.

Compare this result with the given options:

(A) $|-1| = 1$. This matches our result.

(B) $|-2| = 2$. This is the distance from the xz-plane.

(C) $|-3| = 3$. This is the distance from the xy-plane.

(D) $\sqrt{(-2)^2 + (-3)^2} = \sqrt{4+9} = \sqrt{13}$. This is the distance from the x-axis.

The distance of the point $(-1, -2, -3)$ from the yz-plane is $|-1| = 1$.

Thus, the correct option is (A) $|-1| = 1$.

The distance between two points in three-dimensional space, say $P(x_1, y_1, z_1)$ and $Q(x_2, y_2, z_2)$, is a fundamental concept in coordinate geometry.

The formula used to calculate this distance is derived from the Pythagorean theorem extended to three dimensions.

The formula is $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}$.

This formula is universally known as the **distance formula**.

Let's consider the given options:

(A) The **Section formula** is used to find the coordinates of a point that divides a line segment in a given ratio, not the distance between the endpoints.

(B) The **Midpoint formula** is a special case of the section formula (for a ratio of $1:1$) and is used to find the coordinates of the midpoint of a line segment, not the distance between the endpoints.

(C) The **Distance formula** is precisely the formula used to find the distance between two points.

(D) The **Centroid formula** is used to find the coordinates of the centroid of a geometric figure (like a triangle or tetrahedron) based on the coordinates of its vertices.

Therefore, the distance between two points is found using the Distance formula.

The correct option is (C) Distance.

Let the coordinates of the two points be $A(x_1, y_1, z_1) = (1, -2, 3)$ and $B(x_2, y_2, z_2) = (4, 5, -1)$.

Let the point that divides the line segment joining A and B internally in the ratio $m:n$ be $P(x, y, z)$.

The section formula for internal division in 3D is:

$x = \frac{m x_2 + n x_1}{m+n}$

$y = \frac{m y_2 + n y_1}{m+n}$

$z = \frac{m z_2 + n z_1}{m+n}$

In this question, the ratio is $2:1$ internally. So, $m=2$ and $n=1$.

Substitute the values of $(x_1, y_1, z_1)$, $(x_2, y_2, z_2)$, $m$, and $n$ into the formula:

$x = \frac{2 \cdot 4 + 1 \cdot 1}{2+1} = \frac{8 + 1}{3} = \frac{9}{3} = 3$

$y = \frac{2 \cdot 5 + 1 \cdot (-2)}{2+1} = \frac{10 - 2}{3} = \frac{8}{3}$

$z = \frac{2 \cdot (-1) + 1 \cdot 3}{2+1} = \frac{-2 + 3}{3} = \frac{1}{3}$

So, the coordinates of the point are $(3, \frac{8}{3}, \frac{1}{3})$.

Compare this result with the given options:

(A) $(\frac{2 \cdot 4 + 1 \cdot 1}{2+1}, \frac{2 \cdot 5 + 1 \cdot (-2)}{2+1}, \frac{2 \cdot (-1) + 1 \cdot 3}{2+1}) = (\frac{9}{3}, \frac{8}{3}, \frac{1}{3}) = (3, 8/3, 1/3)$. This shows the calculation steps and the final result, which matches our derivation.

(B) $(3, 8/3, 1/3)$. This is the final result, which matches our calculation.

(C) $(2, 1.5, 1) = (2, 3/2, 1)$. This does not match our result.

(D) $(3, 2, 0)$. This does not match our result.

Both options (A) and (B) provide the correct coordinates. Option (A) includes the intermediate steps of the calculation, while option (B) provides just the final coordinates. In the context of objective type questions, either form is acceptable if it correctly identifies the point.

Given that option (A) shows the formula application and simplification leading to the answer, and option (B) is just the final answer, option (A) is a more complete presentation if steps are expected in the options. However, both are numerically correct.

Since the question asks to find the coordinates and option (A) provides the coordinates derived from the correct steps, and option (B) just gives the coordinates, option (A) is a better fit as a complete answer choice description.

The correct coordinates are $(3, \frac{8}{3}, \frac{1}{3})$.

Thus, the correct option is (A) $(3, 8/3, 1/3)$.

Let the given point be $P(x, y, z) = (a, b, c)$.

To find the distance of a point from the x-axis in 3D space, we find the coordinates of the point on the x-axis that is closest to the given point. This closest point is the orthogonal projection of the given point onto the x-axis.

The x-axis consists of all points of the form $(x, 0, 0)$. The projection of the point $(a, b, c)$ onto the x-axis is the point where the y and z coordinates are zero, while the x coordinate is the same as the point's x coordinate. So, the projection of $P(a, b, c)$ onto the x-axis is the point $Q(a, 0, 0)$.

The distance of the point $P(a, b, c)$ from the x-axis is the distance between $P(a, b, c)$ and $Q(a, 0, 0)$.

Using the distance formula between two points $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$, which is $d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2 + (z_2-z_1)^2}$:

Distance $= \sqrt{(a-a)^2 + (0-b)^2 + (0-c)^2}$

Simplify the expression:

Distance $= \sqrt{(0)^2 + (-b)^2 + (-c)^2}

Distance $= \sqrt{0 + b^2 + c^2}

Distance $= \sqrt{b^2 + c^2}$

Compare this result with the given options:

(A) $|a|$. This is the distance from the yz-plane.

(B) $\sqrt{b^2 + c^2}$. This matches our calculated distance.

(C) $\sqrt{a^2 + c^2}$. This is the distance from the y-axis.

(D) $\sqrt{a^2 + b^2}$. This is the distance from the z-axis.

The distance of the point $(a, b, c)$ from the x-axis is $\sqrt{b^2 + c^2}$.

Thus, the correct option is (B) $\sqrt{b^2 + c^2}$.

Let the coordinates of the points be $A(x_A, y_A, z_A) = (1, 2, 3)$, $B(x_B, y_B, z_B) = (4, 5, 6)$, and $C(x_C, y_C, z_C) = (x, y, z)$.

Given that $B$ is the midpoint of the line segment $AC$.

The midpoint formula for a line segment joining $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$ is $\left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}, \frac{z_1+z_2}{2}\right)$.

Since $B$ is the midpoint of $AC$, the coordinates of $B$ are the average of the coordinates of $A$ and $C$.

$(x_B, y_B, z_B) = \left(\frac{x_A+x_C}{2}, \frac{y_A+y_C}{2}, \frac{z_A+z_C}{2}\right)$

Substitute the given coordinates:

$(4, 5, 6) = \left(\frac{1+x}{2}, \frac{2+y}{2}, \frac{3+z}{2}\right)$

Equating the corresponding coordinates:

$\frac{1+x}{2} = 4$

$\frac{2+y}{2} = 5$

$\frac{3+z}{2} = 6$

Solve for $x, y, z$ from the equations:

From (i): $1+x = 4 \cdot 2 = 8 \implies x = 8 - 1 = 7$

From (ii): $2+y = 5 \cdot 2 = 10 \implies y = 10 - 2 = 8$

From (iii): $3+z = 6 \cdot 2 = 12 \implies z = 12 - 3 = 9$

The coordinates of point $C$ are $(7, 8, 9)$.

Compare this result with the given options:

(A) $(7, 8, 9)$. This matches our calculated coordinates.

(B) $(\frac{1+x}{2} = 4, \frac{2+y}{2} = 5, \frac{3+z}{2} = 6) \implies (1+x = 8, 2+y=10, 3+z=12) \implies (x=7, y=8, z=9)$. This option shows the steps using the midpoint formula and correctly derives the coordinates $(7, 8, 9)$.

(C) $(5, 7, 9)$. This does not match our result.

(D) $(7, 9, 11)$. This does not match our result.

Both options (A) and (B) provide the correct coordinates for point C. Option (B) includes the derivation steps, making it a more complete answer choice.

The coordinates of C are $(7, 8, 9)$.

Thus, the correct option is (B) $(7, 8, 9)$ (as it includes the derivation steps matching the structure provided in the option).

Let the coordinates of the sensor be $S(x_S, y_S, z_S) = (2, 3, 4)$ and the coordinates of source $A$ be $A(x_A, y_A, z_A) = (1, 1, 1)$.

We need to find the distance between points $S$ and $A$. The distance between two points in 3D space $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$ is given by the distance formula:

$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}$

Using the coordinates of $S(2, 3, 4)$ and $A(1, 1, 1)$, let $(x_1, y_1, z_1) = S(2, 3, 4)$ and $(x_2, y_2, z_2) = A(1, 1, 1)$.

The distance between $S$ and $A$, denoted as $d_{SA}$, is:

$d_{SA} = \sqrt{(1 - 2)^2 + (1 - 3)^2 + (1 - 4)^2}$

Calculate the differences in the coordinates:

$1 - 2 = -1$

$1 - 3 = -2$

$1 - 4 = -3$

Substitute these differences back into the distance formula:

$d_{SA} = \sqrt{(-1)^2 + (-2)^2 + (-3)^2}

Calculate the squares of the differences:

$(-1)^2 = 1$

$(-2)^2 = 4$

$(-3)^2 = 9$

Sum the squares:

$1 + 4 + 9 = 14$

Take the square root:

$d_{SA} = \sqrt{14}$

Comparing our result with the given options:

(A) $\sqrt{(2-1)^2 + (3-1)^2 + (4-1)^2} = \sqrt{1^2 + 2^2 + 3^2} = \sqrt{1+4+9} = \sqrt{14}$. This option shows the correct calculation steps and the final result.

(B) $\sqrt{14}$. This option provides the correct final result.

(C) $\sqrt{27}$. This is incorrect.

(D) $\sqrt{4+9+16} = \sqrt{29}$. This is incorrect.

Both options (A) and (B) give the correct distance. Option (A) provides the steps leading to the answer, which is useful for verification.

The distance between sensor $S(2, 3, 4)$ and source $A(1, 1, 1)$ is $\sqrt{14}$.

Thus, the correct option is (A) $\sqrt{14}$ (as it includes the calculation steps).

Let the coordinates of the sensor be $S(x_S, y_S, z_S) = (2, 3, 4)$ and the coordinates of source $B$ be $B(x_B, y_B, z_B) = (5, 5, 5)$.

We need to find the distance between points $S$ and $B$. The distance between two points in 3D space $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$ is given by the distance formula:

$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}$

Using the coordinates of $S(2, 3, 4)$ and $B(5, 5, 5)$, let $(x_1, y_1, z_1) = S(2, 3, 4)$ and $(x_2, y_2, z_2) = B(5, 5, 5)$.

The distance between $S$ and $B$, denoted as $d_{SB}$, is:

$d_{SB} = \sqrt{(5 - 2)^2 + (5 - 3)^2 + (5 - 4)^2}$

Calculate the differences in the coordinates and their squares:

$5 - 2 = 3 \implies 3^2 = 9$

$5 - 3 = 2 \implies 2^2 = 4$

$5 - 4 = 1 \implies 1^2 = 1$

Substitute the squares of the differences back into the distance formula:

$d_{SB} = \sqrt{9 + 4 + 1}

Sum the squares:

$9 + 4 + 1 = 14$

Take the square root:

$d_{SB} = \sqrt{14}$

Comparing our result with the given options:

(A) $\sqrt{(5-2)^2 + (5-3)^2 + (5-4)^2} = \sqrt{3^2 + 2^2 + 1^2} = \sqrt{9+4+1} = \sqrt{14}$. This option shows the correct calculation steps and the final result.

(B) $\sqrt{14}$. This option provides the correct final result.

(C) $\sqrt{27}$. This is incorrect.

(D) $\sqrt{9+4+1} = \sqrt{14}$. This option also shows some calculation steps and the correct final result.

Both options (A), (B), and (D) give the correct distance $\sqrt{14}$. Option (A) provides the most complete step-by-step calculation among the choices that show the steps from the differences.

The distance between sensor $S(2, 3, 4)$ and source $B(5, 5, 5)$ is $\sqrt{14}$.

Thus, the correct option is (A) $\sqrt{(5-2)^2 + (5-3)^2 + (5-4)^2} = \sqrt{14}$.

From Question 45, we found the distance between the sensor $S(2, 3, 4)$ and source $A(1, 1, 1)$.

The distance $d_{SA} = \sqrt{14}$.

From Question 46, we found the distance between the sensor $S(2, 3, 4)$ and source $B(5, 5, 5)$.

The distance $d_{SB} = \sqrt{14}$.

Now, we compare the two distances:

$d_{SA} = \sqrt{14}$

$d_{SB} = \sqrt{14}$

Since $d_{SA} = d_{SB} = \sqrt{14}$, the distance from the sensor $S$ to source $A$ is equal to the distance from the sensor $S$ to source $B$.

Therefore, the sensor is equidistant from the two sources.

Let's look at the given options:

(A) Yes, because the distances are equal. This statement is consistent with our finding.

(B) No, because the points are different. The points A and B are different, but the question is about the distance from S to A and S to B being equal. The fact that A and B are distinct points does not preclude S from being equidistant from them.

(C) Yes, but only by chance. The equality of distances is determined by the specific coordinates. While the specific values might be coincidental in a real-world scenario, mathematically, based on the given coordinates, the distances are equal. The reason for the equality (chance or design) is irrelevant to the question of whether they *are* equal based on the calculation.

(D) Cannot be determined from the distances. We have calculated the distances, and based on these values, we can directly determine if they are equal.

Based on the calculated distances, the sensor is equidistant from the two sources because the distances are equal ($\sqrt{14} = \sqrt{14}$).

Thus, the correct option is (A) Yes, because the distances are equal.

Let the given point be $P(x, y, z) = (a, b, c)$.

To find the distance of a point from the y-axis in 3D space, we find the coordinates of the point on the y-axis that is closest to the given point. This closest point is the orthogonal projection of the given point onto the y-axis.

The y-axis consists of all points of the form $(0, y, 0)$. The projection of the point $(a, b, c)$ onto the y-axis is the point where the x and z coordinates are zero, while the y coordinate is the same as the point's y coordinate. So, the projection of $P(a, b, c)$ onto the y-axis is the point $Q(0, b, 0)$.

The distance of the point $P(a, b, c)$ from the y-axis is the distance between $P(a, b, c)$ and $Q(0, b, 0)$.

Using the distance formula between two points $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$, which is $d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2 + (z_2-z_1)^2}$:

Distance $= \sqrt{(0-a)^2 + (b-b)^2 + (0-c)^2}

Simplify the expression:

Distance $= \sqrt{(-a)^2 + (0)^2 + (-c)^2}

Distance $= \sqrt{a^2 + 0 + c^2}

Distance $= \sqrt{a^2 + c^2}$

Compare this result with the given options:

(A) $|b|$. This is related to the distance from the xz-plane.

(B) $\sqrt{a^2 + c^2}$. This matches our calculated distance.

(C) $\sqrt{a^2 + b^2}$. This is the distance from the z-axis.

(D) $\sqrt{b^2 + c^2}$. This is the distance from the x-axis.

The distance of the point $(a, b, c)$ from the y-axis is $\sqrt{a^2 + c^2}$.

Thus, the correct option is (B) $\sqrt{a^2 + c^2}$.

The yz-plane is a coordinate plane in the three-dimensional Cartesian coordinate system.

By definition, the yz-plane is the plane where the x-coordinate of every point is equal to zero.

The equation of the yz-plane is $x = 0$.

To determine if a point lies on the yz-plane, we check if its x-coordinate is 0.

Let's examine the x-coordinate of each given point:

(A) For the point $(0, 5, -2)$, the x-coordinate is $0$. Since $x=0$, this point lies on the yz-plane.

(B) For the point $(0, 0, 4)$, the x-coordinate is $0$. Since $x=0$, this point lies on the yz-plane.

(C) For the point $(0, -1, 0)$, the x-coordinate is $0$. Since $x=0$, this point lies on the yz-plane.

(D) For the point $(3, 0, 0)$, the x-coordinate is $3$. Since $x \neq 0$, this point does not lie on the yz-plane. This point lies on the x-axis.

Points (A), (B), and (C) satisfy the condition $x=0$, so they lie on the yz-plane.

The correct options are (A) $(0, 5, -2)$, (B) $(0, 0, 4)$, and (C) $(0, -1, 0)$.

Let the point be $P(x, y, z)$.

The origin is the point $O(0, 0, 0)$.

The distance of the point $P(x, y, z)$ from the origin $O(0, 0, 0)$ is given by the distance formula:

$d = \sqrt{(x-0)^2 + (y-0)^2 + (z-0)^2}$

$d = \sqrt{x^2 + y^2 + z^2}$

The problem states that the distance from the origin is always 7.

To find the equation of the locus, we square both sides of the equation:

$(\sqrt{x^2 + y^2 + z^2})^2 = 7^2$

$x^2 + y^2 + z^2 = 49$

The equation $x^2 + y^2 + z^2 = r^2$ represents a sphere centered at the origin $(0, 0, 0)$ with radius $r$.

In our case, the equation is $x^2 + y^2 + z^2 = 49$, which is in the form $x^2 + y^2 + z^2 = r^2$ with $r^2 = 49$. Thus, $r = \sqrt{49} = 7$ (since radius must be non-negative).

Therefore, the locus of the point is a sphere centered at the origin with radius 7.

Let's examine the options:

(A) A circle in the xy-plane: A circle in the xy-plane is a 2D locus. The equation $x^2+y^2=r^2$ for $z=0$ represents a circle. This is not the locus in 3D space where the distance from the origin is constant.

(B) A sphere centered at the origin with radius 7: This matches our derived equation $x^2 + y^2 + z^2 = 7^2$.

(C) A line passing through the origin: A line passing through the origin has linear equations, e.g., $\frac{x}{a} = \frac{y}{b} = \frac{z}{c}$. The distance from the origin varies along a line.

(D) A cube: A cube is a solid region defined by inequalities, not the set of points equidistant from the origin.

The locus is a sphere centered at the origin with radius 7.

Thus, the correct option is (B) A sphere centered at the origin with radius 7.

Let the line segment joining $A(x_1, y_1, z_1) = (1, -1, 3)$ and $B(x_2, y_2, z_2) = (2, 4, 5)$ be divided by the xz-plane in the ratio $m:n$.

A point lies on the xz-plane if its y-coordinate is 0. So, the dividing point $P(x, y, z)$ must have $y=0$.

Using the section formula for division in the ratio $m:n$ (which can be internal or external, depending on the sign of the ratio), the y-coordinate of the dividing point is given by:

$y = \frac{m y_2 + n y_1}{m+n}$

Substitute the given values $y_1 = -1$, $y_2 = 4$, and $y = 0$ (since the point is on the xz-plane):

$0 = \frac{m \cdot 4 + n \cdot (-1)}{m+n}$

Since the denominator $m+n$ cannot be zero (otherwise the point is at infinity), the numerator must be zero:

$4m - n = 0$

Rearrange the equation to find the ratio $m:n$:

$4m = n$

$\frac{m}{n} = \frac{1}{4}$

So the ratio $m:n$ is $1:4$.

Since the ratio $\frac{m}{n} = \frac{1}{4}$ is positive, the division is internal.

The ratio is $1:4$ internally.

Compare this result with the given options:

(A) 1 : 4 internally. This matches our result.

(B) 1 : 4 externally. Incorrect because the ratio is positive.

(C) 4 : 1 internally. Incorrect ratio.

(D) 4 : 1 externally. Incorrect ratio and type of division.

The xz-plane divides the line segment joining $A(1, -1, 3)$ and $B(2, 4, 5)$ in the ratio $1:4$ internally.

Thus, the correct option is (A) 1 : 4 internally.

In a three-dimensional Cartesian coordinate system, points are represented by ordered triplets $(x, y, z)$.

The x-axis is defined as the set of all points where the y-coordinate and the z-coordinate are both zero.

Any point on the x-axis can have any real value for its x-coordinate, but its y and z coordinates must be 0.

Therefore, the coordinates of any point on the x-axis are of the form $(x, 0, 0)$.

Let's look at the options to fill in the blanks $(\_\_\_\_, \_\_\_\_)$ in $(x, \_\_\_\_, \_\_\_\_)$:

(A) 0, 0: This matches our understanding that the y and z coordinates are 0 on the x-axis.

(B) y, z: This represents a general point in 3D space, not specifically on the x-axis.

(C) 1, 0: This would mean the point is of the form $(x, 1, 0)$, which lies on a line parallel to the x-axis in the xy-plane, specifically the line $y=1, z=0$.

(D) 0, 1: This would mean the point is of the form $(x, 0, 1)$, which lies on a line parallel to the x-axis in the xz-plane, specifically the line $y=0, z=1$.

The coordinates of any point on the x-axis are of the form $(x, 0, 0)$.

Thus, the correct option is (A) 0, 0.

Given: The vertices of the triangle are $A(1, 0, 0)$, $B(0, 1, 0)$, and $C(0, 0, 1)$.

To Find: The area of triangle $ABC$.

Solution:

We can find the area of the triangle by first determining the lengths of its sides using the distance formula between two points $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$ in 3D space, which is $d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2 + (z_2-z_1)^2}$.

Length of side AB:

$AB = \sqrt{(0-1)^2 + (1-0)^2 + (0-0)^2}$

$AB = \sqrt{(-1)^2 + 1^2 + 0^2}$

$AB = \sqrt{1 + 1 + 0}$

$AB = \sqrt{2}$

Length of side BC:

$BC = \sqrt{(0-0)^2 + (0-1)^2 + (1-0)^2}$

$BC = \sqrt{0^2 + (-1)^2 + 1^2}$

$BC = \sqrt{0 + 1 + 1}$

$BC = \sqrt{2}$

Length of side CA:

$CA = \sqrt{(1-0)^2 + (0-0)^2 + (0-1)^2}$

$CA = \sqrt{1^2 + 0^2 + (-1)^2}$

$CA = \sqrt{1 + 0 + 1}$

$CA = \sqrt{2}$

Since $AB = BC = CA = \sqrt{2}$, the triangle $ABC$ is an equilateral triangle with side length $s = \sqrt{2}$.

The area of an equilateral triangle with side length $s$ is given by the formula:

Area $= \frac{\sqrt{3}}{4} s^2$

Substitute the side length $s = \sqrt{2}$:

Area $= \frac{\sqrt{3}}{4} (\sqrt{2})^2$

Area $= \frac{\sqrt{3}}{4} \times 2$

Area $= \frac{2\sqrt{3}}{4}$

Area $= \frac{\sqrt{3}}{2}$

Alternatively, using vectors:

Let $\vec{AB} = B - A = (0-1, 1-0, 0-0) = (-1, 1, 0)$.

Let $\vec{AC} = C - A = (0-1, 0-0, 1-0) = (-1, 0, 1)$.

The area of triangle ABC is half the magnitude of the cross product of $\vec{AB}$ and $\vec{AC}$.

$\vec{AB} \times \vec{AC} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ -1 & 1 & 0 \\ -1 & 0 & 1 \end{vmatrix}$

$\vec{AB} \times \vec{AC} = \mathbf{i}(1 \cdot 1 - 0 \cdot 0) - \mathbf{j}((-1) \cdot 1 - 0 \cdot (-1)) + \mathbf{k}((-1) \cdot 0 - 1 \cdot (-1))$

$\vec{AB} \times \vec{AC} = \mathbf{i}(1) - \mathbf{j}(-1) + \mathbf{k}(1)$

$\vec{AB} \times \vec{AC} = (1, 1, 1)$

The magnitude is $|\vec{AB} \times \vec{AC}| = \sqrt{1^2 + 1^2 + 1^2} = \sqrt{1+1+1} = \sqrt{3}$.

Area $= \frac{1}{2} |\vec{AB} \times \vec{AC}| = \frac{1}{2} \sqrt{3} = \frac{\sqrt{3}}{2}$.

Both methods yield the area $\frac{\sqrt{3}}{2}$.

Compare the result with the options:

(A) $\frac{\sqrt{3}}{2}$. This matches our calculated area.

(B) $\sqrt{3}/2$ (with calculation steps). This also matches our calculated area and shows the approach using side lengths.

(C) 1. Incorrect.

(D) $\sqrt{3}$. Incorrect.

The area of the triangle is $\frac{\sqrt{3}}{2}$. Both (A) and (B) present this value, with (B) providing the method using side lengths which leads to the result.

Thus, the correct option is (A) $\frac{\sqrt{3}}{2}$ or (B) $\sqrt{3}/2$. Option (B) provides the justification steps in the option text itself.

The yz-plane is one of the three coordinate planes in the three-dimensional Cartesian coordinate system.

The yz-plane is the plane that contains the y-axis and the z-axis.

For any point that lies on the yz-plane, its x-coordinate must be equal to zero.

Consider a point $P(x, y, z)$ in 3D space. If this point lies on the yz-plane, then its x-coordinate must be 0.

Thus, the condition for a point to be on the yz-plane is $x=0$.

The equation of the yz-plane is $x = 0$.

Let's look at the given options:

(A) $x = 0$: This is the equation where the x-coordinate is always zero, which defines the yz-plane.

(B) $y = 0$: This defines the xz-plane.

(C) $z = 0$: This defines the xy-plane.

(D) $y + z = 0$: This defines a plane that passes through the origin but is not one of the coordinate planes.

The equation of the yz-plane is $x=0$.

Thus, the correct option is (A) $x = 0$.

Assertion (A): The distance of the point $(3, -4, 5)$ from the origin is 5.

Let the point be $P(x, y, z) = (3, -4, 5)$ and the origin be $O(0, 0, 0)$.

The distance of the point from the origin is given by $\sqrt{x^2 + y^2 + z^2}$.

Distance $= \sqrt{3^2 + (-4)^2 + 5^2}

Distance $= \sqrt{9 + 16 + 25}

Distance $= \sqrt{50}

Distance $= \sqrt{25 \times 2} = 5\sqrt{2}$

The distance is $5\sqrt{2}$, not 5. So, Assertion (A) is false.

Reason (R): The distance from the origin is $\sqrt{3^2 + (-4)^2 + 5^2} = \sqrt{9 + 16 + 25} = \sqrt{50}$.

The calculation steps provided in Reason (R) are:

$\sqrt{3^2 + (-4)^2 + 5^2} = \sqrt{9 + 16 + 25}$

$9 + 16 + 25 = 25 + 25 = 50$

So, $\sqrt{9 + 16 + 25} = \sqrt{50}$.

The calculation in Reason (R) is correct. The distance from the origin is indeed $\sqrt{50}$. So, Reason (R) is true.

Now we evaluate the options based on our findings:

Assertion (A) is false.

Reason (R) is true.

Comparing with the options:

(A) Both A and R are true and R is the correct explanation of A. (False, since A is false)

(B) Both A and R are true but R is not the correct explanation of A. (False, since A is false)

(C) A is true but R is false. (False, since A is false and R is true)

(D) A is false but R is true. (True, as A is false and R is true)

Thus, the correct option is (D) A is false but R is true.

Let the two given points be $P_1(a, 0, 0)$ and $P_2(-a, 0, 0)$.

Let $P(x, y, z)$ be a point that is equidistant from $P_1$ and $P_2$.

The distance between $P(x, y, z)$ and $P_1(a, 0, 0)$ is $d(P, P_1) = \sqrt{(x-a)^2 + (y-0)^2 + (z-0)^2} = \sqrt{(x-a)^2 + y^2 + z^2}$.

The distance between $P(x, y, z)$ and $P_2(-a, 0, 0)$ is $d(P, P_2) = \sqrt{(x-(-a))^2 + (y-0)^2 + (z-0)^2} = \sqrt{(x+a)^2 + y^2 + z^2}$.

Since the point $P(x, y, z)$ is equidistant from $P_1$ and $P_2$, we have $d(P, P_1) = d(P, P_2)$.

Squaring both sides, we get $d(P, P_1)^2 = d(P, P_2)^2$.

Subtract $y^2 + z^2$ from both sides:

$(x-a)^2 = (x+a)^2$

Expand both sides:

$x^2 - 2ax + a^2 = x^2 + 2ax + a^2$

Subtract $x^2 + a^2$ from both sides:

$-2ax = 2ax$

Add $2ax$ to both sides:

$0 = 4ax$

Assuming $a \neq 0$ (if $a=0$, the points are the same), this equation implies $x=0$.

The locus of the point $P(x, y, z)$ is given by the equation $x=0$.

The equation $x=0$ represents the set of all points in 3D space whose x-coordinate is zero. This set of points forms the yz-plane.

Let's compare this result with the given options:

(A) x-axis: The equation of the x-axis is $y=0, z=0$.

(B) y-axis: The equation of the y-axis is $x=0, z=0$. This is a line within the yz-plane.

(C) z-axis: The equation of the z-axis is $x=0, y=0$. This is a line within the yz-plane.

(D) yz-plane: The equation of the yz-plane is $x=0$. This matches our derived locus.

The locus of a point equidistant from $(a, 0, 0)$ and $(-a, 0, 0)$ (where $a \neq 0$) is the yz-plane.

Thus, the correct option is (D) yz-plane.

Let the line segment joining $A(x_1, y_1, z_1) = (1, 2, 3)$ and $B(x_2, y_2, z_2) = (4, -5, 6)$ be divided by the xz-plane in the ratio $m:n$.

A point lies on the xz-plane if its y-coordinate is 0. So, the dividing point $P(x, y, z)$ must have $y=0$.

Using the section formula for division in the ratio $m:n$ (which can be internal or external), the y-coordinate of the dividing point is given by:

$y = \frac{m y_2 + n y_1}{m+n}$

Substitute the given values $y_1 = 2$, $y_2 = -5$, and $y = 0$ (since the point is on the xz-plane):

$0 = \frac{m \cdot (-5) + n \cdot 2}{m+n}$

Since the denominator $m+n$ cannot be zero, the numerator must be zero:

$-5m + 2n = 0$}

Rearrange the equation to find the ratio $m:n$:

$2n = 5m$

$\frac{m}{n} = \frac{2}{5}$

So the ratio $m:n$ is $2:5$.

Since the ratio $\frac{m}{n} = \frac{2}{5}$ is positive, the division is internal.

The ratio is $2:5$ internally.

Compare this result with the given options:

(A) 2 : 5 internally. This matches our result.

(B) 2 : 5 externally. Incorrect because the ratio is positive.

(C) 5 : 2 internally. Incorrect ratio.

(D) 5 : 2 externally. Incorrect ratio and type of division.

The xz-plane divides the line segment joining $A(1, 2, 3)$ and $B(4, -5, 6)$ in the ratio $2:5$ internally.

Thus, the correct option is (A) 2 : 5 internally.

Given: The two points are the origin $O(0, 0, 0)$ and $P(2a, 2b, 2c)$.

To Find: The midpoint of the line segment joining the origin and the point $(2a, 2b, 2c)$.

Solution:

Let the coordinates of the origin be $(x_1, y_1, z_1) = (0, 0, 0)$.

Let the coordinates of the other point be $(x_2, y_2, z_2) = (2a, 2b, 2c)$.

The midpoint formula for a line segment joining $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$ is:

$M = \left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}, \frac{z_1+z_2}{2}\right)$

Substitute the coordinates of the origin and the point $(2a, 2b, 2c)$ into the midpoint formula:

$M = \left(\frac{0+2a}{2}, \frac{0+2b}{2}, \frac{0+2c}{2}\right)$

Simplify the expressions:

$M = \left(\frac{2a}{2}, \frac{2b}{2}, \frac{2c}{2}\right)$

$M = (a, b, c)$

So, the coordinates of the midpoint are $(a, b, c)$.

Compare this result with the given options:

(A) $(a, b, c)$. This matches our calculated midpoint.

(B) $(2a, 2b, 2c)$. This is the given point, not the midpoint.

(C) $(a/2, b/2, c/2)$. This would be the midpoint of the segment joining the origin and $(a, b, c)$.

(D) $(0, 0, 0)$. This is the origin, not the midpoint.

The midpoint of the line segment joining the origin and the point $(2a, 2b, 2c)$ is $(a, b, c)$.

Thus, the correct option is (A) $(a, b, c)$.

Let the given point be $P(x_0, y_0, z_0) = (1, 2, 3)$.

To find the distance of a point from the z-axis in 3D space, we determine the coordinates of the point on the z-axis that is closest to the given point. This is the orthogonal projection of the given point onto the z-axis.

The z-axis consists of all points of the form $(0, 0, z)$. The projection of the point $(x_0, y_0, z_0)$ onto the z-axis is the point $(0, 0, z_0)$.

For the given point $P(1, 2, 3)$, its projection onto the z-axis is $Q(0, 0, 3)$.

The distance of the point $P(1, 2, 3)$ from the z-axis is the distance between $P(1, 2, 3)$ and $Q(0, 0, 3)$.

Using the distance formula between two points $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$, which is $d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2 + (z_2-z_1)^2}$:

Distance from z-axis $= \sqrt{(1-0)^2 + (2-0)^2 + (3-3)^2}

Simplify the expression:

Distance $= \sqrt{(1)^2 + (2)^2 + (0)^2}

Distance $= \sqrt{1^2 + 2^2 + 0}

Distance $= \sqrt{1 + 4}

Distance $= \sqrt{5}

In general, the distance of a point $(x, y, z)$ from the z-axis is $\sqrt{x^2 + y^2}$.

For the point $(1, 2, 3)$, the distance from the z-axis is $\sqrt{1^2 + 2^2} = \sqrt{1 + 4} = \sqrt{5}$.

Compare this result with the given options:

(A) $|3| = 3$. This is the distance from the xy-plane.

(B) $\sqrt{1^2 + 2^2} = \sqrt{1+4} = \sqrt{5}$. This matches our calculated distance.

(C) $\sqrt{1^2 + 3^2} = \sqrt{1+9} = \sqrt{10}$. This is the distance from the y-axis.

(D) $\sqrt{2^2 + 3^2} = \sqrt{4+9} = \sqrt{13}$. This is the distance from the x-axis.

The distance of the point $(1, 2, 3)$ from the z-axis is $\sqrt{5}$.

Thus, the correct option is (B) $\sqrt{1^2 + 2^2} = \sqrt{5}$.

In a three-dimensional Cartesian coordinate system, a line is represented by the intersection of two planes.

The x-axis is the intersection of the xy-plane and the xz-plane.

The equation of the xy-plane is $z=0$ (since the z-coordinate is zero for all points on this plane).

The equation of the xz-plane is $y=0$ (since the y-coordinate is zero for all points on this plane).

Therefore, the equation of the x-axis is given by the pair of equations $y=0$ and $z=0$. A point $(x, y, z)$ lies on the x-axis if and only if its y-coordinate is 0 and its z-coordinate is 0.

Let's look at the given options:

(A) $y = 0, z = 0$: This pair of equations represents the x-axis.

(B) $x = 0, z = 0$: This pair of equations represents the y-axis (intersection of yz-plane and xy-plane).

(C) $x = 0, y = 0$: This pair of equations represents the z-axis (intersection of yz-plane and xz-plane).

(D) $x=0$: This is the equation of the yz-plane, which is a plane, not a line (the x-axis).

The equation of the x-axis in 3D space is given by $y=0$ and $z=0$.

Thus, the correct option is (A) $y = 0, z = 0$.

Let the point be $P(x, y, z)$.

Let the two given points be $A(a, 0, 0)$ and $B(0, a, 0)$. We assume $a \neq 0$, otherwise the two points are the same.

The distance between two points $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$ is given by the distance formula:

$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}$

The distance of point $P(x, y, z)$ from point $A(a, 0, 0)$ is:

$d(P, A) = \sqrt{(x - a)^2 + (y - 0)^2 + (z - 0)^2}$

$d(P, A) = \sqrt{(x - a)^2 + y^2 + z^2}$

The distance of point $P(x, y, z)$ from point $B(0, a, 0)$ is:

$d(P, B) = \sqrt{(x - 0)^2 + (y - a)^2 + (z - 0)^2}$

$d(P, B) = \sqrt{x^2 + (y - a)^2 + z^2}$

The problem states that the point $P(x, y, z)$ is equidistant from points $A$ and $B$. Therefore, $d(P, A) = d(P, B)$.

To eliminate the square roots, square both sides of the equation:

$(x - a)^2 + y^2 + z^2 = x^2 + (y - a)^2 + z^2$

Expand the squared terms:

$(x^2 - 2ax + a^2) + y^2 + z^2 = x^2 + (y^2 - 2ay + a^2) + z^2$}

$x^2 - 2ax + a^2 + y^2 + z^2 = x^2 + y^2 - 2ay + a^2 + z^2$

Subtract $x^2$, $y^2$, $z^2$, and $a^2$ from both sides of the equation:

$-2ax = -2ay$

Rearrange the terms to one side:

$2ay - 2ax = 0$

$2a(y - x) = 0$

Since we assumed $a \neq 0$, we can divide by $2a$:

$y - x = 0$

$y = x$

or $x - y = 0$

The equation $x = y$ (or $x - y = 0$) describes the locus of the point $P(x, y, z)$. This equation is linear in $x, y, z$ (specifically, $1x - 1y + 0z + 0 = 0$) and does not involve squared terms like $x^2, y^2, z^2$ (which would indicate a sphere or other quadratic surface). A linear equation in $x, y, z$ represents a plane in 3D space.

The locus is the set of all points $(x, y, z)$ such that $x=y$. This is a plane passing through the origin and perpendicular to the vector $(1, -1, 0)$.

Let's examine the given options:

(A) A sphere: The equation of a sphere is of the form $(x-h)^2 + (y-k)^2 + (z-l)^2 = r^2$. Our locus is not a sphere.

(B) A plane: The equation of a plane is of the form $Ax + By + Cz + D = 0$. Our derived equation $x - y = 0$ is in this form (with $A=1, B=-1, C=0, D=0$). This matches.

(C) A line: A line in 3D space is typically represented by the intersection of two planes or by parametric equations.

(D) A point: A point is represented by specific fixed coordinates $(x_0, y_0, z_0)$.

The locus of a point equidistant from two fixed points is the perpendicular bisector plane of the line segment joining the two points. In this case, the line segment is on the xy-plane, and the perpendicular bisector is a plane defined by $x=y$.

Thus, the correct option is (B) A plane.

Let the coordinates of the point that divides the line segment joining $A(x_1, y_1, z_1)$ and $B(x_2, y_2, z_2)$ internally in the ratio $m:n$ be $P(x, y, z)$.

According to the section formula for internal division in three dimensions, the coordinates of the dividing point $P$ are given by:

$x = \frac{m x_2 + n x_1}{m+n}$

$y = \frac{m y_2 + n y_1}{m+n}$

$z = \frac{m z_2 + n z_1}{m+n}$

So, the coordinates of the point are $\left(\frac{m x_2 + n x_1}{m+n}, \frac{m y_2 + n y_1}{m+n}, \frac{m z_2 + n z_1}{m+n}\right)$.

Let's compare this formula with the given options:

(A) $(\frac{nx_1 + mx_2}{m+n}, \frac{ny_1 + my_2}{m+n}, \frac{nz_1 + mz_2}{m+n})$

Using the commutative property of addition, the numerator of the x-coordinate in option (A) is $nx_1 + mx_2 = mx_2 + nx_1$. The denominator is $m+n$. So, option (A) is $(\frac{mx_2 + nx_1}{m+n}, \frac{my_2 + ny_1}{m+n}, \frac{mz_2 + nz_1}{m+n})$. This matches the standard formula derived above.

(B) $(\frac{mx_1 + nx_2}{m+n}, \frac{my_1 + ny_2}{m+n}, \frac{mz_1 + nz_2}{m+n})$

In this option, the ratio $m$ is multiplied by the coordinates of the first point $(x_1, y_1, z_1)$, and the ratio $n$ is multiplied by the coordinates of the second point $(x_2, y_2, z_2)$. This is incorrect for dividing the segment from A to B in the ratio $m:n$. It represents division in the ratio $n:m$ or from B to A in the ratio $m:n$.

(C) This formula uses subtraction, which is for external division.

(D) This formula uses subtraction, which is for external division.

The coordinates of the point that divides the line segment joining $A(x_1, y_1, z_1)$ and $B(x_2, y_2, z_2)$ internally in the ratio $m:n$ is $(\frac{mx_2 + nx_1}{m+n}, \frac{my_2 + ny_1}{m+n}, \frac{mz_2 + nz_1}{m+n})$.

Option (A) is an equivalent representation of this formula due to the commutative property of addition in the numerator.

Thus, the correct option is (A) $(\frac{nx_1 + mx_2}{m+n}, \frac{ny_1 + my_2}{m+n}, \frac{nz_1 + mz_2}{m+n})$.

Given:

Points $P(x, y, z), Q(1, 2, 3), R(-1, -2, -3)$ are collinear.

To Find:

The ratio in which $Q$ divides $PR$.

Solution:

Let the origin be $O(0, 0, 0)$. We are given the points $Q(1, 2, 3)$ and $R(-1, -2, -3)$.

Let's find the distance of $Q$ and $R$ from the origin $O$.

$OQ = \sqrt{(1-0)^2 + (2-0)^2 + (3-0)^2} = \sqrt{1^2 + 2^2 + 3^2} = \sqrt{1 + 4 + 9} = \sqrt{14}$

$OR = \sqrt{(-1-0)^2 + (-2-0)^2 + (-3-0)^2} = \sqrt{(-1)^2 + (-2)^2 + (-3)^2} = \sqrt{1 + 4 + 9} = \sqrt{14}$

Also, we notice that the coordinates of $R$ are the negation of the coordinates of $Q$. This means the position vector of $R$ is the negative of the position vector of $Q$, i.e., $\vec{OR} = - \vec{OQ}$.

This implies that the origin $O(0, 0, 0)$, $Q(1, 2, 3)$, and $R(-1, -2, -3)$ are collinear, and $O$ is the midpoint of the line segment $QR$.

The distance $QR$ is:

$QR = \sqrt{(-1-1)^2 + (-2-2)^2 + (-3-3)^2} = \sqrt{(-2)^2 + (-4)^2 + (-6)^2} = \sqrt{4 + 16 + 36} = \sqrt{56}$

$\sqrt{56} = \sqrt{4 \times 14} = 2\sqrt{14}$.

We can verify that $OQ + OR = \sqrt{14} + \sqrt{14} = 2\sqrt{14} = QR$, confirming $O$ is on the segment $QR$ and $OQ = OR$. Thus, $O$ is the midpoint of $QR$.

We are given that points $P(x, y, z), Q(1, 2, 3), R(-1, -2, -3)$ are collinear. Since $Q$ and $R$ define a line that passes through the origin $O$, the point $P(x, y, z)$ must also lie on this line.

The question asks for the ratio in which $Q$ divides $PR$. Let this ratio be $m:n$. According to the section formula, the coordinates of $Q$ are given by:

$Q = \left( \frac{nx + m(-1)}{m+n}, \frac{ny + m(-2)}{m+n}, \frac{nz + m(-3)}{m+n} \right)$

$(1, 2, 3) = \left( \frac{nx - m}{m+n}, \frac{ny - 2m}{m+n}, \frac{nz - 3m}{m+n} \right)$

Equating the coordinates, we get:

$1 = \frac{nx - m}{m+n} \implies n+m = nx - m \implies nx - 2m = n$

$2 = \frac{ny - 2m}{m+n} \implies 2n+2m = ny - 2m \implies ny - 4m = 2n$

$3 = \frac{nz - 3m}{m+n} \implies 3n+3m = nz - 3m \implies nz - 6m = 3n$

Since $P$ is collinear with $O, Q, R$, its coordinates $(x, y, z)$ must be a scalar multiple of $(1, 2, 3)$. Let $P = s(1, 2, 3) = (s, 2s, 3s)$ for some scalar $s$. Then $x=s, y=2s, z=3s$. Substituting these into the equations above:

$n(s) - 2m = n \implies ns - n = 2m \implies n(s-1) = 2m$

$n(2s) - 4m = 2n \implies 2ns - 4m = 2n \implies ns - 2m = n \implies n(s-1) = 2m$

$n(3s) - 6m = 3n \implies 3ns - 6m = 3n \implies ns - 2m = n \implies n(s-1) = 2m$

All equations yield $n(s-1) = 2m$. The ratio in which $Q$ divides $PR$ is $m:n$. From this equation, we get $\frac{m}{n} = \frac{s-1}{2}$.

The ratio depends on the parameter $s$, which defines the position of $P$ on the line. However, the options provided are fixed numerical ratios, implying the ratio in question is independent of the specific choice of $P(x,y,z)$ (as long as it is collinear with $Q$ and $R$).

Given the special relationship between $Q$, $R$, and the origin $O$ (where $O$ is the midpoint of $QR$), and the nature of the multiple-choice options, it is probable that the question implicitly refers to a ratio derived from the fixed points $O, Q, R$. A natural interpretation that yields a fixed ratio among the options is the ratio of the distance from $O$ to $Q$ and the distance from $Q$ to $R$.

We have already calculated $OQ = \sqrt{14}$ and $QR = 2\sqrt{14}$.

The ratio $OQ:QR = \sqrt{14} : 2\sqrt{14} = 1:2$.

This ratio matches option (B).

While the question phrasing "the ratio in which $Q$ divides $PR$" normally refers to the section ratio $m:n$ which depends on $P$, in the context of a multiple-choice question with fixed options and the special geometry of $Q$ and $R$ relative to the origin, the ratio $OQ:QR$ is the most likely intended answer that is constant for any collinear $P$ (as it is independent of $P$).

The final answer is $\boxed{1:2}$.

The correct option is (B).

Solution:

To find the distance of a point $P(x, y, z)$ from the x-axis, we need to find the coordinates of the point on the x-axis that is closest to $P$.

Any point on the x-axis has coordinates of the form $(x', 0, 0)$.

The point on the x-axis closest to $P(x, y, z)$ will have the same x-coordinate as $P$, and its y and z coordinates will be 0. Let's call this point $P'(x, 0, 0)$.

The distance between $P(x, y, z)$ and $P'(x, 0, 0)$ is given by the distance formula in three dimensions:

$D = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}$

Here, $(x_1, y_1, z_1) = (x, y, z)$ and $(x_2, y_2, z_2) = (x, 0, 0)$.

$D = \sqrt{(x - x)^2 + (0 - y)^2 + (0 - z)^2}$

$D = \sqrt{(0)^2 + (-y)^2 + (-z)^2}$

$D = \sqrt{0 + y^2 + z^2}$

$D = \sqrt{y^2 + z^2}$

Thus, the distance of the point $(x, y, z)$ from the x-axis is $\sqrt{y^2 + z^2}$.

Comparing this result with the given options, we see that it matches option (B).

The correct option is (B).

Solution:

Let's analyze the Assertion (A):

The three coordinate planes (xy-plane, yz-plane, and xz-plane) intersect at the origin and are mutually perpendicular. Each plane divides the three-dimensional space into two parts. The combination of these three planes divides the space into $2 \times 2 \times 2 = 8$ regions. These regions are called octants.

Thus, Assertion (A) is True.

Let's analyze the Reason (R):

Each octant is specifically defined by the signs of the x, y, and z coordinates of the points within it. For example, the first octant is where $x > 0, y > 0, z > 0$. The other seven octants correspond to the other seven combinations of positive and negative signs for the three coordinates.

Thus, Reason (R) is True.

Now, let's determine if Reason (R) is the correct explanation for Assertion (A).

The coordinate planes are defined by $x=0$, $y=0$, and $z=0$. These planes are the boundaries where the signs of the coordinates change. The regions between these planes, where the signs of $x, y,$ and $z$ are constant, are precisely the octants. The 8 possible combinations of signs for $(x, y, z)$ (excluding the planes themselves) uniquely define the 8 octants. Therefore, the definition of octants based on the signs of coordinates directly explains how the coordinate planes divide the space into these 8 regions.

Thus, Reason (R) is the correct explanation for Assertion (A).

Both Assertion (A) and Reason (R) are true, and Reason (R) correctly explains Assertion (A).

The correct option is (A).

Solution:

A point lies on the z-axis if and only if its x-coordinate and y-coordinate are both zero.

The general form of a point on the z-axis is $(0, 0, k)$, where $k$ is any real number.

Let's examine each option:

(A) $(2, 0, 0)$: Here, the x-coordinate is $2$, which is not zero. This point lies on the x-axis.

(B) $(0, 3, 0)$: Here, the y-coordinate is $3$, which is not zero. This point lies on the y-axis.

(C) $(0, 0, -4)$: Here, the x-coordinate is $0$ and the y-coordinate is $0$. The z-coordinate is $-4$. This point fits the form $(0, 0, k)$.

(D) $(1, 1, 1)$: Here, both the x-coordinate and y-coordinate are not zero.

Therefore, the point that lies on the z-axis is $(0, 0, -4)$.

The correct option is (C).

In three-dimensional space, the coordinate planes (xy-plane, yz-plane, and xz-plane) divide the space into eight regions called octants.

The sign of the x, y, and z coordinates determines the octant in which a point lies. The octants are typically numbered based on the signs as follows:

Octant I: (+, +, +)

Octant II: (-, +, +)

Octant III: (-, -, +)

Octant IV: (+, -, +)

Octant V: (+, +, -)

Octant VI: (-, +, -)

Octant VII: (-, -, -)

Octant VIII: (+, -, -)

For the point (1, 2, 3):

The x-coordinate is 1 (positive).

The y-coordinate is 2 (positive).

The z-coordinate is 3 (positive).

Since all three coordinates are positive (+, +, +), the point (1, 2, 3) lies in the first octant.

For the point (-1, 1, 2):

The x-coordinate is -1 (negative).

The y-coordinate is 1 (positive).

The z-coordinate is 2 (positive).

The signs of the coordinates are (-, +, +). This combination corresponds to the second octant.

Therefore, the point (-1, 1, 2) lies in the second octant.

Given:

Point 1: $(x_1, y_1, z_1) = (2, 3, 5)$

Point 2: $(x_2, y_2, z_2) = (4, 3, 1)$

To Find:

The distance between the two points.

Solution:

The distance between two points $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$ in three-dimensional space is given by the distance formula:

$\text{Distance} = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2 + (z_2-z_1)^2}$

Substitute the coordinates of the given points into the formula:

$x_1 = 2, y_1 = 3, z_1 = 5$

$x_2 = 4, y_2 = 3, z_2 = 1$

Distance $= \sqrt{(4-2)^2 + (3-3)^2 + (1-5)^2}$

Distance $= \sqrt{(2)^2 + (0)^2 + (-4)^2}$

Distance $= \sqrt{4 + 0 + 16}$

Distance $= \sqrt{20}$

To simplify $\sqrt{20}$, we find the prime factorization of 20:

$\sqrt{20} = \sqrt{4 \times 5} = \sqrt{4} \times \sqrt{5} = 2\sqrt{5}$

The distance between the points (2, 3, 5) and (4, 3, 1) is $\underline{2\sqrt{5}}$ units.

Given:

Point 1: $P_1(x_1, y_1, z_1) = (-2, 3, 5)$

Point 2: $P_2(x_2, y_2, z_2) = (1, -4, 6)$

Ratio of internal division: $m:n = 2:3$, so $m=2$ and $n=3$.

To Find:

The coordinates of the point that divides the line segment joining $P_1$ and $P_2$ internally in the ratio $2:3$.

Solution:

Let the point be $P(x, y, z)$. The coordinates of a point that divides the line segment joining $P_1(x_1, y_1, z_1)$ and $P_2(x_2, y_2, z_2)$ internally in the ratio $m:n$ are given by the section formula:

$x = \frac{mx_2 + nx_1}{m+n}$

$y = \frac{my_2 + ny_1}{m+n}$

$z = \frac{mz_2 + nz_1}{m+n}$

Substitute the given values into the formula:

$x = \frac{(2)(1) + (3)(-2)}{2+3} = \frac{2 - 6}{5} = \frac{-4}{5}$

$y = \frac{(2)(-4) + (3)(3)}{2+3} = \frac{-8 + 9}{5} = \frac{1}{5}$

$z = \frac{(2)(6) + (3)(5)}{2+3} = \frac{12 + 15}{5} = \frac{27}{5}$

The coordinates of the point are $\left(-\frac{4}{5}, \frac{1}{5}, \frac{27}{5}\right)$.

Given:

The vertices of the triangle are:

$A(x_1, y_1, z_1) = (0, 0, 0)$

$B(x_2, y_2, z_2) = (a, 0, 0)$

$C(x_3, y_3, z_3) = (0, b, 0)$

To Find:

The coordinates of the centroid of the triangle ABC.

Solution:

The coordinates of the centroid of a triangle with vertices $(x_1, y_1, z_1)$, $(x_2, y_2, z_2)$, and $(x_3, y_3, z_3)$ are given by the formula:

$G(x, y, z) = \left(\frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3}, \frac{z_1 + z_2 + z_3}{3}\right)$

Substitute the coordinates of the given vertices into the formula:

$x = \frac{0 + a + 0}{3} = \frac{a}{3}$

$y = \frac{0 + 0 + b}{3} = \frac{b}{3}$

$z = \frac{0 + 0 + 0}{3} = \frac{0}{3} = 0$

Thus, the coordinates of the centroid of the triangle are $\left(\frac{a}{3}, \frac{b}{3}, 0\right)$.

Given:

The three points are A(0, 7, 10), B(-1, 6, 6), and C(-4, 9, 6).

Solution:

To show that the points form a right-angled triangle, we can calculate the squares of the lengths of the sides of the triangle formed by these points. We will use the distance formula in three dimensions, which states that the distance $d$ between two points $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$ is given by:

$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}$

The square of the distance between the two points is:

$d^2 = (x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2$

Let's calculate the square of the lengths of the sides AB, BC, and AC.

For side AB, using points A(0, 7, 10) and B(-1, 6, 6):

$AB^2 = (-1 - 0)^2 + (6 - 7)^2 + (6 - 10)^2$

$AB^2 = (-1)^2 + (-1)^2 + (-4)^2$

$AB^2 = 1 + 1 + 16$

$AB^2 = 18$

For side BC, using points B(-1, 6, 6) and C(-4, 9, 6):

$BC^2 = (-4 - (-1))^2 + (9 - 6)^2 + (6 - 6)^2$

$BC^2 = (-4 + 1)^2 + (3)^2 + (0)^2$

$BC^2 = (-3)^2 + 9 + 0$

$BC^2 = 9 + 9$

$BC^2 = 18$

For side AC, using points A(0, 7, 10) and C(-4, 9, 6):

$AC^2 = (-4 - 0)^2 + (9 - 7)^2 + (6 - 10)^2$

$AC^2 = (-4)^2 + (2)^2 + (-4)^2$

$AC^2 = 16 + 4 + 16$

$AC^2 = 36$

Now, we check if the square of the longest side is equal to the sum of the squares of the other two sides using the Pythagorean theorem. The squares of the side lengths are 18, 18, and 36.

We observe that:

$AB^2 + BC^2 = 18 + 18 = 36$

And

$AC^2 = 36$

Since $AB^2 + BC^2 = AC^2$, the triangle formed by the points A, B, and C satisfies the Pythagorean theorem.

Therefore, the triangle ABC is a right-angled triangle, with the right angle at the vertex opposite to the longest side AC, which is vertex B.

Conclusion:

The points (0, 7, 10), (-1, 6, 6), and (-4, 9, 6) are indeed the vertices of a right-angled triangle.

Given:

The two points are A(1, -2, 3) and B(3, 4, -5).

The ratio of external division is $m:n = 2:3$.

To Find:

The coordinates of the point that divides the line segment AB externally in the ratio 2:3.

Solution:

Let the coordinates of the point be P(x, y, z). The formula for the coordinates of a point that divides the line segment joining $A(x_1, y_1, z_1)$ and $B(x_2, y_2, z_2)$ externally in the ratio $m:n$ is:

$x = \frac{m x_2 - n x_1}{m - n}$

$y = \frac{m y_2 - n y_1}{m - n}$

$z = \frac{m z_2 - n z_1}{m - n}$

Here, $(x_1, y_1, z_1) = (1, -2, 3)$, $(x_2, y_2, z_2) = (3, 4, -5)$, $m = 2$, and $n = 3$.

Substituting these values into the formula, we get:

$x = \frac{(2)(3) - (3)(1)}{2 - 3}$

$x = \frac{6 - 3}{-1}$

$x = \frac{3}{-1}$

$x = -3$

$y = \frac{(2)(4) - (3)(-2)}{2 - 3}$

$y = \frac{8 - (-6)}{-1}$

$y = \frac{8 + 6}{-1}$

$y = \frac{14}{-1}$

$y = -14$

$z = \frac{(2)(-5) - (3)(3)}{2 - 3}$

$z = \frac{-10 - 9}{-1}$

$z = \frac{-19}{-1}$

$z = 19$

Therefore, the coordinates of the point are (-3, -14, 19).

Conclusion:

The coordinates of the point which divides the line segment joining the points (1, -2, 3) and (3, 4, -5) externally in the ratio 2:3 are (-3, -14, 19).

Given:

The point is P(a, b, c).

We need to find the distance of this point from the xz-plane.

To Find:

The distance of the point (a, b, c) from the xz-plane.

Solution:

In a three-dimensional coordinate system, the xz-plane is defined by the set of all points where the y-coordinate is zero. The equation of the xz-plane is $y = 0$.

The distance of a point $(x_0, y_0, z_0)$ from a plane with the equation $Ax + By + Cz + D = 0$ is given by the formula:

Distance $= \frac{|Ax_0 + By_0 + Cz_0 + D|}{\sqrt{A^2 + B^2 + C^2}}$

The equation of the xz-plane can be written as $0 \cdot x + 1 \cdot y + 0 \cdot z + 0 = 0$.

So, for the xz-plane, we have $A = 0$, $B = 1$, $C = 0$, and $D = 0$.

The given point is $(x_0, y_0, z_0) = (a, b, c)$.

Substituting these values into the distance formula:

Distance $= \frac{|(0)(a) + (1)(b) + (0)(c) + 0|}{\sqrt{0^2 + 1^2 + 0^2}}$

Distance $= \frac{|0 + b + 0 + 0|}{\sqrt{0 + 1 + 0}}$

Distance $= \frac{|b|}{\sqrt{1}}$

Distance $= |b|$

Alternatively, the distance of a point from any coordinate plane is the absolute value of the coordinate perpendicular to that plane. For the xz-plane, the perpendicular axis is the y-axis. Therefore, the distance of the point (a, b, c) from the xz-plane is the absolute value of its y-coordinate.

Distance $= |b|$

Conclusion:

The distance of the point (a, b, c) from the xz-plane is $|b|$.

Given:

The two points are A(-2, 4, 7) and B(3, -5, 8).

The dividing plane is the YZ-plane.

To Find:

The ratio in which the YZ-plane divides the line segment joining the points A and B.

Solution:

Let the YZ-plane divide the line segment joining the points A($x_1$, $y_1$, $z_1$) = (-2, 4, 7) and B($x_2$, $y_2$, $z_2$) = (3, -5, 8) in the ratio $m:n$.

Let the point of division be P(x, y, z).

The coordinates of the point of division P are given by the section formula:

$x = \frac{m x_2 + n x_1}{m + n}$

$y = \frac{m y_2 + n y_1}{m + n}$

$z = \frac{m z_2 + n z_1}{m + n}$

The equation of the YZ-plane is $x = 0$. This means that any point lying on the YZ-plane has its x-coordinate equal to 0.

Since the point P lies on the YZ-plane, its x-coordinate must be 0.

Using the section formula for the x-coordinate and setting it to 0:

$0 = \frac{m (3) + n (-2)}{m + n}$

$0 = \frac{3m - 2n}{m + n}$

For this equation to hold true, the numerator must be zero (assuming $m+n \neq 0$, which is true since A and B are distinct points):

$3m - 2n = 0$

$3m = 2n$

To find the ratio $m:n$, we can write:

$\frac{m}{n} = \frac{2}{3}$

So, the ratio $m:n$ is $2:3$. Since the ratio is positive, the division is internal.

Conclusion:

The YZ-plane divides the line segment joining the points (-2, 4, 7) and (3, -5, 8) in the ratio 2:3 internally.

Given:

The three collinear points are P(3, 2, -4), Q(5, 4, -6), and R(9, 8, -10).

To Find:

The ratio in which point Q divides the line segment PR.

Solution:

Let the point Q divide the line segment joining P($x_1, y_1, z_1$) = (3, 2, -4) and R($x_2, y_2, z_2$) = (9, 8, -10) in the ratio $k:1$.

Using the section formula for internal division, the coordinates of the point Q(x, y, z) are given by:

$x = \frac{k x_2 + 1 x_1}{k + 1}$

$y = \frac{k y_2 + 1 y_1}{k + 1}$

$z = \frac{k z_2 + 1 z_1}{k + 1}$

The coordinates of Q are given as (5, 4, -6). We can use any one coordinate to find the value of $k$. Let's use the x-coordinate:

$5 = \frac{k (9) + 1 (3)}{k + 1}$

$5 (k + 1) = 9k + 3$

$5k + 5 = 9k + 3$

$5 - 3 = 9k - 5k$

$2 = 4k$

$k = \frac{2}{4}$

$k = \frac{1}{2}$

So, the ratio $k:1$ is $\frac{1}{2}:1$. Multiplying both parts by 2 to get integer ratio, we get $1:2$.

We can verify this using the other coordinates.

Using the y-coordinate:

$4 = \frac{k (8) + 1 (2)}{k + 1}$

$4 (k + 1) = 8k + 2$

$4k + 4 = 8k + 2$

$4 - 2 = 8k - 4k$

$2 = 4k$

$k = \frac{1}{2}$

Using the z-coordinate:

$-6 = \frac{k (-10) + 1 (-4)}{k + 1}$

$-6 (k + 1) = -10k - 4$

$-6k - 6 = -10k - 4$

$-6 + 4 = -10k + 6k$

$-2 = -4k$

$k = \frac{-2}{-4}$

$k = \frac{1}{2}$

All coordinates give the same value of $k = \frac{1}{2}$. Since $k$ is positive, Q divides PR internally.

The ratio is $k:1 = \frac{1}{2}:1$, which is equivalent to $1:2$.

Conclusion:

The point Q divides the line segment PR in the ratio 1:2 internally.

Given:

The point is P(1, -1, 3).

The origin is O(0, 0, 0).

To Find:

The distance of the point P(1, -1, 3) from the origin O(0, 0, 0).

Solution:

Let the point be $P(x, y, z) = (1, -1, 3)$ and the origin be $O(0, 0, 0)$.

The distance between two points $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$ in three-dimensional space is given by the distance formula:

$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}$

In this case, $(x_1, y_1, z_1) = (0, 0, 0)$ and $(x_2, y_2, z_2) = (1, -1, 3)$.

Substituting the coordinates into the formula:

Distance OP $= \sqrt{(1 - 0)^2 + (-1 - 0)^2 + (3 - 0)^2}$

Distance OP $= \sqrt{(1)^2 + (-1)^2 + (3)^2}$

Distance OP $= \sqrt{1 + 1 + 9}$

Distance OP $= \sqrt{11}$

The distance of the point (1, -1, 3) from the origin is $\sqrt{11}$.

Conclusion:

The distance of the point (1, -1, 3) from the origin is $\sqrt{11}$.

Given:

The two points are A(4, -5, 6) and B(1, 2, 3).

The ratio of internal division is $m:n = 1:2$.

To Find:

The coordinates of the point that divides the line segment AB internally in the ratio 1:2.

Solution:

Let the coordinates of the point be P(x, y, z). The formula for the coordinates of a point that divides the line segment joining $A(x_1, y_1, z_1)$ and $B(x_2, y_2, z_2)$ internally in the ratio $m:n$ is:

$x = \frac{m x_2 + n x_1}{m + n}$

$y = \frac{m y_2 + n y_1}{m + n}$

$z = \frac{m z_2 + n z_1}{m + n}$

Here, $(x_1, y_1, z_1) = (4, -5, 6)$, $(x_2, y_2, z_2) = (1, 2, 3)$, $m = 1$, and $n = 2$.

Substituting these values into the formula, we get:

$x = \frac{(1)(1) + (2)(4)}{1 + 2}$

$x = \frac{1 + 8}{3}$

$x = \frac{9}{3}$

$x = 3$

$y = \frac{(1)(2) + (2)(-5)}{1 + 2}$

$y = \frac{2 - 10}{3}$

$y = \frac{-8}{3}$

$z = \frac{(1)(3) + (2)(6)}{1 + 2}$

$z = \frac{3 + 12}{3}$

$z = \frac{15}{3}$

$z = 5$

Therefore, the coordinates of the point are $(3, -\frac{8}{3}, 5)$.

Conclusion:

The coordinates of the point which divides the line segment joining the points (4, -5, 6) and (1, 2, 3) internally in the ratio 1:2 are $(3, -\frac{8}{3}, 5)$.

Given:

Two points A(1, 2, 3) and B(3, 2, -1).

The condition that the sum of the squares of the distances from a point P to A and B is 20.

To Find:

The equation of the set of all points P(x, y, z) that satisfy the given condition.

Solution:

Let P(x, y, z) be any point in the set. The distance between two points $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$ is given by $\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}$. The square of the distance is $(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2$.

The square of the distance between P(x, y, z) and A(1, 2, 3) is:

$PA^2 = (x - 1)^2 + (y - 2)^2 + (z - 3)^2$

Expanding this, we get:

$PA^2 = (x^2 - 2x + 1) + (y^2 - 4y + 4) + (z^2 - 6z + 9)$

$PA^2 = x^2 - 2x + y^2 - 4y + z^2 - 6z + 1 + 4 + 9$

$PA^2 = x^2 + y^2 + z^2 - 2x - 4y - 6z + 14$

The square of the distance between P(x, y, z) and B(3, 2, -1) is:

$PB^2 = (x - 3)^2 + (y - 2)^2 + (z - (-1))^2$

$PB^2 = (x - 3)^2 + (y - 2)^2 + (z + 1)^2$

Expanding this, we get:

$PB^2 = (x^2 - 6x + 9) + (y^2 - 4y + 4) + (z^2 + 2z + 1)$

$PB^2 = x^2 - 6x + y^2 - 4y + z^2 + 2z + 9 + 4 + 1$

$PB^2 = x^2 + y^2 + z^2 - 6x - 4y + 2z + 14$

The given condition is $PA^2 + PB^2 = 20$. Substituting the expressions for $PA^2$ and $PB^2$:

$(x^2 + y^2 + z^2 - 2x - 4y - 6z + 14) + (x^2 + y^2 + z^2 - 6x - 4y + 2z + 14) = 20$

Combine the like terms:

$(x^2 + x^2) + (y^2 + y^2) + (z^2 + z^2) + (-2x - 6x) + (-4y - 4y) + (-6z + 2z) + (14 + 14) = 20$

$2x^2 + 2y^2 + 2z^2 - 8x - 8y - 4z + 28 = 20$

Subtract 20 from both sides:

$2x^2 + 2y^2 + 2z^2 - 8x - 8y - 4z + 28 - 20 = 0$

$2x^2 + 2y^2 + 2z^2 - 8x - 8y - 4z + 8 = 0$

Divide the entire equation by 2:

$x^2 + y^2 + z^2 - 4x - 4y - 2z + 4 = 0$

This is the equation of the set of points that satisfy the given condition.

Conclusion:

The equation of the set of points is $x^2 + y^2 + z^2 - 4x - 4y - 2z + 4 = 0$.

Given:

The point is P(-4, 3, 5).

We need to find the distance of this point from the x-axis, y-axis, and z-axis.

To Find:

The distances of the point (-4, 3, 5) from the coordinate axes.

Solution:

The distance of a point $(x_0, y_0, z_0)$ from the coordinate axes can be found using the distance formula to the point on the axis closest to the given point.

The point on the x-axis closest to $(x_0, y_0, z_0)$ is $(x_0, 0, 0)$.

The point on the y-axis closest to $(x_0, y_0, z_0)$ is $(0, y_0, 0)$.

The point on the z-axis closest to $(x_0, y_0, z_0)$ is $(0, 0, z_0)$.

The distance formula between two points $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$ is $\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}$.

Given point P(-4, 3, 5), so $(x_0, y_0, z_0) = (-4, 3, 5)$.

Distance from the x-axis:

The closest point on the x-axis is (-4, 0, 0). The distance is:

$d_x = \sqrt{(-4 - (-4))^2 + (3 - 0)^2 + (5 - 0)^2}$

$d_x = \sqrt{(0)^2 + (3)^2 + (5)^2}$

$d_x = \sqrt{0 + 9 + 25}$

$d_x = \sqrt{34}$

Alternatively, the distance from the x-axis is $\sqrt{y_0^2 + z_0^2}$.

$d_x = \sqrt{3^2 + 5^2} = \sqrt{9 + 25} = \sqrt{34}$.

Distance from the y-axis:

The closest point on the y-axis is (0, 3, 0). The distance is:

$d_y = \sqrt{(-4 - 0)^2 + (3 - 3)^2 + (5 - 0)^2}$

$d_y = \sqrt{(-4)^2 + (0)^2 + (5)^2}$

$d_y = \sqrt{16 + 0 + 25}$

$d_y = \sqrt{41}$

Alternatively, the distance from the y-axis is $\sqrt{x_0^2 + z_0^2}$.

$d_y = \sqrt{(-4)^2 + 5^2} = \sqrt{16 + 25} = \sqrt{41}$.

Distance from the z-axis:

The closest point on the z-axis is (0, 0, 5). The distance is:

$d_z = \sqrt{(-4 - 0)^2 + (3 - 0)^2 + (5 - 5)^2}$

$d_z = \sqrt{(-4)^2 + (3)^2 + (0)^2}$

$d_z = \sqrt{16 + 9 + 0}$

$d_z = \sqrt{25}$

$d_z = 5$

Alternatively, the distance from the z-axis is $\sqrt{x_0^2 + y_0^2}$.

$d_z = \sqrt{(-4)^2 + 3^2} = \sqrt{16 + 9} = \sqrt{25} = 5$.

Conclusion:

The distance of the point P(-4, 3, 5) from the x-axis is $\sqrt{34}$.

The distance of the point P(-4, 3, 5) from the y-axis is $\sqrt{41}$.

The distance of the point P(-4, 3, 5) from the z-axis is 5.

Given:

The two points are A(4, 8, 10) and B(6, 10, -8).

The dividing plane is the XY-plane.

To Find:

The ratio in which the XY-plane divides the line segment joining the points A and B.

Solution:

Let the XY-plane divide the line segment joining the points A($x_1$, $y_1$, $z_1$) = (4, 8, 10) and B($x_2$, $y_2$, $z_2$) = (6, 10, -8) in the ratio $k:1$.

Let the point of division be P(x, y, z).

The coordinates of the point of division P are given by the section formula:

$x = \frac{k x_2 + 1 x_1}{k + 1}$

$y = \frac{k y_2 + 1 y_1}{k + 1}$

$z = \frac{k z_2 + 1 z_1}{k + 1}$

The equation of the XY-plane is $z = 0$. This means that any point lying on the XY-plane has its z-coordinate equal to 0.

Since the point P lies on the XY-plane, its z-coordinate must be 0.

Using the section formula for the z-coordinate and setting it to 0:

$0 = \frac{k (-8) + 1 (10)}{k + 1}$

$0 = \frac{-8k + 10}{k + 1}$

For this equation to hold true, the numerator must be zero (assuming $k+1 \neq 0$, which is true since A and B are distinct points):

$-8k + 10 = 0$

$10 = 8k$

To find the ratio $k:1$, we solve for $k$:

$k = \frac{10}{8}$

$k = \frac{5}{4}$

So, the ratio $k:1$ is $\frac{5}{4}:1$. Multiplying both parts by 4 to get integer ratio, we get $5:4$. Since the value of $k$ is positive, the division is internal.

Conclusion:

The XY-plane divides the line segment joining the points (4, 8, 10) and (6, 10, -8) in the ratio 5:4 internally.

Given:

The three points are A(1, -1, 3), B(2, -4, 6), and C(3, -7, 9).

To Show:

The points A, B, and C are collinear.

Solution:

To show that three points A, B, and C are collinear, we can calculate the distances between each pair of points and check if the sum of the distances of two segments equals the distance of the third segment. We use the distance formula between two points $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$ in three dimensions:

$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}$

Let's calculate the distances AB, BC, and AC.

Distance AB: Using points A(1, -1, 3) and B(2, -4, 6)

$AB = \sqrt{(2 - 1)^2 + (-4 - (-1))^2 + (6 - 3)^2}$

$AB = \sqrt{(1)^2 + (-4 + 1)^2 + (3)^2}$

$AB = \sqrt{(1)^2 + (-3)^2 + (3)^2}$

$AB = \sqrt{1 + 9 + 9}$

$AB = \sqrt{19}$

Distance BC: Using points B(2, -4, 6) and C(3, -7, 9)

$BC = \sqrt{(3 - 2)^2 + (-7 - (-4))^2 + (9 - 6)^2}$

$BC = \sqrt{(1)^2 + (-7 + 4)^2 + (3)^2}$

$BC = \sqrt{(1)^2 + (-3)^2 + (3)^2}$

$BC = \sqrt{1 + 9 + 9}$

$BC = \sqrt{19}$

Distance AC: Using points A(1, -1, 3) and C(3, -7, 9)

$AC = \sqrt{(3 - 1)^2 + (-7 - (-1))^2 + (9 - 3)^2}$

$AC = \sqrt{(2)^2 + (-7 + 1)^2 + (6)^2}$

$AC = \sqrt{(2)^2 + (-6)^2 + (6)^2}$

$AC = \sqrt{4 + 36 + 36}$

$AC = \sqrt{76}$

$AC = \sqrt{4 \times 19}$

$AC = 2\sqrt{19}$

Now, let's check if the sum of any two distances equals the third distance.

We see that $AB = \sqrt{19}$ and $BC = \sqrt{19}$.

Their sum is $AB + BC = \sqrt{19} + \sqrt{19} = 2\sqrt{19}$.

The distance AC is $AC = 2\sqrt{19}$.

Since $AB + BC = AC$, the points A, B, and C are collinear, and point B lies between A and C.

Conclusion:

As the sum of the distances AB and BC is equal to the distance AC ($AB + BC = AC$), the points A(1, -1, 3), B(2, -4, 6), and C(3, -7, 9) are collinear.

Given:

The two points are A(-3, 1, 2) and B(5, 4, -6).

The ratio of external division is $m:n = 2:1$.

To Find:

The coordinates of the point that divides the line segment AB externally in the ratio 2:1.

Solution:

Let the coordinates of the point be P(x, y, z). The formula for the coordinates of a point that divides the line segment joining $A(x_1, y_1, z_1)$ and $B(x_2, y_2, z_2)$ externally in the ratio $m:n$ is:

$x = \frac{m x_2 - n x_1}{m - n}$

$y = \frac{m y_2 - n y_1}{m - n}$

$z = \frac{m z_2 - n z_1}{m - n}$

Here, $(x_1, y_1, z_1) = (-3, 1, 2)$, $(x_2, y_2, z_2) = (5, 4, -6)$, $m = 2$, and $n = 1$.

Substituting these values into the formula, we get:

$x = \frac{(2)(5) - (1)(-3)}{2 - 1}$

$x = \frac{10 - (-3)}{1}$

$x = \frac{10 + 3}{1}$

$x = 13$

$y = \frac{(2)(4) - (1)(1)}{2 - 1}$

$y = \frac{8 - 1}{1}$

$y = \frac{7}{1}$

$y = 7$

$z = \frac{(2)(-6) - (1)(2)}{2 - 1}$

$z = \frac{-12 - 2}{1}$

$z = \frac{-14}{1}$

$z = -14$

Therefore, the coordinates of the point are (13, 7, -14).

Conclusion:

The coordinates of the point which divides the line segment joining the points (-3, 1, 2) and (5, 4, -6) externally in the ratio 2:1 are (13, 7, -14).

Given:

The coordinates of three vertices of a parallelogram ABCD are A(1, 2, 3), B(4, 5, 6), and C(7, 8, 9).

To Find:

The coordinates of the fourth vertex D(x, y, z).

Solution:

In a parallelogram, the diagonals bisect each other. This means that the midpoint of the diagonal AC is the same as the midpoint of the diagonal BD.

Let the coordinates of the fourth vertex be D(x, y, z).

The midpoint of a line segment joining two points $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$ is given by the formula:

Midpoint $= \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}, \frac{z_1 + z_2}{2}\right)$

First, let's find the midpoint of the diagonal AC. Using points A(1, 2, 3) and C(7, 8, 9):

Midpoint of AC $= \left(\frac{1 + 7}{2}, \frac{2 + 8}{2}, \frac{3 + 9}{2}\right)$

Midpoint of AC $= \left(\frac{8}{2}, \frac{10}{2}, \frac{12}{2}\right)$

Midpoint of AC $= (4, 5, 6)$

Next, let's find the midpoint of the diagonal BD. Using points B(4, 5, 6) and D(x, y, z):

Midpoint of BD $= \left(\frac{4 + x}{2}, \frac{5 + y}{2}, \frac{6 + z}{2}\right)$

Since the midpoints of AC and BD are the same, we can equate the coordinates:

$\left(\frac{4 + x}{2}, \frac{5 + y}{2}, \frac{6 + z}{2}\right) = (4, 5, 6)$

Equating the corresponding coordinates:

$\frac{4 + x}{2} = 4$

$4 + x = 4 \times 2$

$4 + x = 8$

$x = 8 - 4$

$x = 4$

$\frac{5 + y}{2} = 5$

$5 + y = 5 \times 2$

$5 + y = 10$

$y = 10 - 5$

$y = 5$

$\frac{6 + z}{2} = 6$

$6 + z = 6 \times 2$

$6 + z = 12$

$z = 12 - 6$

$z = 6$

Therefore, the coordinates of the fourth vertex D are (4, 5, 6).

Conclusion:

The coordinates of the fourth vertex D of the parallelogram ABCD are (4, 5, 6).

Given:

The point is P(1, 2, -4).

We need to find the distance of this point from the y-axis.

To Find:

The distance of the point (1, 2, -4) from the y-axis.

Solution:

The distance of a point $(x_0, y_0, z_0)$ from the y-axis is given by the formula:

Distance $= \sqrt{x_0^2 + z_0^2}$

Here, the given point is (1, 2, -4).

So, $x_0 = 1$, $y_0 = 2$, and $z_0 = -4$.

Substituting the values of $x_0$ and $z_0$ into the formula:

Distance from y-axis $= \sqrt{(1)^2 + (-4)^2}$

Distance from y-axis $= \sqrt{1 + 16}$

Distance from y-axis $= \sqrt{17}$

Conclusion:

The distance of the point (1, 2, -4) from the y-axis is $\sqrt{17}$.

Given:

The two points are A(2, 4, -3) and B(-3, 5, 6).

The dividing plane is the XZ-plane.

To Find:

The ratio in which the XZ-plane divides the line segment joining the points A and B.

Solution:

Let the XZ-plane divide the line segment joining the points A($x_1$, $y_1$, $z_1$) = (2, 4, -3) and B($x_2$, $y_2$, $z_2$) = (-3, 5, 6) in the ratio $m:n$.

Let the point of division be P(x, y, z).

The coordinates of the point of division P are given by the section formula:

$x = \frac{m x_2 + n x_1}{m + n}$

$y = \frac{m y_2 + n y_1}{m + n}$

$z = \frac{m z_2 + n z_1}{m + n}$

The equation of the XZ-plane is $y = 0$. This means that any point lying on the XZ-plane has its y-coordinate equal to 0.

Since the point P lies on the XZ-plane, its y-coordinate must be 0.

Using the section formula for the y-coordinate and setting it to 0:

$0 = \frac{m (5) + n (4)}{m + n}$

$0 = \frac{5m + 4n}{m + n}$

For this equation to hold true, the numerator must be zero (assuming $m+n \neq 0$, which is true since A and B are distinct points):

$5m + 4n = 0$

$5m = -4n$

To find the ratio $m:n$, we can write:

$\frac{m}{n} = -\frac{4}{5}$

So, the ratio $m:n$ is $-4:5$ or $4:-5$. The negative sign indicates that the division is external.

The ratio is $4:5$ externally.

Conclusion:

The XZ-plane divides the line segment joining the points (2, 4, -3) and (-3, 5, 6) in the ratio 4:5 externally.

Given:

The point is P(x, y, z).

The origin is O(0, 0, 0).

The distance of point P from the origin is 10 units.

To Write:

The equation representing the locus of point P.

Solution:

The locus of a point is the set of all points that satisfy a given condition.

The given condition is that the distance between the point P(x, y, z) and the origin O(0, 0, 0) is 10 units.

Using the distance formula between two points $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$, which is $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}$.

Let $(x_1, y_1, z_1) = (0, 0, 0)$ (Origin) and $(x_2, y_2, z_2) = (x, y, z)$ (Point P).

The distance between P and the origin, OP, is given as 10 units.

$OP = \sqrt{(x - 0)^2 + (y - 0)^2 + (z - 0)^2}$

$OP = \sqrt{x^2 + y^2 + z^2}$

According to the problem, $OP = 10$.

So, $\sqrt{x^2 + y^2 + z^2} = 10$

To eliminate the square root, we square both sides of the equation:

$(\sqrt{x^2 + y^2 + z^2})^2 = 10^2$

$x^2 + y^2 + z^2 = 100$

This equation represents the locus of all points P(x, y, z) that are at a distance of 10 units from the origin. This is the equation of a sphere centered at the origin with radius 10.

Conclusion:

The equation representing the locus of point P is $x^2 + y^2 + z^2 = 100$.

Given:

The two points are A(5, -3, 2) and B(3, 1, -4).

To Find:

The coordinates of the midpoint of the line segment AB.

Solution:

Let the coordinates of the midpoint be M(x, y, z). The formula for the coordinates of the midpoint of a line segment joining two points $A(x_1, y_1, z_1)$ and $B(x_2, y_2, z_2)$ is:

$x = \frac{x_1 + x_2}{2}$

$y = \frac{y_1 + y_2}{2}$

$z = \frac{z_1 + z_2}{2}$

Here, $(x_1, y_1, z_1) = (5, -3, 2)$ and $(x_2, y_2, z_2) = (3, 1, -4)$.

Substituting these values into the formula, we get:

$x = \frac{5 + 3}{2}$

$x = \frac{8}{2}$

$x = 4$

$y = \frac{-3 + 1}{2}$

$y = \frac{-2}{2}$

$y = -1$

$z = \frac{2 + (-4)}{2}$

$z = \frac{2 - 4}{2}$

$z = \frac{-2}{2}$

$z = -1$

Therefore, the coordinates of the midpoint are (4, -1, -1).

Conclusion:

The coordinates of the midpoint of the line segment joining the points (5, -3, 2) and (3, 1, -4) are (4, -1, -1).

Given:

The point is (4, -2, -5).

To Determine:

The octant in which the given point lies.

Solution:

The three-dimensional space is divided into eight octants by the three coordinate planes (XY-plane, YZ-plane, and XZ-plane).

The octant is determined by the signs of the coordinates (x, y, z) of the point.

For the given point (4, -2, -5):

The x-coordinate is 4, which is positive (+).

The y-coordinate is -2, which is negative (-).

The z-coordinate is -5, which is negative (-).

So the signs of the coordinates are (+, -, -).

The octants are defined by the signs as follows:

Octant I: (+, +, +)

Octant II: (-, +, +)

Octant III: (-, -, +)

Octant IV: (+, -, +)

Octant V: (+, +, -)

Octant VI: (-, +, -)

Octant VII: (-, -, -)

Octant VIII: (+, -, -)

The point (4, -2, -5) has signs (+, -, -).

Comparing the signs (+, -, -) with the octant definitions, we see that the point lies in the Eighth Octant.

Conclusion:

The point (4, -2, -5) lies in the Eighth Octant.

Given:

The four points are A(1, 2, 3), B(-1, -2, -1), C(2, 3, 2), and D(4, 7, 6).

To Show:

1. The points A, B, C, and D are the vertices of a parallelogram ABCD.

2. Determine if the parallelogram is a rectangle.

Solution:

To show that the points A, B, C, and D form a parallelogram ABCD, we can show that the midpoints of the diagonals AC and BD coincide. If the midpoints are the same, the diagonals bisect each other, which is a property of a parallelogram.

The midpoint of a line segment joining points $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$ is given by the formula:

$M = \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}, \frac{z_1 + z_2}{2}\right)$

Let's find the midpoint of diagonal AC, using points A(1, 2, 3) and C(2, 3, 2):

Midpoint of AC $= \left(\frac{1 + 2}{2}, \frac{2 + 3}{2}, \frac{3 + 2}{2}\right)$

Midpoint of AC $= \left(\frac{3}{2}, \frac{5}{2}, \frac{5}{2}\right)$

Now let's find the midpoint of diagonal BD, using points B(-1, -2, -1) and D(4, 7, 6):

Midpoint of BD $= \left(\frac{-1 + 4}{2}, \frac{-2 + 7}{2}, \frac{-1 + 6}{2}\right)$

Midpoint of BD $= \left(\frac{3}{2}, \frac{5}{2}, \frac{5}{2}\right)$

Since the midpoint of AC is the same as the midpoint of BD $(\frac{3}{2}, \frac{5}{2}, \frac{5}{2})$, the diagonals AC and BD bisect each other. Therefore, the points A, B, C, and D are the vertices of a parallelogram ABCD.

Next, to determine if the parallelogram ABCD is a rectangle, we can check if the lengths of the diagonals are equal. In a rectangle, the diagonals are equal in length. We use the distance formula between two points $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$:

$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}$

Let's calculate the length of diagonal AC, using points A(1, 2, 3) and C(2, 3, 2):

$AC = \sqrt{(2 - 1)^2 + (3 - 2)^2 + (2 - 3)^2}$

$AC = \sqrt{(1)^2 + (1)^2 + (-1)^2}$

$AC = \sqrt{1 + 1 + 1}$

$AC = \sqrt{3}$

Now let's calculate the length of diagonal BD, using points B(-1, -2, -1) and D(4, 7, 6):

$BD = \sqrt{(4 - (-1))^2 + (7 - (-2))^2 + (6 - (-1))^2}$

$BD = \sqrt{(4 + 1)^2 + (7 + 2)^2 + (6 + 1)^2}$

$BD = \sqrt{(5)^2 + (9)^2 + (7)^2}$

$BD = \sqrt{25 + 81 + 49}$

$BD = \sqrt{155}$

Since the length of diagonal AC ($\sqrt{3}$) is not equal to the length of diagonal BD ($\sqrt{155}$), the parallelogram ABCD is not a rectangle.

Conclusion:

The points A(1, 2, 3), B(-1, -2, -1), C(2, 3, 2), and D(4, 7, 6) are the vertices of a parallelogram ABCD because the midpoints of its diagonals AC and BD coincide.

The parallelogram ABCD is not a rectangle because the lengths of its diagonals AC and BD are not equal ($AC = \sqrt{3}$ and $BD = \sqrt{155}$).

Given:

The vertices of the triangle are A(1, 2, 3), B(-2, -1, 3), and C(2, -3, -1).

To Find:

1. The length of the median from vertex A to the side BC.

2. The coordinates of the centroid of the triangle ABC.

Solution:

1. Length of the median from vertex A to the side BC:

A median of a triangle is a line segment joining a vertex to the midpoint of the opposite side.

Let M be the midpoint of the side BC. The coordinates of the midpoint M of the line segment joining $B(x_1, y_1, z_1)$ and $C(x_2, y_2, z_2)$ are given by:

$M = \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}, \frac{z_1 + z_2}{2}\right)$

Using B(-2, -1, 3) and C(2, -3, -1):

Coordinates of M $= \left(\frac{-2 + 2}{2}, \frac{-1 + (-3)}{2}, \frac{3 + (-1)}{2}\right)$

Coordinates of M $= \left(\frac{0}{2}, \frac{-4}{2}, \frac{2}{2}\right)$

Coordinates of M $= (0, -2, 1)$

The median from vertex A to side BC is the line segment AM.

The length of the median AM is the distance between point A(1, 2, 3) and point M(0, -2, 1).

The distance between two points $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$ is given by the formula:

$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}$

Length of AM $= \sqrt{(0 - 1)^2 + (-2 - 2)^2 + (1 - 3)^2}$

Length of AM $= \sqrt{(-1)^2 + (-4)^2 + (-2)^2}$

Length of AM $= \sqrt{1 + 16 + 4}$

Length of AM $= \sqrt{21}$

2. Coordinates of the centroid of the triangle ABC:

The centroid of a triangle with vertices $A(x_1, y_1, z_1)$, $B(x_2, y_2, z_2)$, and $C(x_3, y_3, z_3)$ is given by the formula:

$G = \left(\frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3}, \frac{z_1 + z_2 + z_3}{3}\right)$

Using A(1, 2, 3), B(-2, -1, 3), and C(2, -3, -1):

Coordinates of Centroid G $= \left(\frac{1 + (-2) + 2}{3}, \frac{2 + (-1) + (-3)}{3}, \frac{3 + 3 + (-1)}{3}\right)$

Coordinates of Centroid G $= \left(\frac{1 - 2 + 2}{3}, \frac{2 - 1 - 3}{3}, \frac{3 + 3 - 1}{3}\right)$

Coordinates of Centroid G $= \left(\frac{1}{3}, \frac{-2}{3}, \frac{5}{3}\right)$

Conclusion:

The length of the median from vertex A to the side BC is $\sqrt{21}$ units.

The coordinates of the centroid of the triangle ABC are $\left(\frac{1}{3}, -\frac{2}{3}, \frac{5}{3}\right)$.

Given:

The equation of the sphere is $x^2 + y^2 + z^2 = 50$.

The two points are A(1, -2, 3) and B(3, 4, -5).

To Find:

The ratio in which the sphere divides the line segment joining points A and B.

Solution:

Let the sphere divide the line segment joining points A($x_1$, $y_1$, $z_1$) = (1, -2, 3) and B($x_2$, $y_2$, $z_2$) = (3, 4, -5) in the ratio $k:1$.

Let the point of division be P(x, y, z).

Using the section formula, the coordinates of the point of division P are given by:

$x = \frac{k x_2 + 1 x_1}{k + 1} = \frac{k(3) + 1(1)}{k + 1} = \frac{3k + 1}{k + 1}$

$y = \frac{k y_2 + 1 y_1}{k + 1} = \frac{k(4) + 1(-2)}{k + 1} = \frac{4k - 2}{k + 1}$

$z = \frac{k z_2 + 1 z_1}{k + 1} = \frac{k(-5) + 1(3)}{k + 1} = \frac{-5k + 3}{k + 1}$

Since the point P(x, y, z) lies on the sphere $x^2 + y^2 + z^2 = 50$, its coordinates must satisfy the equation of the sphere.

Substitute the coordinates of P into the sphere's equation:

$\left(\frac{3k + 1}{k + 1}\right)^2 + \left(\frac{4k - 2}{k + 1}\right)^2 + \left(\frac{-5k + 3}{k + 1}\right)^2 = 50$

$\frac{(3k + 1)^2}{(k + 1)^2} + \frac{(4k - 2)^2}{(k + 1)^2} + \frac{(-5k + 3)^2}{(k + 1)^2} = 50$

$\frac{(9k^2 + 6k + 1) + (16k^2 - 16k + 4) + (25k^2 - 30k + 9)}{(k + 1)^2} = 50$

Combine the terms in the numerator:

$\frac{(9k^2 + 16k^2 + 25k^2) + (6k - 16k - 30k) + (1 + 4 + 9)}{(k + 1)^2} = 50$

$\frac{50k^2 - 40k + 14}{(k + 1)^2} = 50$

Assuming $k \neq -1$, multiply both sides by $(k+1)^2$:

$50k^2 - 40k + 14 = 50(k + 1)^2$

$50k^2 - 40k + 14 = 50(k^2 + 2k + 1)$

$50k^2 - 40k + 14 = 50k^2 + 100k + 50$

Subtract $50k^2$ from both sides:

$-40k + 14 = 100k + 50$

Rearrange the terms to solve for k:

$14 - 50 = 100k + 40k$

$-36 = 140k$

$k = \frac{-36}{140}$

Simplify the fraction:

$k = \frac{-9}{35}$

The ratio is $k:1 = -\frac{9}{35}:1$, which simplifies to $-9:35$. The negative sign indicates that the division is external.

The ratio in which the sphere divides the line segment AB is $9:35$ externally.

Conclusion:

The sphere $x^2 + y^2 + z^2 = 50$ divides the line segment joining the points A(1, -2, 3) and B(3, 4, -5) in the ratio 9:35 externally.

Given:

Two points A(3, 4, -5) and B(-2, 1, 4).

The condition that the distance of a point P from A is equal to its distance from B.

To Find:

The equation of the set of all points P(x, y, z) that satisfy the given condition.

Solution:

Let P(x, y, z) be any point in the set. The distance between two points $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$ is given by the distance formula:

$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}$

The condition given is that the distance PA is equal to the distance PB, i.e., $PA = PB$.

Squaring both sides, we get $PA^2 = PB^2$.

The square of the distance between P(x, y, z) and A(3, 4, -5) is:

$PA^2 = (x - 3)^2 + (y - 4)^2 + (z - (-5))^2$

$PA^2 = (x - 3)^2 + (y - 4)^2 + (z + 5)^2$

Expanding this, we get:

$PA^2 = (x^2 - 6x + 9) + (y^2 - 8y + 16) + (z^2 + 10z + 25)$

$PA^2 = x^2 + y^2 + z^2 - 6x - 8y + 10z + 50$

The square of the distance between P(x, y, z) and B(-2, 1, 4) is:

$PB^2 = (x - (-2))^2 + (y - 1)^2 + (z - 4)^2$

$PB^2 = (x + 2)^2 + (y - 1)^2 + (z - 4)^2$

Expanding this, we get:

$PB^2 = (x^2 + 4x + 4) + (y^2 - 2y + 1) + (z^2 - 8z + 16)$

$PB^2 = x^2 + y^2 + z^2 + 4x - 2y - 8z + 21$

Now, set $PA^2 = PB^2$:

$x^2 + y^2 + z^2 - 6x - 8y + 10z + 50 = x^2 + y^2 + z^2 + 4x - 2y - 8z + 21$

Subtract $x^2 + y^2 + z^2$ from both sides:

$-6x - 8y + 10z + 50 = 4x - 2y - 8z + 21$

Move all terms to one side to form the equation of the locus:

$-6x - 4x - 8y + 2y + 10z + 8z + 50 - 21 = 0$

Combine the like terms:

$-10x - 6y + 18z + 29 = 0$

Multiplying by -1 (optional):

$10x + 6y - 18z - 29 = 0$

This equation represents the set of all points P(x, y, z) that are equidistant from points A and B. This locus is a plane that is the perpendicular bisector of the line segment AB.

Conclusion:

The equation of the set of points P such that the distance of P from A(3, 4, -5) is equal to its distance from B(-2, 1, 4) is $10x + 6y - 18z - 29 = 0$.

Given:

The three points are A(1, -2, 3), B(4, -3, 6), and C(3, 0, -2).

To Show:

That the points A, B, and C form a right-angled triangle at vertex A.

Solution:

To show that a triangle is right-angled at a specific vertex, we can use the converse of the Pythagorean theorem. For a right angle at vertex A, the square of the length of the side opposite to A (which is BC) must be equal to the sum of the squares of the lengths of the other two sides (AB and AC). That is, $AB^2 + AC^2 = BC^2$.

We will calculate the squares of the lengths of the sides AB, AC, and BC using the distance formula in three dimensions. The square of the distance $d^2$ between two points $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$ is given by:

$d^2 = (x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2$

Let's calculate the square of the length of side AB, using points A(1, -2, 3) and B(4, -3, 6):

$AB^2 = (4 - 1)^2 + (-3 - (-2))^2 + (6 - 3)^2$

$AB^2 = (3)^2 + (-3 + 2)^2 + (3)^2$

$AB^2 = (3)^2 + (-1)^2 + (3)^2$

$AB^2 = 9 + 1 + 9$

$AB^2 = 19$

Now, let's calculate the square of the length of side AC, using points A(1, -2, 3) and C(3, 0, -2):

$AC^2 = (3 - 1)^2 + (0 - (-2))^2 + (-2 - 3)^2$

$AC^2 = (2)^2 + (0 + 2)^2 + (-5)^2$

$AC^2 = (2)^2 + (2)^2 + (-5)^2$

$AC^2 = 4 + 4 + 25$

$AC^2 = 33$

Finally, let's calculate the square of the length of side BC, using points B(4, -3, 6) and C(3, 0, -2):

$BC^2 = (3 - 4)^2 + (0 - (-3))^2 + (-2 - 6)^2$

$BC^2 = (-1)^2 + (0 + 3)^2 + (-8)^2$

$BC^2 = (-1)^2 + (3)^2 + (-8)^2$

$BC^2 = 1 + 9 + 64$

$BC^2 = 74$

For the triangle ABC to be right-angled at A, the Pythagorean theorem $AB^2 + AC^2 = BC^2$ must hold true.

Let's calculate the sum of the squares of sides AB and AC:

$AB^2 + AC^2 = 19 + 33 = 52$

Now, we compare this sum with the square of the length of side BC:

$BC^2 = 74$

We see that $AB^2 + AC^2 = 52$, which is not equal to $BC^2 = 74$.

$AB^2 + AC^2 \neq BC^2$

Since the sum of the squares of the lengths of sides AB and AC is not equal to the square of the length of side BC, the triangle formed by the given points does not have a right angle at A.

Conclusion:

Based on the calculations ($AB^2 = 19$, $AC^2 = 33$, $BC^2 = 74$), $AB^2 + AC^2 \neq BC^2$. Therefore, the points A(1, -2, 3), B(4, -3, 6), and C(3, 0, -2) do not form a right-angled triangle at A.

Given:

The equation of the plane is $2x - 3y + 4z + 5 = 0$.

The two points are A(3, 4, -5) and B(-2, 1, 4).

To Find:

The ratio in which the plane divides the line segment joining points A and B.

Solution:

Let the plane $2x - 3y + 4z + 5 = 0$ divide the line segment joining the points A($x_1$, $y_1$, $z_1$) = (3, 4, -5) and B($x_2$, $y_2$, $z_2$) = (-2, 1, 4) in the ratio $k:1$.

Let the point of division be P(x, y, z).

Using the section formula, the coordinates of the point of division P are given by:

$x = \frac{k x_2 + 1 x_1}{k + 1} = \frac{k(-2) + 1(3)}{k + 1} = \frac{-2k + 3}{k + 1}$

$y = \frac{k y_2 + 1 y_1}{k + 1} = \frac{k(1) + 1(4)}{k + 1} = \frac{k + 4}{k + 1}$

$z = \frac{k z_2 + 1 z_1}{k + 1} = \frac{k(4) + 1(-5)}{k + 1} = \frac{4k - 5}{k + 1}$

Since the point P(x, y, z) lies on the plane $2x - 3y + 4z + 5 = 0$, its coordinates must satisfy the equation of the plane.

Substitute the coordinates of P into the plane's equation:

$2\left(\frac{-2k + 3}{k + 1}\right) - 3\left(\frac{k + 4}{k + 1}\right) + 4\left(\frac{4k - 5}{k + 1}\right) + 5 = 0$

Multiply the entire equation by $(k+1)$ to eliminate the denominators (assuming $k+1 \neq 0$):

$2(-2k + 3) - 3(k + 4) + 4(4k - 5) + 5(k + 1) = 0$

Expand and simplify the equation:

$-4k + 6 - 3k - 12 + 16k - 20 + 5k + 5 = 0$

Combine the terms with $k$ and the constant terms:

$(-4k - 3k + 16k + 5k) + (6 - 12 - 20 + 5) = 0$

$14k - 21 = 0$

Solve for $k$:

$14k = 21$

$k = \frac{21}{14}$

$k = \frac{3}{2}$

The ratio is $k:1 = \frac{3}{2}:1$. Multiplying both parts by 2 gives the integer ratio $3:2$.

Since the value of $k$ is positive, the division is internal.

Conclusion:

The plane $2x - 3y + 4z + 5 = 0$ divides the line segment joining the points A(3, 4, -5) and B(-2, 1, 4) in the ratio 3:2 internally.

Given:

The vertices of the triangle are A(-2, 3, 5), B(1, 2, 3), and C(7, 0, -1).

To Find:

1. The lengths of the sides AB, BC, and AC.

2. The type of triangle formed by these vertices.

Solution:

We will find the lengths of the sides of the triangle using the distance formula in three dimensions. The distance $d$ between two points $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$ is given by:

$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}$

Let's calculate the length of side AB, using points A(-2, 3, 5) and B(1, 2, 3):

$AB = \sqrt{(1 - (-2))^2 + (2 - 3)^2 + (3 - 5)^2}$

$AB = \sqrt{(1 + 2)^2 + (-1)^2 + (-2)^2}$

$AB = \sqrt{(3)^2 + 1 + 4}$

$AB = \sqrt{9 + 1 + 4}$

$AB = \sqrt{14}$

Now, let's calculate the length of side BC, using points B(1, 2, 3) and C(7, 0, -1):

$BC = \sqrt{(7 - 1)^2 + (0 - 2)^2 + (-1 - 3)^2}$

$BC = \sqrt{(6)^2 + (-2)^2 + (-4)^2}$

$BC = \sqrt{36 + 4 + 16}$

$BC = \sqrt{56}$

$BC = \sqrt{4 \times 14}$

$BC = 2\sqrt{14}$

Finally, let's calculate the length of side AC, using points A(-2, 3, 5) and C(7, 0, -1):

$AC = \sqrt{(7 - (-2))^2 + (0 - 3)^2 + (-1 - 5)^2}$

$AC = \sqrt{(7 + 2)^2 + (-3)^2 + (-6)^2}$

$AC = \sqrt{(9)^2 + 9 + 36}$

$AC = \sqrt{81 + 9 + 36}$

$AC = \sqrt{126}$

$AC = \sqrt{9 \times 14}$

$AC = 3\sqrt{14}$

The lengths of the sides are $AB = \sqrt{14}$, $BC = 2\sqrt{14}$, and $AC = 3\sqrt{14}$.

Now, let's determine the type of triangle based on these lengths.

Since all three side lengths are different ($\sqrt{14} \neq 2\sqrt{14} \neq 3\sqrt{14}$), the triangle is a scalene triangle.

Let's check if it is a right-angled triangle by verifying the Pythagorean theorem, $a^2 + b^2 = c^2$. The squares of the side lengths are:

$AB^2 = 14$

$BC^2 = 56$

$AC^2 = 126$

The longest side is AC. Let's check if $AB^2 + BC^2 = AC^2$:

$AB^2 + BC^2 = 14 + 56 = 70$

$AC^2 = 126$

Since $AB^2 + BC^2 \neq AC^2$, the triangle is not right-angled at B.

Let's check other combinations:

$AB^2 + AC^2 = 14 + 126 = 140 \neq BC^2 (56)$

$BC^2 + AC^2 = 56 + 126 = 182 \neq AB^2 (14)$

The triangle is not a right-angled triangle.

Conclusion:

The lengths of the sides of the triangle are $AB = \sqrt{14}$, $BC = 2\sqrt{14}$, and $AC = 3\sqrt{14}$.

Since all sides have different lengths and the Pythagorean theorem does not hold, the triangle ABC is a scalene triangle.

Given:

The two points are P(4, 2, -6) and Q(10, -16, 6).

To Find:

The coordinates of the two points that trisect the line segment PQ.

Solution:

To trisect a line segment PQ, we need to find two points, let's call them R and S, such that R divides PQ in the ratio 1:2 and S divides PQ in the ratio 2:1.

We will use the section formula for internal division. The coordinates of a point that divides the line segment joining $P(x_1, y_1, z_1)$ and $Q(x_2, y_2, z_2)$ internally in the ratio $m:n$ are:

$x = \frac{m x_2 + n x_1}{m + n}$

$y = \frac{m y_2 + n y_1}{m + n}$

$z = \frac{m z_2 + n z_1}{m + n}$

Here, $P(x_1, y_1, z_1) = (4, 2, -6)$ and $Q(x_2, y_2, z_2) = (10, -16, 6)$.

Let's find the coordinates of the first point R, which divides PQ in the ratio 1:2. Here, $m=1$ and $n=2$.

$x_R = \frac{(1)(10) + (2)(4)}{1 + 2} = \frac{10 + 8}{3} = \frac{18}{3} = 6$

$y_R = \frac{(1)(-16) + (2)(2)}{1 + 2} = \frac{-16 + 4}{3} = \frac{-12}{3} = -4$

$z_R = \frac{(1)(6) + (2)(-6)}{1 + 2} = \frac{6 - 12}{3} = \frac{-6}{3} = -2$

The coordinates of the first trisecting point R are (6, -4, -2).

Now, let's find the coordinates of the second point S, which divides PQ in the ratio 2:1. Here, $m=2$ and $n=1$.

$x_S = \frac{(2)(10) + (1)(4)}{2 + 1} = \frac{20 + 4}{3} = \frac{24}{3} = 8$

$y_S = \frac{(2)(-16) + (1)(2)}{2 + 1} = \frac{-32 + 2}{3} = \frac{-30}{3} = -10$

$z_S = \frac{(2)(6) + (1)(-6)}{2 + 1} = \frac{12 - 6}{3} = \frac{6}{3} = 2$

The coordinates of the second trisecting point S are (8, -10, 2).

Conclusion:

The coordinates of the points which trisect the line segment joining P(4, 2, -6) and Q(10, -16, 6) are (6, -4, -2) and (8, -10, 2).

Given:

The four points are A(2, 3, -4), B(-1, 2, -3), C(-4, 1, -2), and D(5, 4, -5).

To Show:

That the points A, B, C, and D are coplanar.

Solution:

To show that four points are coplanar, we can check if three vectors formed by these points are coplanar. If three vectors $\vec{u}$, $\vec{v}$, and $\vec{w}$ starting from a common point are coplanar, then their scalar triple product is zero, i.e., $\vec{u} \cdot (\vec{v} \times \vec{w}) = 0$.

Alternatively, if we can show that three of the points are collinear, and the fourth point lies on the line formed by these three, then all four points are collinear, and thus coplanar.

Let's form vectors $\vec{AB}$, $\vec{BC}$, and $\vec{CD}$.

Vector $\vec{AB}$ from A(2, 3, -4) to B(-1, 2, -3):

$\vec{AB} = (-1 - 2, 2 - 3, -3 - (-4))$

$\vec{AB} = (-3, -1, -3 + 4)$

$\vec{AB} = (-3, -1, 1)$

Vector $\vec{BC}$ from B(-1, 2, -3) to C(-4, 1, -2):

$\vec{BC} = (-4 - (-1), 1 - 2, -2 - (-3))$

$\vec{BC} = (-4 + 1, -1, -2 + 3)$

$\vec{BC} = (-3, -1, 1)$

We observe that $\vec{AB} = \vec{BC}$. Since the vectors $\vec{AB}$ and $\vec{BC}$ are equal and they share a common point B, the points A, B, and C are collinear.

Now, let's check if the fourth point D(5, 4, -5) also lies on the line passing through A, B, and C. We can check if the vector $\vec{CD}$ is parallel to $\vec{AB}$ (or $\vec{BC}$).

Vector $\vec{CD}$ from C(-4, 1, -2) to D(5, 4, -5):

$\vec{CD} = (5 - (-4), 4 - 1, -5 - (-2))$

$\vec{CD} = (5 + 4, 3, -5 + 2)$

$\vec{CD} = (9, 3, -3)$

Compare $\vec{CD}$ with $\vec{AB} = (-3, -1, 1)$.

We can see that $\vec{CD} = -3 \times (-3, -1, 1) = (9, 3, -3)$.

So, $\vec{CD} = -3 \vec{AB}$. Since $\vec{CD}$ is a scalar multiple of $\vec{AB}$ and they share a common line (implicitly through B and C), the point D is also collinear with A, B, and C.

Since all four points A, B, C, and D are collinear, they all lie on the same straight line. A line always lies within a plane.

Therefore, the four points are coplanar.

Conclusion:

As shown by calculating the vectors $\vec{AB}$, $\vec{BC}$, and $\vec{CD}$, the points A, B, and C are collinear ($\vec{AB} = \vec{BC}$) and the point D is also collinear with them ($\vec{CD}$ is a scalar multiple of $\vec{AB}$). Since the four points are collinear, they are coplanar.

Given:

The vertices of the triangle are A(1, 2, 3), B(2, 3, 1), and C(3, 1, 2).

To Find:

The length of the perimeter of the triangle ABC.

Solution:

The perimeter of a triangle is the sum of the lengths of its three sides.

We will calculate the lengths of the sides AB, BC, and AC using the distance formula in three dimensions. The distance $d$ between two points $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$ is given by:

$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}$

Let's calculate the length of side AB, using points A(1, 2, 3) and B(2, 3, 1):

$AB = \sqrt{(2 - 1)^2 + (3 - 2)^2 + (1 - 3)^2}$

$AB = \sqrt{(1)^2 + (1)^2 + (-2)^2}$

$AB = \sqrt{1 + 1 + 4}$

$AB = \sqrt{6}$

Now, let's calculate the length of side BC, using points B(2, 3, 1) and C(3, 1, 2):

$BC = \sqrt{(3 - 2)^2 + (1 - 3)^2 + (2 - 1)^2}$

$BC = \sqrt{(1)^2 + (-2)^2 + (1)^2}$

$BC = \sqrt{1 + 4 + 1}$

$BC = \sqrt{6}$

Finally, let's calculate the length of side AC, using points A(1, 2, 3) and C(3, 1, 2):

$AC = \sqrt{(3 - 1)^2 + (1 - 2)^2 + (2 - 3)^2}$

$AC = \sqrt{(2)^2 + (-1)^2 + (-1)^2}$

$AC = \sqrt{4 + 1 + 1}$

$AC = \sqrt{6}$

The lengths of the sides are $AB = \sqrt{6}$, $BC = \sqrt{6}$, and $AC = \sqrt{6}$.

The perimeter of triangle ABC is the sum of the lengths of its sides:

Perimeter = AB + BC + AC

Perimeter = $\sqrt{6} + \sqrt{6} + \sqrt{6}$

Perimeter = $3\sqrt{6}$

Conclusion:

The length of the perimeter of the triangle whose vertices are A(1, 2, 3), B(2, 3, 1), and C(3, 1, 2) is $3\sqrt{6}$ units.

Given:

The four points are O(0, 0, 0), A(2, 0, 0), B(0, 2, 0), and C(0, 0, 2).

To Find:

The coordinates of the point that is equidistant from the four given points.

Solution:

Let the coordinates of the required point be P(x, y, z).

The distance of P from the point O(0, 0, 0) is given by the distance formula:

$PO = \sqrt{(x - 0)^2 + (y - 0)^2 + (z - 0)^2} = \sqrt{x^2 + y^2 + z^2}$

The square of the distance is $PO^2 = x^2 + y^2 + z^2$.

The distance of P from the point A(2, 0, 0) is:

$PA = \sqrt{(x - 2)^2 + (y - 0)^2 + (z - 0)^2} = \sqrt{(x - 2)^2 + y^2 + z^2}$

The square of the distance is $PA^2 = (x - 2)^2 + y^2 + z^2 = x^2 - 4x + 4 + y^2 + z^2$.

The distance of P from the point B(0, 2, 0) is:

$PB = \sqrt{(x - 0)^2 + (y - 2)^2 + (z - 0)^2} = \sqrt{x^2 + (y - 2)^2 + z^2}$

The square of the distance is $PB^2 = x^2 + (y - 2)^2 + z^2 = x^2 + y^2 - 4y + 4 + z^2$.

The distance of P from the point C(0, 0, 2) is:

$PC = \sqrt{(x - 0)^2 + (y - 0)^2 + (z - 2)^2} = \sqrt{x^2 + y^2 + (z - 2)^2}$

The square of the distance is $PC^2 = x^2 + y^2 + (z - 2)^2 = x^2 + y^2 + z^2 - 4z + 4$.

Since the point P is equidistant from O, A, B, and C, we have $PO^2 = PA^2 = PB^2 = PC^2$.

Equating $PO^2$ and $PA^2$:

$x^2 + y^2 + z^2 = x^2 - 4x + 4 + y^2 + z^2$

$0 = -4x + 4$

$4x = 4$

$x = 1$

Equating $PO^2$ and $PB^2$:

$x^2 + y^2 + z^2 = x^2 + y^2 - 4y + 4 + z^2$

$0 = -4y + 4$

$4y = 4$

$y = 1$

Equating $PO^2$ and $PC^2$:

$x^2 + y^2 + z^2 = x^2 + y^2 + z^2 - 4z + 4$

$0 = -4z + 4$

$4z = 4$

$z = 1$

Thus, the coordinates of the point P are (1, 1, 1).

We can verify that the distance from P(1, 1, 1) to each of the points is $\sqrt{1^2 + 1^2 + 1^2} = \sqrt{3}$ units.

Conclusion:

The coordinates of the point equidistant from the four points (0, 0, 0), (2, 0, 0), (0, 2, 0), and (0, 0, 2) are (1, 1, 1).

Given:

The vertices of the triangle are A(5, 4, 6), B(1, -1, 3), and C(4, 3, 2).

To Find:

The coordinates of the point where the angle bisector of $\angle \text{BAC}$ meets the side BC.

Solution:

Let the angle bisector of $\angle \text{BAC}$ meet the side BC at point D. According to the Angle Bisector Theorem, the angle bisector of a vertex divides the opposite side in the ratio of the lengths of the other two sides.

Thus, point D divides the line segment BC in the ratio AB:AC internally, i.e., $\frac{BD}{DC} = \frac{AB}{AC}$.

First, we need to find the lengths of the sides AB and AC using the distance formula between two points $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$:

$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}$

Length of side AB (using A(5, 4, 6) and B(1, -1, 3)):

$AB = \sqrt{(1 - 5)^2 + (-1 - 4)^2 + (3 - 6)^2}$

$AB = \sqrt{(-4)^2 + (-5)^2 + (-3)^2}$

$AB = \sqrt{16 + 25 + 9}$

$AB = \sqrt{50} = \sqrt{25 \times 2} = 5\sqrt{2}$

Length of side AC (using A(5, 4, 6) and C(4, 3, 2)):

$AC = \sqrt{(4 - 5)^2 + (3 - 4)^2 + (2 - 6)^2}$

$AC = \sqrt{(-1)^2 + (-1)^2 + (-4)^2}$

$AC = \sqrt{1 + 1 + 16}$

$AC = \sqrt{18} = \sqrt{9 \times 2} = 3\sqrt{2}$

The ratio in which D divides BC is $BD:DC = AB:AC = 5\sqrt{2}:3\sqrt{2} = 5:3$.

So, point D divides BC internally in the ratio $m:n = 5:3$.

Now, we use the section formula to find the coordinates of point D. For a point dividing the line segment joining $B(x_1, y_1, z_1)$ and $C(x_2, y_2, z_2)$ internally in the ratio $m:n$, the coordinates $(x, y, z)$ are:

$x = \frac{m x_2 + n x_1}{m + n}$

$y = \frac{m y_2 + n y_1}{m + n}$

$z = \frac{m z_2 + n z_1}{m + n}$

Here, $B(x_1, y_1, z_1) = (1, -1, 3)$, $C(x_2, y_2, z_2) = (4, 3, 2)$, $m = 5$, and $n = 3$.

Coordinates of D(x, y, z):

$x = \frac{(5)(4) + (3)(1)}{5 + 3} = \frac{20 + 3}{8} = \frac{23}{8}$

$y = \frac{(5)(3) + (3)(-1)}{5 + 3} = \frac{15 - 3}{8} = \frac{12}{8} = \frac{3}{2}$

$z = \frac{(5)(2) + (3)(3)}{5 + 3} = \frac{10 + 9}{8} = \frac{19}{8}$

The coordinates of point D are $\left(\frac{23}{8}, \frac{3}{2}, \frac{19}{8}\right)$.

Conclusion:

The coordinates of the point where the angle bisector of $\angle \text{BAC}$ meets the side BC are $\left(\frac{23}{8}, \frac{3}{2}, \frac{19}{8}\right)$.