Chapter 12 Introduction to Three Dimensional Geometry (Concepts)

Cartesian Coordinate System in Three Dimensions

While the two-dimensional Cartesian plane is used to describe points on a flat surface using an ordered pair $(x, y)$, our world is three-dimensional. To locate points in space, we need to extend this system by adding a third dimension. The three-dimensional Cartesian coordinate system is the framework that allows us to uniquely define the position of any point in space using an ordered triplet of real numbers, $(x, y, z)$.

---

Coordinate Axes and Coordinate Planes

The foundation of the 3D coordinate system is a set of three mutually perpendicular lines, known as the coordinate axes, that intersect at a single point.

• The horizontal axis is usually called the x-axis.
• The second horizontal axis, perpendicular to the x-axis, is the y-axis.
• The vertical axis, perpendicular to both the x and y axes, is the z-axis.

The point where these three axes meet is called the origin, denoted by O. Its coordinates are $(0, 0, 0)$. The positive directions of the axes are typically arranged according to the right-hand rule: if you align the fingers of your right hand with the positive x-axis and curl them towards the positive y-axis, your thumb will point in the direction of the positive z-axis.

These three axes, taken in pairs, define three infinite planes called the coordinate planes. These planes serve as the primary references for locating points.

• The xy-plane is the flat surface containing the x-axis and the y-axis. For any point on this plane, its z-coordinate is always zero. The equation of the xy-plane is $z = 0$.
• The yz-plane is the flat surface containing the y-axis and the z-axis. For any point on this plane, its x-coordinate is always zero. The equation of the yz-plane is $x = 0$.
• The xz-plane is the flat surface containing the x-axis and the z-axis. For any point on this plane, its y-coordinate is always zero. The equation of the xz-plane is $y = 0$.

---

Coordinates of a Point in Space

The location of any point P in 3D space is given by an ordered triplet of numbers, $(x, y, z)$, called its coordinates. Each coordinate represents the signed perpendicular distance from the point to one of the coordinate planes.

• The x-coordinate is the perpendicular distance of P from the yz-plane.
• The y-coordinate is the perpendicular distance of P from the xz-plane.
• The z-coordinate is the perpendicular distance of P from the xy-plane.

For example, to locate the point P(2, 3, 4), you would start at the origin, move 2 units along the positive x-axis, then 3 units parallel to the positive y-axis, and finally 4 units parallel to the positive z-axis.

Answer:

---

Octants

Just as the two coordinate axes divide a 2D plane into four quadrants, the three coordinate planes divide 3D space into eight distinct regions. Each of these regions is called an octant.

The specific octant in which a point lies is determined by the signs (+ or –) of its x, y, and z coordinates. The first octant is the one where all three coordinates are positive. The naming convention for the other octants can vary, but the sign combinations are fixed.

Octant	x	y	z	Example Point
I	+	+	+	(2, 3, 5)
II	–	+	+	(-2, 3, 5)
III	–	–	+	(-2, -3, 5)
IV	+	–	+	(2, -3, 5)
V	+	+	–	(2, 3, -5)
VI	–	+	–	(-2, 3, -5)
VII	–	–	–	(-2, -3, -5)
VIII	+	–	–	(2, -3, -5)

Points that lie on a coordinate plane (where one coordinate is zero) or on a coordinate axis (where two coordinates are zero) are on the boundaries of the octants and do not belong to any single octant.

---

Answer:

Distance Formula in Three Dimensions

Just as we can calculate the distance between two points on a 2D plane, we can also find the distance between two points in 3D space. The formula is a straightforward and intuitive extension of the two-dimensional distance formula, incorporating the third coordinate, z. It is derived from the Pythagorean theorem.

---

Derivation of the 3D Distance Formula

Let's find the distance between two points in space, $P(x_1, y_1, z_1)$ and $Q(x_2, y_2, z_2)$.

We can visualize the distance as the length of the main diagonal of a rectangular box (a cuboid) where P and Q are opposite corners. The sides of this box are parallel to the coordinate axes.

Step 1: Find the length of the diagonal on the base of the box.

Consider the point R with coordinates $(x_2, y_2, z_1)$. The points P and R lie on the same horizontal plane ($z=z_1$). The distance between them is a 2D distance problem.

Using the 2D distance formula, the square of the distance PR is:

$(PR)^2 = (x_2 - x_1)^2 + (y_2 - y_1)^2$

Step 2: Use the vertical height to form a right triangle.

Now, consider the triangle $\triangle PRQ$. The segment PR is horizontal, and the segment RQ is vertical (connecting $(x_2, y_2, z_1)$ to $(x_2, y_2, z_2)$). Therefore, $\triangle PRQ$ is a right-angled triangle with the right angle at R.

The length of the vertical side RQ is the difference in the z-coordinates: $|z_2 - z_1|$.

The square of this length is $(RQ)^2 = (z_2 - z_1)^2$.

Step 3: Apply the Pythagorean theorem to the right triangle.

The distance we want, PQ, is the hypotenuse of the right triangle $\triangle PRQ$.

$(PQ)^2 = (PR)^2 + (RQ)^2$

Substitute the expressions we found for $(PR)^2$ and $(RQ)^2$.

$(PQ)^2 = [(x_2 - x_1)^2 + (y_2 - y_1)^2] + (z_2 - z_1)^2$

Taking the square root of both sides gives the distance formula in three dimensions:

Distance of a Point from the Origin

A special case of this formula is the distance of any point $P(x, y, z)$ from the origin $O(0, 0, 0)$.

Substituting $(x_1, y_1, z_1) = (0, 0, 0)$ and $(x_2, y_2, z_2) = (x, y, z)$ into the formula, we get:

---

Answer:

---

Answer:

Section Formula in Three Dimensions

The section formula, which we used in two-dimensional geometry to find a point that divides a line segment in a specific ratio, can be extended to work in three-dimensional space. The principle remains the same, but we now have to calculate a z-coordinate in addition to the x and y-coordinates.

This formula is fundamental for problems involving ratios, midpoints, centroids of triangles, and other geometric figures in 3D space.

---

Section Formula for Internal Division

Imagine a line segment in space connecting two points, $P(x_1, y_1, z_1)$ and $Q(x_2, y_2, z_2)$. Let's say we want to find the coordinates of a point $R(x, y, z)$ that lies on this segment and divides it internally in the ratio $m:n$.

This means that the point R is located between P and Q, and the ratio of the length of the segment PR to the length of the segment RQ is $\frac{m}{n}$.

Derivation

Let's derive the formula using similar triangles. From points P, R, and Q, draw planes parallel to the yz-plane, intersecting the x-axis at points A, C, and B, respectively. The coordinates of these points on the x-axis will be $(x_1, 0, 0)$, $(x, 0, 0)$, and $(x_2, 0, 0)$.

The planes divide the line segment PQ in the same ratio as they divide the segment AB on the x-axis. Therefore, the ratio of the length of AC to the length of CB is also $m:n$.

$\frac{AC}{CB} = \frac{m}{n}$

The lengths of these segments on the x-axis are the differences in their x-coordinates:

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

Now, we solve for $x$ by cross-multiplying:

$n(x - x_1) = m(x_2 - x)$

$nx - nx_1 = mx_2 - mx$

$nx + mx = mx_2 + nx_1$

$x(m+n) = mx_2 + nx_1$

$x = \frac{mx_2 + nx_1}{m+n}$

By a completely analogous argument (by drawing planes parallel to the xz-plane and xy-plane), we can derive the formulas for the y and z coordinates. The coordinates of the point R are:

Midpoint Formula in 3D

A special and very common case of internal division is finding the midpoint of a line segment. The midpoint divides the segment in the ratio $1:1$. By setting $m=1$ and $n=1$ in the internal division formula, we get the midpoint formula:

$x = \frac{1 \cdot x_2 + 1 \cdot x_1}{1+1} = \frac{x_1 + x_2}{2}$

Similarly for y and z, the midpoint is:

The coordinates of the midpoint are simply the average of the corresponding coordinates of the endpoints.

---

Section Formula for External Division

If the point R lies on the line passing through P and Q, but outside the segment PQ, it is said to divide the segment externally. The derivation is identical, except the ratio $\frac{PR}{RQ}$ is considered negative, which can be achieved by replacing $n$ with $-n$ in the internal formula.

The coordinates of a point R that divides the segment PQ externally in the ratio $m:n$ are:

---

Answer:

---

Answer:

---

Answer:

---

Centroid of a Triangle and Tetrahedron

The concept of finding a "center of mass" or centroid by averaging coordinates extends from 2D to 3D for various geometric figures.

Centroid of a Triangle in 3D

The centroid of a triangle is the point of intersection of its medians. A median is a line segment joining a vertex to the midpoint of the opposite side. The centroid is the geometric center of the triangle and would be its center of mass if the triangle were a thin, uniform plate.

Derivation

Let the vertices of a triangle in 3D space be $A(x_1, y_1, z_1)$, $B(x_2, y_2, z_2)$, and $C(x_3, y_3, z_3)$.

Let AD be the median from vertex A to the side BC. By definition, D is the midpoint of BC.

Step 1: Find the coordinates of the midpoint D.

Using the midpoint formula, the coordinates of D are:

$D = \left( \frac{x_2 + x_3}{2}, \frac{y_2 + y_3}{2}, \frac{z_2 + z_3}{2} \right)$

Step 2: Use the property that the centroid divides the median in a 2:1 ratio.

The centroid, G, lies on the median AD and divides it internally in the ratio $AG:GD = 2:1$.

We can now use the section formula to find the coordinates of G, with $A(x_1, y_1, z_1)$ as the first point, $D$ as the second point, and the ratio $m:n = 2:1$.

The x-coordinate of G is:

$x_G = \frac{2 \cdot x_D + 1 \cdot x_1}{2+1} = \frac{2\left(\frac{x_2 + x_3}{2}\right) + x_1}{3} = \frac{(x_2 + x_3) + x_1}{3} = \frac{x_1 + x_2 + x_3}{3}$

Similarly, the y-coordinate of G is:

$y_G = \frac{2 \cdot y_D + 1 \cdot y_1}{2+1} = \frac{2\left(\frac{y_2 + y_3}{2}\right) + y_1}{3} = \frac{(y_2 + y_3) + y_1}{3} = \frac{y_1 + y_2 + y_3}{3}$

And the z-coordinate of G is:

$z_G = \frac{2 \cdot z_D + 1 \cdot z_1}{2+1} = \frac{2\left(\frac{z_2 + z_3}{2}\right) + z_1}{3} = \frac{(z_2 + z_3) + z_1}{3} = \frac{z_1 + z_2 + z_3}{3}$

This gives us the formula for the centroid, which is simply the average of the coordinates of the vertices:

Centroid of a Tetrahedron

A tetrahedron is a pyramid with a triangular base and three other triangular faces that meet at a common vertex. It is the simplest of all ordinary convex polyhedra and has four vertices.

The centroid of a tetrahedron is the point where the segments connecting each vertex to the centroid of the opposite triangular face intersect. It is the geometric center and the center of mass of the tetrahedron.

Derivation

Let the four vertices of the tetrahedron be $A(x_1, y_1, z_1)$, $B(x_2, y_2, z_2)$, $C(x_3, y_3, z_3)$, and $D(x_4, y_4, z_4)$.

Consider the line segment connecting vertex A to the centroid of the opposite face, the triangle BCD. Let's call the centroid of triangle BCD as $G_1$.

Step 1: Find the coordinates of the centroid $G_1$ of the opposite face.

Using the formula for the centroid of a triangle on the vertices B, C, and D, we get:

$G_1 = \left( \frac{x_2 + x_3 + x_4}{3}, \frac{y_2 + y_3 + y_4}{3}, \frac{z_2 + z_3 + z_4}{3} \right)$

Step 2: Use the property that the centroid of the tetrahedron divides this segment in a 3:1 ratio.

The centroid of the tetrahedron, G, lies on the segment $AG_1$ and divides it internally in the ratio $AG:GG_1 = 3:1$.

We can now use the section formula to find the coordinates of G, with $A(x_1, y_1, z_1)$ as the first point, $G_1$ as the second point, and the ratio $m:n = 3:1$.

The x-coordinate of G is:

$x_G = \frac{3 \cdot x_{G_1} + 1 \cdot x_1}{3+1} = \frac{3\left(\frac{x_2 + x_3 + x_4}{3}\right) + x_1}{4} = \frac{(x_2 + x_3 + x_4) + x_1}{4} = \frac{x_1 + x_2 + x_3 + x_4}{4}$

Similarly, the y and z coordinates will follow the same pattern:

$y_G = \frac{y_1 + y_2 + y_3 + y_4}{4}$

$z_G = \frac{z_1 + z_2 + z_3 + z_4}{4}$

This gives us the formula for the centroid of a tetrahedron, which is the average of the coordinates of its four vertices: