Chapter 11 Three Dimensional Geometry (Concepts)

Introduction to Three Dimensional Geometry

Three-dimensional geometry, often abbreviated as 3D geometry, is a branch of geometry that studies figures in three-dimensional space. Unlike two-dimensional geometry which deals with shapes on a flat plane, 3D geometry explores the properties, measurements, and relationships of points, lines, planes, and solids in space. It provides a mathematical framework for describing the physical world around us.

Coordinate System in Space

To locate the position of a point in three-dimensional space, we use a rectangular coordinate system. This system is formed by three mutually perpendicular straight lines that intersect at a single point. These lines are traditionally designated as the X-axis, Y-axis, and Z-axis. The point of intersection is called the origin, denoted by O.

These three axes determine three mutually perpendicular planes:

• The plane containing the X and Y axes is called the XY-plane.
• The plane containing the Y and Z axes is called the YZ-plane.
• The plane containing the Z and X axes is called the ZX-plane or XZ-plane.

These three coordinate planes divide the entire space into eight regions, known as octants.

The position of any point P in space is uniquely determined by its distances from these three planes. These distances, taken with appropriate signs depending on the octant the point lies in, are called the coordinates of the point P. If the perpendicular distances from point P to the YZ, ZX, and XY planes are $x$, $y$, and $z$ respectively, then the coordinates of P are written as an ordered triplet $(x, y, z)$.

Coordinates of points on axes and planes:

• The coordinates of the origin O are $(0, 0, 0)$.
• A point on the X-axis has coordinates $(x, 0, 0)$. (Only the x-coordinate is non-zero).
• A point on the Y-axis has coordinates $(0, y, 0)$. (Only the y-coordinate is non-zero).
• A point on the Z-axis has coordinates $(0, 0, z)$. (Only the z-coordinate is non-zero).
• A point on the XY-plane has coordinates $(x, y, 0)$. (The z-coordinate is zero).
• A point on the YZ-plane has coordinates $(0, y, z)$. (The x-coordinate is zero).
• A point on the ZX-plane has coordinates $(x, 0, z)$. (The y-coordinate is zero).

Distance Formula in 3D

The distance between any two points in three-dimensional space can be found using the distance formula, which is a direct extension of the Pythagorean theorem applied in three dimensions.

Let P be a point with coordinates $(x_1, y_1, z_1)$ and Q be another point with coordinates $(x_2, y_2, z_2)$. The distance between points P and Q, denoted as PQ, is given by:

This formula can be visualised by constructing a rectangular box with diagonal PQ and edges parallel to the coordinate axes. The lengths of the edges would be $|x_2 - x_1|$, $|y_2 - y_1|$, and $|z_2 - z_1|$. Applying the Pythagorean theorem twice (first on a face diagonal, then on the space diagonal PQ) leads to this formula.

A special case of the distance formula is the distance of a point P$(x, y, z)$ from the origin O$(0, 0, 0)$. Using the coordinates of the origin as $(x_1, y_1, z_1) = (0, 0, 0)$ and the point P as $(x_2, y_2, z_2) = (x, y, z)$ in the distance formula, we get:

Section Formula in 3D

The section formula is used to find the coordinates of a point that divides the line segment joining two given points in a specific ratio. This is a fundamental tool in coordinate geometry, extended here to three dimensions.

Let P$(x_1, y_1, z_1)$ and Q$(x_2, y_2, z_2)$ be two points in space.

Internal Division

Let R$(x, y, z)$ be a point that divides the line segment PQ internally in the ratio $m:n$. This means that R lies between P and Q such that the ratio of the length PR to RQ is $m:n$. The coordinates of the point R are given by:

This formula can be derived using the concept of collinear points and ratios of vectors, or by applying similar triangles property on the projections of the points onto the coordinate axes.

External Division

Let R$(x, y, z)$ be a point that divides the line segment PQ externally in the ratio $m:n$. This means that R lies on the line extending PQ, outside the segment PQ, such that PR : RQ = $m:n$. The coordinates of the point R are given by:

Note the change in sign in the numerator and denominator compared to the internal division formula.

Midpoint Formula

The midpoint of a line segment is a special case of internal division where the ratio is $1:1$ (i.e., $m=1, n=1$). Substituting $m=1$ and $n=1$ into the internal division formula (iii), we get the coordinates of the midpoint of PQ as:

The midpoint formula gives the average of the coordinates of the two given points.

Direction Cosines and Direction Ratios of a Line

In 3D geometry, the orientation or direction of a line in space is described by quantities called direction cosines or direction ratios. These concepts are crucial for defining lines and planes in vector and Cartesian forms.

Direction Cosines

Consider a directed line L in space passing through the origin O. This line makes certain angles with the positive directions of the X, Y, and Z axes. Let these angles be $\alpha, \beta$, and $\gamma$ respectively. These angles are called the direction angles of the line L. The cosines of these angles, i.e., $\cos\alpha, \cos\beta$, and $\cos\gamma$, are called the direction cosines of the line L. They are usually denoted by $l, m, n$.

The direction cosines of a line are not independent; they are related by the fundamental identity:

This relation can be derived by considering a point P$(x, y, z)$ on the line L at a distance $r$ from the origin. Then $x = lr$, $y = mr$, $z = nr$. From the distance formula from the origin, $r = \sqrt{x^2+y^2+z^2}$. Substituting the values of $x, y, z$ gives $r = \sqrt{(lr)^2+(mr)^2+(nr)^2} = \sqrt{r^2(l^2+m^2+n^2)}$, which implies $l^2+m^2+n^2=1$.

If the line does not pass through the origin, we can consider a line parallel to it passing through the origin. The direction cosines of the original line are the same as the direction cosines of this parallel line through the origin because their direction angles with the axes are the same.

Direction Ratios

Any three numbers $a, b, c$ which are proportional to the direction cosines $l, m, n$ of a line are called the direction ratios of the line.

If $a, b, c$ are direction ratios, then there exists a constant $k$ such that:

This gives $l = ak, m = bk, n = ck$. Substituting these into the relation $l^2 + m^2 + n^2 = 1$:

Thus, if $a, b, c$ are direction ratios of a line, its direction cosines $l, m, n$ can be found using the formulas:

There are two sets of direction cosines for a line, corresponding to the two opposite directions on the line. The choice of positive or negative sign depends on the direction of the line being considered.

Direction Ratios and Direction Cosines of a Line Segment Joining Two Points

Let P$(x_1, y_1, z_1)$ and Q$(x_2, y_2, z_2)$ be two points in space.

The numbers $(x_2 - x_1)$, $(y_2 - y_1)$, and $(z_2 - z_1)$ are the differences in the coordinates. These numbers are proportional to the projections of the line segment PQ on the X, Y, and Z axes, respectively. Hence, they are the direction ratios of the line segment PQ.

Let $a = x_2 - x_1$, $b = y_2 - y_1$, and $c = z_2 - z_1$. The distance PQ is $\sqrt{(x_2-x_1)^2 + (y_2-y_1)^2 + (z_2-z_1)^2} = \sqrt{a^2+b^2+c^2}$.

The direction cosines of the line segment PQ, directed from P to Q, are obtained by dividing the direction ratios by the magnitude of the line segment (distance PQ):

where $\text{PQ} = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2 + (z_2-z_1)^2}$.

Answer:

Straight Line (in Space)

In three-dimensional geometry, a straight line is a fundamental geometric object. Its position and direction in space can be uniquely determined if we know either:

1. A point on the line and its direction (parallel to a given vector).
2. Two distinct points through which the line passes.

We can represent the equation of a straight line in space in two primary forms: the Vector Form and the Cartesian Form.

Equation of a Line Passing Through a Given Point and Parallel to a Given Vector

Let's find the equation of a line L that passes through a given point A and is parallel to a given vector $\vec{b}$.

Assume the origin is O. Let the position vector of the given point A with respect to the origin be $\vec{a}$.

Let $\vec{b}$ be the vector parallel to the line L. This vector defines the direction of the line.

Now, consider an arbitrary point P on the line L. Let its position vector with respect to the origin be $\vec{r}$.

Since the point P lies on the line passing through A and is parallel to $\vec{b}$, the vector $\vec{\text{AP}}$ must be parallel to the vector $\vec{b}$.

This means $\vec{\text{AP}}$ can be expressed as a scalar multiple of $\vec{b}$. Let $\lambda$ be a scalar (a real number).

From the triangle law of vector addition in $\triangle \text{OAP}$, we have:

Substituting the position vectors and the expression for $\vec{\text{AP}}$:

This is the vector equation of a straight line passing through the point with position vector $\vec{a}$ and parallel to the vector $\vec{b}$.

The scalar $\lambda$ is called the parameter. As $\lambda$ takes different real values, the point P with position vector $\vec{r}$ traces out all the points on the line.

Cartesian Form

To derive the Cartesian equation from the vector form, we express the position vectors and the parallel vector in terms of their components along the coordinate axes.

Let the coordinates of the given point A be $(x_1, y_1, z_1)$. Then its position vector is:

Let the direction ratios of the line (which are the components of the parallel vector $\vec{b}$) be $a, b, c$. Then the vector $\vec{b}$ is:

Let the coordinates of the arbitrary point P on the line be $(x, y, z)$. Its position vector is:

Substitute these component forms into the vector equation $\vec{r} = \vec{a} + \lambda \vec{b}$:

Equating the coefficients of $\hat{i}, \hat{j}, \hat{k}$ on both sides (since the vectors are equal):

These three equations are the parametric equations of the line in Cartesian form. They represent the coordinates $(x, y, z)$ of any point on the line in terms of the parameter $\lambda$.

If $a, b, c$ are all non-zero, we can solve for $\lambda$ from each equation:

Since all these expressions are equal to the same parameter $\lambda$, we can equate them:

This is the Cartesian equation of a straight line passing through the point $(x_1, y_1, z_1)$ and having direction ratios $a, b, c$.

Note: If any of $a, b,$ or $c$ is zero, the Cartesian equation takes a slightly different form. For example, if $a=0$, the line is perpendicular to the x-axis (i.e., parallel to the YZ-plane). The equations would become $\frac{y - y_1}{b} = \frac{z - z_1}{c}$ and $x = x_1$. We would write it as $\frac{x - x_1}{0} = \frac{y - y_1}{b} = \frac{z - z_1}{c}$, where $\frac{x-x_1}{0}$ implicitly means $x - x_1 = 0$, or $x=x_1$.

Equation of a Line Passing Through Two Given Points

Now let's find the equation of a line L that passes through two distinct points A and B.

Let the position vector of point A be $\vec{a}$ and the position vector of point B be $\vec{b}$ with respect to the origin O.

Any line passing through points A and B is uniquely determined. The line passes through point A (we could also choose B) and is parallel to the vector $\vec{\text{AB}}$.

The vector $\vec{\text{AB}}$ is given by:

Now, this problem is similar to the previous case: find the equation of a line passing through point A (position vector $\vec{a}$) and parallel to the vector $(\vec{b} - \vec{a})$.

Using the vector form $\vec{r} = \vec{a} + \lambda \vec{b}$ from equation (i), we replace the parallel vector $\vec{b}$ with $(\vec{b} - \vec{a})$:

This is the vector equation of a straight line passing through two points with position vectors $\vec{a}$ and $\vec{b}$.

As the parameter $\lambda$ varies, the point P with position vector $\vec{r}$ moves along the line passing through A and B. Note that if $\lambda=0$, $\vec{r}=\vec{a}$ (point A), and if $\lambda=1$, $\vec{r} = \vec{a} + (\vec{b} - \vec{a}) = \vec{b}$ (point B).

Cartesian Form

Let the coordinates of the two given points be P$_1(x_1, y_1, z_1)$ and P$_2(x_2, y_2, z_2)$.

The position vector of P$_1$ is $\vec{a} = x_1\hat{i} + y_1\hat{j} + z_1\hat{k}$.

The position vector of P$_2$ is $\vec{b} = x_2\hat{i} + y_2\hat{j} + z_2\hat{k}$.

The vector parallel to the line is $\vec{\text{P}_1\text{P}_2} = \vec{b} - \vec{a}$. Its components are the direction ratios of the line:

So, the direction ratios of the line are $a = (x_2 - x_1)$, $b = (y_2 - y_1)$, and $c = (z_2 - z_1)$.

Let the coordinates of an arbitrary point P on the line be $(x, y, z)$, with position vector $\vec{r} = x\hat{i} + y\hat{j} + z\hat{k}$.

Using the Cartesian form of the line equation $\frac{x - x_1}{a} = \frac{y - y_1}{b} = \frac{z - z_1}{c}$ from equation (v), and substituting the direction ratios $a, b, c$ with $(x_2 - x_1), (y_2 - y_1), (z_2 - z_1)$ respectively, we get:

This is the Cartesian equation of a straight line passing through two points $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$.

Similar to the previous case, if any denominator is zero, it means the numerator must also be zero. For example, if $x_2 - x_1 = 0$, then the line is parallel to the YZ-plane, and the equation is $\frac{y - y_1}{y_2 - y_1} = \frac{z - z_1}{z_2 - z_1}$ and $x = x_1$.

Answer:

Angle Between Two Lines

In three-dimensional space, two non-parallel lines will either intersect at a single point or be skew lines (non-intersecting and non-parallel). Regardless of whether they intersect or are skew, the angle between two lines is defined as the angle between their corresponding direction vectors. If the lines are parallel, the angle is considered to be $0^\circ$ or $180^\circ$. Conventionally, we consider the acute angle between the two lines, which lies in the range $[0, \pi/2]$.

Angle Between Two Lines Given in Vector Form

Let the equations of two lines L$_1$ and L$_2$ be given in vector form as:

Here, $\vec{b}_1$ and $\vec{b}_2$ are the direction vectors of the lines L$_1$ and L$_2$, respectively. The angle between the two lines is defined as the angle between their direction vectors $\vec{b}_1$ and $\vec{b}_2$.

Let $\theta$ be the angle between the vectors $\vec{b}_1$ and $\vec{b}_2$. Using the definition of the dot product of two vectors, we have:

Solving for $\cos\theta$:

Since the angle between two lines is conventionally taken as the acute angle, we consider the absolute value of $\cos\theta$:

where $0 \leq \theta \leq \pi/2$.

Angle Between Two Lines Given in Cartesian Form

Let the equations of two lines L$_1$ and L$_2$ be given in Cartesian form as:

The direction ratios of the line L$_1$ are $a_1, b_1, c_1$. A vector parallel to this line is $\vec{b}_1 = a_1\hat{i} + b_1\hat{j} + c_1\hat{k}$.

The direction ratios of the line L$_2$ are $a_2, b_2, c_2$. A vector parallel to this line is $\vec{b}_2 = a_2\hat{i} + b_2\hat{j} + c_2\hat{k}$.

The angle $\theta$ between the two lines is the angle between their direction vectors $\vec{b}_1$ and $\vec{b}_2$. Using the dot product formula from equation (i) in component form:

The magnitudes of the vectors are:

Substituting these into the formula $\cos\theta = \frac{|\vec{b}_1 \cdot \vec{b}_2|}{|\vec{b}_1| |\vec{b}_2|}$:

This formula gives the cosine of the acute angle between the two lines in terms of their direction ratios.

Conditions for Parallel and Perpendicular Lines

Using the formula for the angle between two lines, we can derive the conditions for them to be parallel or perpendicular.

1. Parallel Lines

Two lines are parallel if the angle between them is $0^\circ$ or $180^\circ$. This occurs when their direction vectors $\vec{b}_1$ and $\vec{b}_2$ are parallel.

Vectors $\vec{b}_1$ and $\vec{b}_2$ are parallel if and only if one is a scalar multiple of the other, i.e., $\vec{b}_1 = k \vec{b}_2$ for some non-zero scalar $k$.

In terms of components (direction ratios $a_1, b_1, c_1$ and $a_2, b_2, c_2$), this means:

Assuming $a_2, b_2, c_2$ are not all zero, this is equivalent to the condition that their direction ratios are proportional:

Also, if $\theta = 0^\circ$ or $180^\circ$, then $\cos\theta = \pm 1$. From formula (i), $\frac{|\vec{b}_1 \cdot \vec{b}_2|}{|\vec{b}_1| |\vec{b}_2|} = 1$, which means $|\vec{b}_1 \cdot \vec{b}_2| = |\vec{b}_1| |\vec{b}_2|$. This only happens when $\vec{b}_1$ and $\vec{b}_2$ are parallel.

2. Perpendicular Lines

Two lines are perpendicular if the angle between them is $90^\circ$ or $\pi/2$. This occurs when their direction vectors $\vec{b}_1$ and $\vec{b}_2$ are perpendicular.

Two vectors are perpendicular if and only if their dot product is zero:

In terms of components (direction ratios $a_1, b_1, c_1$ and $a_2, b_2, c_2$), the condition for perpendicularity is:

Also, if $\theta = 90^\circ = \pi/2$, then $\cos\theta = \cos(\pi/2) = 0$. From formula (i) or (ii), this directly implies $|\vec{b}_1 \cdot \vec{b}_2| = 0$ or $|a_1 a_2 + b_1 b_2 + c_1 c_2| = 0$, which simplifies to $\vec{b}_1 \cdot \vec{b}_2 = 0$ or $a_1 a_2 + b_1 b_2 + c_1 c_2 = 0$.

Answer:

Shortest Distance between Two Lines

In three-dimensional space, unlike in a plane where two lines are either parallel or intersecting, two lines can also be skew. The distance between two lines refers to the length of the shortest line segment connecting them. The nature of this shortest distance depends on the relationship between the two lines:

• If two lines intersect, they have a common point. The shortest distance between them is $\textbf{0}$.
• If two lines are parallel, the distance between them is the constant perpendicular distance from any point on one line to the other line.
• If two lines are skew, they are neither parallel nor intersecting. They lie in different planes. The shortest distance between skew lines is the length of the unique line segment that is perpendicular to both lines.

Shortest Distance between Skew Lines

Skew lines are lines in three-dimensional space that are neither parallel nor intersecting. They exist in different planes.

Let the equations of the two skew lines L$_1$ and L$_2$ be given in vector form:

Here, $\vec{a}_1$ is the position vector of a point A on L$_1$, and $\vec{a}_2$ is the position vector of a point C on L$_2$. $\vec{b}_1$ is the direction vector of L$_1$, and $\vec{b}_2$ is the direction vector of L$_2$. Since the lines are skew, $\vec{b}_1$ is not parallel to $\vec{b}_2$, which implies $\vec{b}_1 \times \vec{b}_2 \neq \vec{0}$.

The shortest distance between the skew lines is the length of the line segment PQ, where P is a point on L$_1$ and Q is a point on L$_2$, such that PQ is perpendicular to both L$_1$ and L$_2$. This segment PQ is perpendicular to both direction vectors $\vec{b}_1$ and $\vec{b}_2$. Therefore, the direction of the shortest distance vector $\vec{\text{PQ}}$ is parallel to the vector $\vec{b}_1 \times \vec{b}_2$.

Consider the vector connecting any point A on L$_1$ to any point C on L$_2$. This vector is $\vec{\text{AC}} = \vec{a}_2 - \vec{a}_1$.

The shortest distance $d$ between the lines is the length of the projection of the vector $\vec{\text{AC}} = \vec{a}_2 - \vec{a}_1$ onto the direction perpendicular to both lines, which is the direction of $\vec{b}_1 \times \vec{b}_2$.

The formula for the scalar projection of a vector $\vec{u}$ onto a vector $\vec{v}$ is $\frac{\vec{u} \cdot \vec{v}}{|\vec{v}|}$.

In our case, $\vec{u} = \vec{a}_2 - \vec{a}_1$ and $\vec{v} = \vec{b}_1 \times \vec{b}_2$. So, the shortest distance $d$ is the magnitude of this projection:

The numerator $|(\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2)|$ is the magnitude of the scalar triple product of the vectors $(\vec{a}_2 - \vec{a}_1)$, $\vec{b}_1$, and $\vec{b}_2$.

Cartesian Form

Let the two skew lines be given in Cartesian form:

From these equations, we can identify the vectors for the formula (i):

The vector connecting the points is $\vec{a}_2 - \vec{a}_1 = (x_2 - x_1)\hat{i} + (y_2 - y_1)\hat{j} + (z_2 - z_1)\hat{k}$.

The cross product of the direction vectors is $\vec{b}_1 \times \vec{b}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \end{vmatrix} = (b_1c_2 - b_2c_1)\hat{i} - (a_1c_2 - a_2c_1)\hat{j} + (a_1b_2 - a_2b_1)\hat{k}$.

The magnitude is $|\vec{b}_1 \times \vec{b}_2| = \sqrt{(b_1c_2 - b_2c_1)^2 + (a_1c_2 - a_2c_1)^2 + (a_1b_2 - a_2b_1)^2}$.

The dot product $(\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2)$ is the scalar triple product, which can be calculated as the determinant:

Using these in formula (i), the shortest distance $d$ between the two skew lines in Cartesian form is:

Note: If the lines intersect, the shortest distance is 0. This will be reflected in the formula (ii) as the determinant in the numerator will be zero (since the vectors $(\vec{a}_2 - \vec{a}_1)$, $\vec{b}_1$, and $\vec{b}_2$ will be coplanar).

Shortest Distance between Parallel Lines

Let the equations of the two parallel lines L$_1$ and L$_2$ be given in vector form:

Here, the direction vector $\vec{b}$ is the same for both lines, indicating they are parallel. $\vec{a}_1$ is the position vector of a point A on L$_1$, and $\vec{a}_2$ is the position vector of a point C on L$_2$.

The shortest distance $d$ between the parallel lines is the perpendicular distance from any point on one line to the other line. Let's find the distance from point A (with position vector $\vec{a}_1$) on L$_1$ to the line L$_2$.

Consider the vector $\vec{\text{AC}} = \vec{a}_2 - \vec{a}_1$ connecting a point on L$_1$ to a point on L$_2$.

Imagine a parallelogram formed by the vectors $\vec{\text{AC}} = \vec{a}_2 - \vec{a}_1$ and $\vec{b}$. The shortest distance $d$ between the parallel lines is the height of this parallelogram with the base along the direction of $\vec{b}$.

The area of this parallelogram is given by the magnitude of the cross product of the adjacent vectors:

The area of a parallelogram is also given by the product of its base and height. Taking $|\vec{b}|$ as the base length (since the base is along the direction of $\vec{b}$), the height is the shortest distance $d$.

Equating the two expressions for the area:

Solving for $d$:

This is the formula for the shortest distance between two parallel lines in vector form.

To find the distance in Cartesian form, we can identify $\vec{a}_1, \vec{a}_2,$ and $\vec{b}$ from the Cartesian equations and use formula (iii). If the Cartesian equations are $\frac{x - x_1}{a} = \frac{y - y_1}{b} = \frac{z - z_1}{c}$ and $\frac{x - x_2}{a} = \frac{y - y_2}{b} = \frac{z - z_2}{c}$ (note the same denominators $a, b, c$ as direction ratios), then $\vec{a}_1 = x_1\hat{i} + y_1\hat{j} + z_1\hat{k}$, $\vec{a}_2 = x_2\hat{i} + y_2\hat{j} + z_2\hat{k}$, and $\vec{b} = a\hat{i} + b\hat{j} + c\hat{k}$.

Answer:

Plane

A plane is a fundamental concept in three-dimensional geometry. It is a flat, two-dimensional surface that extends infinitely in all directions. In a 3D Cartesian coordinate system, a plane is represented by a single linear equation involving the variables $x, y,$ and $z$.

The equation of a plane can be determined under various conditions, such as knowing:

1. The perpendicular distance from the origin and the direction of the normal to the plane.
2. A point on the plane and the direction of the normal to the plane.
3. Three non-collinear points lying on the plane.
4. A line and a point not on the line that lie on the plane.
5. Two intersecting lines lying on the plane.
6. Two parallel lines lying on the plane.

We will discuss some of the most common forms of the equation of a plane.

Equation of a Plane in Normal Form

The equation of a plane is said to be in normal form if it is defined by the perpendicular distance from the origin to the plane and the direction of the normal vector pointing from the origin to the plane.

Let $p$ be the perpendicular distance of the plane from the origin O$(0, 0, 0)$.

Let $\vec{ON}$ be the perpendicular (normal) from the origin to the plane. Let $\hat{n}$ be the unit vector in the direction of $\vec{ON}$. The direction cosines of $\vec{ON}$ are $l, m, n$.

By definition of the unit vector and $p$, we have $\vec{ON} = p \hat{n}$.

Consider any arbitrary point P on the plane. Let its position vector with respect to the origin be $\vec{r}$.

The vector $\vec{\text{NP}}$ lies in the plane. Since $\vec{ON}$ is perpendicular to the plane, it must be perpendicular to every vector lying in the plane, including $\vec{\text{NP}}$.

Thus, $\vec{\text{NP}} \cdot \vec{ON} = 0$.

We know that $\vec{\text{NP}} = \vec{\text{OP}} - \vec{\text{ON}} = \vec{r} - p\hat{n}$.

Substituting this into the dot product equation:

This is not the standard derivation. A simpler way is to consider the projection. The projection of the position vector $\vec{r}$ of any point P on the plane onto the unit normal vector $\hat{n}$ must be equal to the distance $p$ from the origin.

So, we have:

This is the vector equation of the plane in normal form.

Cartesian Form

Let the coordinates of the arbitrary point P on the plane be $(x, y, z)$, so $\vec{r} = x\hat{i} + y\hat{j} + z\hat{k}$.

The unit normal vector $\hat{n}$ has direction cosines $l, m, n$. So, $\hat{n} = l\hat{i} + m\hat{j} + n\hat{k}$.

Substitute these component forms into the vector equation $\vec{r} \cdot \hat{n} = p$:

Taking the dot product:

This is the Cartesian equation of the plane in normal form, where $l, m, n$ are the direction cosines of the normal to the plane and $p$ is the perpendicular distance of the plane from the origin.

General Equation of a Plane

The general equation of a linear equation in three variables $x, y, z$ always represents a plane in 3D space. The most general form of the Cartesian equation of a plane is:

where A, B, C, and D are constants, and A, B, C are not all simultaneously zero.

The coefficients A, B, and C in the general equation are the direction ratios of a vector normal to the plane. That is, the vector $\vec{N} = A\hat{i} + B\hat{j} + C\hat{k}$ is normal to the plane Ax + By + Cz + D = 0.

Conversion from General Form to Normal Form

To convert the general equation Ax + By + Cz + D = 0 into the normal form $lx + my + nz = p$, we need to find the direction cosines of the normal and the perpendicular distance from the origin.

The direction ratios of the normal are A, B, C. The direction cosines $l, m, n$ are obtained by dividing the direction ratios by $\pm \sqrt{A^2+B^2+C^2}$. The sign is chosen such that the constant term $p$ on the right side is positive (representing distance).

Rearrange the general equation: Ax + By + Cz = -D.

Divide both sides by $\sqrt{A^2+B^2+C^2}$:

If $-D/\sqrt{A^2+B^2+C^2}$ is positive, then $l = \frac{A}{\sqrt{A^2+B^2+C^2}}$, $m = \frac{B}{\sqrt{A^2+B^2+C^2}}$, $n = \frac{C}{\sqrt{A^2+B^2+C^2}}$, and $p = \frac{-D}{\sqrt{A^2+B^2+C^2}}$.

If $-D/\sqrt{A^2+B^2+C^2}$ is negative, we multiply the entire equation by -1 to make the right side positive:

In this case, $l = \frac{-A}{\sqrt{A^2+B^2+C^2}}$, $m = \frac{-B}{\sqrt{A^2+B^2+C^2}}$, $n = \frac{-C}{\sqrt{A^2+B^2+C^2}}$, and $p = \frac{D}{\sqrt{A^2+B^2+C^2}}$.

In general, the direction cosines of the normal and the distance $p$ are given by:

where the sign of $\sqrt{A^2+B^2+C^2}$ is chosen to make the constant term positive after dividing.

Equation of a Plane Passing Through a Given Point and Perpendicular to a Given Vector

Suppose we want to find the equation of a plane that passes through a specific point and is perpendicular to a given vector.

Let A be the given point with position vector $\vec{a}$ relative to the origin O.

Let $\vec{N}$ be the given vector that is perpendicular to the plane (this vector is called the normal vector to the plane).

Consider an arbitrary point P on the plane with position vector $\vec{r}$.

The vector $\vec{\text{AP}}$ connects the fixed point A on the plane to the arbitrary point P on the plane. This vector must lie entirely within the plane.

Since the vector $\vec{N}$ is perpendicular to the plane, it must be perpendicular to every vector lying in the plane. Therefore, $\vec{N}$ is perpendicular to $\vec{\text{AP}}$.

Using the property that the dot product of two perpendicular vectors is zero:

Substituting the expression for $\vec{\text{AP}}$:

This is the vector equation of a plane passing through a point with position vector $\vec{a}$ and perpendicular to the vector $\vec{N}$. This can also be written as $\vec{r} \cdot \vec{N} = \vec{a} \cdot \vec{N}$.

Cartesian Form

Let the coordinates of the given point A be $(x_1, y_1, z_1)$. Its position vector is $\vec{a} = x_1\hat{i} + y_1\hat{j} + z_1\hat{k}$.

Let the given normal vector be $\vec{N} = A\hat{i} + B\hat{j} + C\hat{k}$. (Here A, B, C are the direction ratios of the normal).

Let the coordinates of the arbitrary point P on the plane be $(x, y, z)$. Its position vector is $\vec{r} = x\hat{i} + y\hat{j} + z\hat{k}$.

The vector $\vec{r} - \vec{a}$ in component form is:

Substitute the component forms into the vector equation $(\vec{r} - \vec{a}) \cdot \vec{N} = 0$:

Taking the dot product:

This is the Cartesian equation of a plane passing through the point $(x_1, y_1, z_1)$ and having A, B, C as the direction ratios of its normal.

Expanding this equation, we get Ax + By + Cz - Ax$_1$ - By$_1$ - Cz$_1$ = 0. This is in the general form Ax + By + Cz + D = 0, where D = -Ax$_1$ - By$_1$ - Cz$_1$.

Equation of a Plane Passing Through Three Non-Collinear Points

A unique plane is determined by three points that do not lie on the same straight line (non-collinear points).

Let the three non-collinear points be A, B, and C with position vectors $\vec{a}, \vec{b}$, and $\vec{c}$ respectively.

Consider an arbitrary point P on the plane with position vector $\vec{r}$.

The three vectors formed by connecting these points, starting from one point (say A), and the arbitrary point P will lie in the same plane. These vectors are:

Since points A, B, C, and P are coplanar (they lie on the same plane), the vectors $\vec{\text{AP}}, \vec{\text{AB}},$ and $\vec{\text{AC}}$ are also coplanar.

The condition for three vectors to be coplanar is that their scalar triple product is zero.

Substituting the expressions for the vectors:

This is the vector equation of a plane passing through three non-collinear points A, B, and C with position vectors $\vec{a}, \vec{b},$ and $\vec{c}$.

Cartesian Form

Let the coordinates of the three non-collinear points be P$_1(x_1, y_1, z_1)$, P$_2(x_2, y_2, z_2)$, and P$_3(x_3, y_3, z_3)$.

Let P$(x, y, z)$ be an arbitrary point on the plane.

The corresponding position vectors are $\vec{a} = x_1\hat{i} + y_1\hat{j} + z_1\hat{k}$, $\vec{b} = x_2\hat{i} + y_2\hat{j} + z_2\hat{k}$, $\vec{c} = x_3\hat{i} + y_3\hat{j} + z_3\hat{k}$, and $\vec{r} = x\hat{i} + y\hat{j} + z\hat{k}$.

The vectors $\vec{\text{P}_1\text{P}} = \vec{r} - \vec{a}$, $\vec{\text{P}_1\text{P}_2} = \vec{b} - \vec{a}$, and $\vec{\text{P}_1\text{P}_3} = \vec{c} - \vec{a}$ are coplanar.

The scalar triple product $(\vec{r} - \vec{a}) \cdot ((\vec{b} - \vec{a}) \times (\vec{c} - \vec{a})) = 0$ can be written in determinant form using the components of the vectors:

Expanding this determinant gives the Cartesian equation of the plane passing through the three given points.

Equation of a Plane Passing Through the Intersection of Two Planes

Consider two distinct planes with equations:

If these two planes intersect, their intersection forms a straight line. There are infinitely many planes that can pass through this line of intersection. This collection of planes is called a pencil of planes.

The equation of any plane passing through the intersection of planes P$_1$ and P$_2$ can be represented by the linear combination of their equations:

where $\lambda$ is any real number (parameter).

This equation is linear in $x, y,$ and $z$, so it represents a plane.

If a point $(x_0, y_0, z_0)$ lies on the intersection of P$_1$ and P$_2$, then $A_1x_0 + B_1y_0 + C_1z_0 + D_1 = 0$ and $A_2x_0 + B_2y_0 + C_2z_0 + D_2 = 0$. Substituting these into equation (viii), we get $0 + \lambda(0) = 0$, which is always true. This shows that any point on the line of intersection of P$_1$ and P$_2$ satisfies equation (viii), meaning the plane represented by equation (viii) passes through the line of intersection.

By varying the value of $\lambda$, we get different planes passing through the same line of intersection.

To find the equation of a specific plane from this family, we usually need an additional condition, such as the plane passing through a given point not on the line of intersection. This condition allows us to determine the unique value of $\lambda$ for that specific plane.

Note: Equation (viii) does not represent the plane $A_2x + B_2y + C_2z + D_2 = 0$. This plane is obtained by taking the limit as $\lambda \to \infty$. However, if we are looking for a plane passing through the intersection and a specific point, and that point happens to be on the plane $P_1$ but not $P_2$, then $\lambda$ would be 0 and we get $P_1=0$. If the point is on $P_2$ but not $P_1$, we can write the equation as $P_2 + \lambda' P_1 = 0$ and find $\lambda'$. Alternatively, one can check if the given point satisfies either $P_1=0$ or $P_2=0$. If it satisfies $P_2=0$, then $P_2=0$ is the required plane (unless the point is on the line of intersection itself). A more robust way is to simply plug the point into $(A_1x + B_1y + C_1z + D_1) + \lambda (A_2x + B_2y + C_2z + D_2) = 0$ and solve for $\lambda$. If the point satisfies $P_2=0$ and not $P_1=0$, this method will yield $P_1 + \lambda(0) = 0 \implies P_1=0$, which is not the correct plane. Therefore, if the given point satisfies $P_2=0$, the required plane is $P_2=0$ itself, unless it also satisfies $P_1=0$ (meaning the point is on the line of intersection). A point not on the line of intersection cannot satisfy both equations simultaneously. So, if the point satisfies $P_2=0$ and not $P_1=0$, the plane $P_2=0$ is the answer. If it satisfies $P_1=0$ and not $P_2=0$, the plane $P_1=0$ is the answer. If it satisfies neither, then use equation (viii) directly.

Angle Between Two Planes

The angle between two planes in space is defined as the angle between their normal vectors. If two planes are parallel, the angle between them is $0^\circ$. If they are perpendicular, the angle is $90^\circ$. For intersecting planes, the angle between them is taken as the acute angle formed by the intersection.

Consider two planes. Let $\vec{N}_1$ be a vector normal to the first plane and $\vec{N}_2$ be a vector normal to the second plane. The angle $\theta$ between the two planes is defined as the acute angle between their normal vectors $\vec{N}_1$ and $\vec{N}_2$.

Angle Between Two Planes Given in Cartesian Form

Let the equations of the two planes P$_1$ and P$_2$ be given in Cartesian form:

From the general form of the plane equation, the coefficients of $x, y,$ and $z$ are the direction ratios of the normal vector to the plane.

A normal vector to Plane P$_1$ is $\vec{N}_1 = A_1\hat{i} + B_1\hat{j} + C_1\hat{k}$.

A normal vector to Plane P$_2$ is $\vec{N}_2 = A_2\hat{i} + B_2\hat{j} + C_2\hat{k}$.

The angle $\theta$ between the two planes is the acute angle between their normal vectors $\vec{N}_1$ and $\vec{N}_2$. Using the dot product formula for the angle between two vectors:

The angle $\theta$ between the planes is the acute angle between the normals, so $\theta = \phi$ if $\phi \leq \pi/2$, and $\theta = \pi - \phi$ if $\phi > \pi/2$. This is equivalent to taking the absolute value of $\cos\phi$.

In component form:

Substituting these into formula (i):

This formula gives the cosine of the acute angle $\theta$ between the two planes.

Conditions for Parallel and Perpendicular Planes

Based on the angle formula, we can state the conditions for planes to be parallel or perpendicular.

1. Parallel Planes

Two planes are parallel if and only if their normal vectors are parallel. This means $\vec{N}_1 = k \vec{N}_2$ for some non-zero scalar $k$.

In terms of direction ratios (coefficients A, B, C), this implies that the direction ratios are proportional:

If, in addition to the coefficients of $x, y, z$ being proportional, the constant terms are also proportional, i.e., $\frac{A_1}{A_2} = \frac{B_1}{B_2} = \frac{C_1}{C_2} = \frac{D_1}{D_2}$, then the planes are not just parallel but are coincident (the same plane).

If planes are parallel, the angle between them is $0^\circ$, so $\cos 0^\circ = 1$. From formula (i), $\frac{|\vec{N}_1 \cdot \vec{N}_2|}{|\vec{N}_1| |\vec{N}_2|} = 1$, which means $|\vec{N}_1 \cdot \vec{N}_2| = |\vec{N}_1| |\vec{N}_2|$. This occurs when $\vec{N}_1$ and $\vec{N}_2$ are parallel.

2. Perpendicular Planes

Two planes are perpendicular if and only if their normal vectors are perpendicular. This means the angle between their normal vectors is $90^\circ$ or $\pi/2$.

The condition for two vectors to be perpendicular is that their dot product is zero:

In component form (using direction ratios A, B, C):

If planes are perpendicular, the angle between them is $90^\circ$, so $\cos 90^\circ = 0$. From formula (i), $\frac{|\vec{N}_1 \cdot \vec{N}_2|}{|\vec{N}_1| |\vec{N}_2|} = 0$, which means $|\vec{N}_1 \cdot \vec{N}_2| = 0$, leading to $\vec{N}_1 \cdot \vec{N}_2 = 0$ or $A_1 A_2 + B_1 B_2 + C_1 C_2 = 0$.

Angle Between a Line and a Plane

The angle between a line and a plane is defined as the complement of the angle between the line and the normal to the plane.

Let the equation of the line be $\vec{r} = \vec{a} + \lambda \vec{b}$. The direction vector of the line is $\vec{b}$.

Let the equation of the plane be $\vec{r} \cdot \vec{N} = p$ or $Ax + By + Cz + D = 0$, where the normal vector to the plane is $\vec{N} = A\hat{i} + B\hat{j} + C\hat{k}$.

Let $\phi$ be the angle between the direction vector of the line $\vec{b}$ and the normal vector of the plane $\vec{N}$. Using the dot product formula:

The angle $\theta$ between the line and the plane is the complement of the angle $\phi$ between the line and the normal.

So, $\theta = \frac{\pi}{2} - \phi$. Taking the sine of both sides:

Therefore, the angle $\theta$ between the line and the plane is given by:

In Cartesian form, if the line is $\frac{x-x_1}{l} = \frac{y-y_1}{m} = \frac{z-z_1}{n}$ (direction ratios $l, m, n$) and the plane is $Ax+By+Cz+D=0$ (normal direction ratios A, B, C), then $\vec{b} = l\hat{i} + m\hat{j} + n\hat{k}$ and $\vec{N} = A\hat{i} + B\hat{j} + C\hat{k}$.

The dot product $\vec{b} \cdot \vec{N} = Al + Bm + Cn$.

The magnitudes are $|\vec{b}| = \sqrt{l^2 + m^2 + n^2}$ and $|\vec{N}| = \sqrt{A^2 + B^2 + C^2}$.

Substituting these into formula (vi):

This formula gives the sine of the angle between the line and the plane.

Conditions for a Line to be Parallel or Perpendicular to a Plane

We can derive conditions for a line and a plane to be parallel or perpendicular using the angle formula.

1. Line is Parallel to a Plane

A line is parallel to a plane if the angle between them is $0^\circ$. This means $\sin\theta = \sin 0^\circ = 0$.

From formula (vi), $\frac{|\vec{b} \cdot \vec{N}|}{|\vec{b}| |\vec{N}|} = 0$, which implies $|\vec{b} \cdot \vec{N}| = 0$, so $\vec{b} \cdot \vec{N} = 0$.

Thus, a line is parallel to a plane if and only if its direction vector is perpendicular to the normal vector of the plane.

In Cartesian form (from formula vii): $Al + Bm + Cn = 0$.

This means the sum of the products of the corresponding direction ratios of the line and the normal to the plane is zero.

Note: If the line lies in the plane, it is also considered parallel. The condition $Al + Bm + Cn = 0$ holds. Additionally, any point on the line must satisfy the plane's equation.

2. Line is Perpendicular to a Plane

A line is perpendicular to a plane if the angle between them is $90^\circ$ or $\pi/2$. This means $\sin\theta = \sin(\pi/2) = 1$.

From formula (vi), $\frac{|\vec{b} \cdot \vec{N}|}{|\vec{b}| |\vec{N}|} = 1$. This occurs when the vectors $\vec{b}$ and $\vec{N}$ are parallel.

Thus, a line is perpendicular to a plane if and only if its direction vector is parallel to the normal vector of the plane. This means $\vec{b} = k \vec{N}$ for some non-zero scalar $k$.

In Cartesian form, the direction ratios of the line $(l, m, n)$ must be proportional to the direction ratios of the normal $(A, B, C)$:

Answer:

Distance of a Point from a Plane

A common problem in three-dimensional geometry is finding the perpendicular distance from a given point to a given plane. This distance is the shortest distance between the point and any point on the plane.

Distance of a Point from a Plane (Cartesian Form)

Let the equation of the plane be given in the general Cartesian form:

Here, A, B, and C are the direction ratios of the normal vector to the plane, and D is a constant.

Let P be a point with coordinates $(x_1, y_1, z_1)$. We want to find the perpendicular distance $d$ from point P to the given plane.

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

Explanation of the formula:

• The numerator, $|Ax_1 + By_1 + Cz_1 + D|$, is the absolute value of the expression Ax + By + Cz + D evaluated at the coordinates of the given point $(x_1, y_1, z_1)$. This value is related to how "far off" the point is from satisfying the plane's equation.
• The denominator, $\sqrt{A^2 + B^2 + C^2}$, is the magnitude of the normal vector $\vec{N} = A\hat{i} + B\hat{j} + C\hat{k}$. Dividing by this magnitude normalizes the coefficients A, B, C, effectively converting them into direction cosines of the normal vector. The term $\sqrt{A^2 + B^2 + C^2}$ acts as a scaling factor to ensure the distance is perpendicular distance.

The absolute value in the numerator ensures that the distance $d$ is always non-negative, as distance is a scalar magnitude.

Distance of the Origin from a Plane

A special case is finding the distance of the origin O$(0, 0, 0)$ from the plane Ax + By + Cz + D = 0. Here, $(x_1, y_1, z_1) = (0, 0, 0)$.

Substituting $(0, 0, 0)$ into the distance formula (i):

This formula matches the perpendicular distance $p$ from the origin to the plane in the normal form $lx + my + nz = p$, where $p = \frac{|-D|}{\sqrt{A^2 + B^2 + C^2}} = \frac{|D|}{\sqrt{A^2 + B^2 + C^2}}$.

Distance Between Two Parallel Planes

If two planes are parallel, the distance between them is constant. This distance is the perpendicular distance from any point on one plane to the other plane.

Let the equations of the two parallel planes be:

Since the planes are parallel, their normal vectors are parallel. This means the coefficients of $x, y,$ and $z$ are proportional: $\frac{A_1}{A_2} = \frac{B_1}{B_2} = \frac{C_1}{C_2} = k$ for some constant $k$. If $k=D_1/D_2$, the planes are coincident. Assuming they are distinct parallel planes, $k \neq D_1/D_2$.

To make the normal vectors identical, we can rewrite the second equation by multiplying it by $k$:

Since $kA_2 = A_1, kB_2 = B_1, kC_2 = C_1$, the equation becomes:

Now we have two parallel planes with the same normal vector direction ratios (A$_1$, B$_1$, C$_1$):

The distance between these two planes is the distance from any point on Plane 1 to Plane 2'.

Let P$_1(x_0, y_0, z_0)$ be any point on Plane 1. Then $A_1x_0 + B_1y_0 + C_1z_0 + D_1 = 0$, which means $A_1x_0 + B_1y_0 + C_1z_0 = -D_1$.

Using the point-plane distance formula (i) for point P$_1(x_0, y_0, z_0)$ and plane Plane 2' (A$_1$x + B$_1$y + C$_1$z + kD$_2$ = 0):

Substitute $A_1x_0 + B_1y_0 + C_1z_0 = -D_1$:

Substitute $k = A_1/A_2$ (assuming $A_2 \neq 0$; similar substitution can be made if $B_2 \neq 0$ or $C_2 \neq 0$):

This form is less symmetric. A simpler formula when the coefficients of x, y, z are identical (say, $Ax + By + Cz + D_1 = 0$ and $Ax + By + Cz + D_2 = 0$) is:

If the coefficients are proportional like $A_1x + B_1y + C_1z + D_1 = 0$ and $A_2x + B_2y + C_2z + D_2 = 0$ with $\frac{A_1}{A_2} = \frac{B_1}{B_2} = \frac{C_1}{C_2} = k$, we can divide the second equation by $k$ to get $\frac{A_2}{k}x + \frac{B_2}{k}y + \frac{C_2}{k}z + \frac{D_2}{k} = 0$, which is $A_1x + B_1y + C_1z + \frac{D_2}{k} = 0$.

Then the distance is $\frac{|D_1 - (D_2/k)|}{\sqrt{A_1^2 + B_1^2 + C_1^2}}$. Substituting $k=A_1/A_2$:

A more general approach that avoids choosing a specific coordinate (like $A_2$) and works directly from the proportional form $\frac{A_1}{A_2} = \frac{B_1}{B_2} = \frac{C_1}{C_2} = k$ is to write Plane 2 as $A_1x + B_1y + C_1z + k D_2 = 0$ by multiplying the second equation by $k$. Then the formula becomes $\frac{|D_1 - kD_2|}{\sqrt{A_1^2 + B_1^2 + C_1^2}}$. This is consistent.

Answer: