Equation of a Plane Perpendicular to a given Vector & Passing through a Given Point

Given any three points that do not all lie on the same line, there is a unique plane that passes through these points. A plane is also determined by a line and any point that does not lie on the line. There are many ways we can determine the equation of a plane on the basis of different conditions. In real life, we use planes to measure the circumference, area, and volume.

This Story also Contains

1. What is a Plane?
2. Plane perpendicular to a given vector and passing through a given point
3. Equation of a plane perpendicular to a given vector and passing through a given point in Vector Form
4. Equation of a plane perpendicular to a given vector and passing through a given point in Cartesian form
5. Solved Examples Based on Equation of a plane perpendicular to a given vector and passing through a given point

In this article, we will cover the concept of the Equation of a plane perpendicular to a given vector and passing through a given point. This topic falls under the broader category of Three Dimensional Geometry, which is a crucial chapter in Class 12 Mathematics. This is very important not only for board exams but also for competitive exams, which even include the Joint Entrance Examination Main and other entrance exams: SRM Joint Engineering Entrance, BITSAT, WBJEE, and BCECE. A total of nineteen questions have been asked on this topic in JEE Main from 2013 to 2023 including two in 2019, two in 2020, five in 2021, and six in 2023.

What is a Plane?

We know that a line is determined by two points. In other words, for any two distinct points, there is exactly one line that passes through those points, whether in two dimensions or three. Similarly, given any three points that do not all lie on the same line, there is a unique plane that passes through these points. Just as a line is determined by two points, a plane is determined by three.

This may be the simplest way to characterize a plane, but we can use other descriptions as well. For example, given two distinct, intersecting lines, there is exactly one plane containing both lines. A plane is also determined by a line and any point that does not lie on the line.

Plane perpendicular to a given vector and passing through a given point

Imagine a pair of orthogonal vectors that share an initial point. Visualize grabbing one of the vectors and twisting it. As you twist, the other vector spins around and sweeps out a plane. The equation of Plane describes this concept mathematically.

Equation of a plane perpendicular to a given vector and passing through a given point in Vector Form

Let $\overrightarrow{\mathbf{n}}=a \hat{\mathbf{i}}+b \hat{\mathbf{j}}+c \hat{\mathbf{k}}$ be a vector and $\mathrm{P}\left(\mathrm{x}_0, \mathrm{y}_0, \mathrm{z}_0\right)$ be a point. Then the set of all points $Q(\mathrm{x}, \mathrm{y}, \mathrm{z})$ such that $\overrightarrow{P Q}$ orthogonal to $\overrightarrow{\mathbf{n}}$ forms a plane.

We say that $\overrightarrow{\mathbf{n}}$ is a normal vector, or perpendicular to the plane.
Remember, the dot product of orthogonal vectors is zero. This fact generates the vector equation of a plane:
$\overrightarrow{\mathbf{n}} \cdot \overrightarrow{P Q}=0$
If the position vector of point $P$ is $\overrightarrow{\mathbf{P}}$ and the plosition vector of point $Q$ is $\overrightarrow{\mathbf{q}}$ , then

$(\overrightarrow{\mathbf{q}}-\overrightarrow{\mathbf{p}}) \cdot \overrightarrow{\mathbf{n}}=\mathbf{0} \quad($ As $\overrightarrow{P Q}=\overrightarrow{\mathbf{q}}-\overrightarrow{\mathbf{p}})$

This is the vector equation of the plane.

Equation of a plane perpendicular to a given vector and passing through a given point in Cartesian form

Position vector of point P and point Q is $\overrightarrow{\mathbf{p}}=x_0 \hat{i}+y_0 \hat{j}+z_0 \hat{k}$ and $\overrightarrow{\mathbf{q}}=x \hat{i}+y \hat{j}+z \hat{k}$ respectively and vector $\overrightarrow{\mathbf{n}}$ is $a \hat{i}+b \hat{j}+c \hat{k}$ Then,

$
\begin{array}{lc}
& (\overrightarrow{\mathbf{q}}-\overrightarrow{\mathbf{p}}) \cdot \overrightarrow{\mathbf{n}}=\mathbf{0} \\
\Rightarrow & \left((x \hat{i}+y \hat{j}+z \hat{k})-\left(x_0 \hat{i}+y_0 \hat{j}+z_0 \hat{k}\right)\right) \cdot(a \hat{i}+b \hat{j}+c \hat{k})=0 \\
\Rightarrow & {\left[\left(x-x_0\right) \hat{i}+\left(y-y_0\right) \hat{j}+\left(z-z_0\right) \hat{k}\right] \cdot(\mathrm{a} \hat{i}+\mathrm{b} \hat{j}+c \hat{k})=0} \\
\text { i.e. } & \mathbf{a}\left(x-x_0\right)+\mathbf{b}\left(y-y_0\right)+\mathbf{c}\left(z-z_0\right)=\mathbf{0}
\end{array}
$

Thus, the coefficients of x, y, and z in the cartesian equation of a plane are the direction ratios of the normal to the plane.

Recommended Video Based on Equation of a Plane Perpendicular to a Given Vector and Passing Through a Given Point

Solved Examples Based on Equation of a plane perpendicular to a given vector and passing through a given point

Example 1: Let $P$ be the plane, passing through the point $(1,-1,-5)$ and perpendicular to the line joining the points $(4,1,-3)$ and $(2,4,3)$ Then the distance of $P$ from the point $(3,-2,2)$ is

Solution: Let $\mathrm{A}(4,1,-3) \& \mathrm{~B}(2,4,3)$

$
\overrightarrow{\mathrm{n}}=\overrightarrow{\mathrm{AB}}=(-2,3,6)
$

Plane P is
$
\begin{aligned}
& -2(\mathrm{x}-1)+3(\mathrm{y}+1)+6(\mathrm{z}+5)=0 \\
& -2 \mathrm{x}+2+3 \mathrm{y}+3+6 \mathrm{z}+30=0 \\
& 2 \mathrm{x}-3 \mathrm{y}-6 \mathrm{z}=35
\end{aligned}
$

Distance from the point $(3,-2,2)$ is

$
\begin{aligned}
& =\frac{|6+6-12-35|}{\sqrt{2^2+3^2+6^2}} \\
& =\frac{35}{7}=5
\end{aligned}
$

Hence, the answer is 5

Example 2: The plane, passing through the points $(0,-1,2)$ and $(-1,2,1)$ and parallel to the line passing through $(5,1,-7)$ and $(1,-1,-1)$, also passes through the point:

Solution: The plane passing through $(0,-1,0)$ and $(-1,2,1)$ The vector in the plane $\langle-1,3,-1\rangle$ vector parallel to the plane is $\langle 4,2,-6\rangle$

Normal vector to plane $(\vec{n})=\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ -1 & 3 & -1 \\ 4 & 2 & -6\end{array}\right|$

$
\begin{aligned}
& =\hat{\mathrm{i}}(16)-\hat{\mathrm{j}}(10)+\hat{\mathrm{k}}(-14) \\
& \overrightarrow{\mathrm{n}}=\langle 8,5,7\rangle
\end{aligned}
$

Equation of plane

$
\begin{aligned}
& 8(x-0)+5(y+1)+7(z-2)=0 \\
& \Rightarrow 8 x+5 y+7 z=9
\end{aligned}
$

From the given options point $(-2,5,0)$ lies on the plane.
Hence, the answer is $(-2,5,0)$

Example 3: If the equation of the plane containing the line $x+2 y+3 z-4=0$ $2 x+y-z+5$ and perpendicular to the plane $\overrightarrow{\mathrm{r}}=(\hat{\mathrm{i}}-\hat{\mathrm{j}})+\lambda(\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}})+\mu(\hat{\mathrm{i}}-2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}})$ is $a x+b y+c z=4$, then $(a-b+c)$ is equal to:

Solution:
D.R's of line $\vec{n}_1=-5 \hat{\mathbf{i}}+7 \hat{\mathbf{j}}-3 \hat{\mathrm{k}}$
D. R's of normal of the second plane

$
\begin{aligned}
& \overrightarrow{\mathrm{n}}_2=5 \hat{i}-2 \hat{j}-3 \hat{k} \\
& \overrightarrow{\mathrm{n}}_1 \times \overrightarrow{\mathrm{n}}_2=-27 \hat{i}-30 \hat{j}-25 \hat{k} \\
& \text { A point on the required plane is }\left(0,-\frac{11}{5}, \frac{14}{5}\right)
\end{aligned}
$

The equation of the required plane is

$
\begin{aligned}
& 27 \mathrm{x}+30 \mathrm{y}+25 \mathrm{z}=4 \\
& \therefore \mathrm{a}-\mathrm{b}+\mathrm{c}=22
\end{aligned}
$

Hence, the answer is 22

Example 4: If the equation of the plane passing through the point $(1,1,2)$ and perpendicular to the line
$x-3 y+2 z-1=0=4 x-y+z_{\text {is }} A x+B y+C z=1$ then $140(C-B+A)$ is equal to

Solution: give line is $x-3 y+2 z-1=0=4 x-y+z$
Direction of line $\vec{a}=\left|\begin{array}{ccc}\hat{\mathrm{i}} & \hat{\mathrm{j}} & \mathrm{k} \\ 1 & -3 & 2 \\ 4 & -1 & 1\end{array}\right|=\hat{\mathrm{i}}(-1)-\hat{j}(-7)+k(11)$ $\Rightarrow \quad \overrightarrow{\mathrm{a}}=\langle-1,7,11\rangle$

$\therefore$ Line is $\perp^{\mathrm{I}}$ to the plane then the direction of the line is parallel to the normal of the plane.

$
\overrightarrow{\mathrm{n}}=\langle-1,7,11\rangle
$

The equation of the plane is

$\begin{aligned} & -1(x-1)+7(y-1)+11(z-2)=0 \\ & -x+7 y+11 z+1-7-22=0 \\ & \Rightarrow \quad-x+7 y+11 z=28 \\ & \Rightarrow \quad-\frac{1}{28} x+\frac{7}{28} y+\frac{11}{28} \mathrm{z}=1 \\ & A=-\frac{1}{28}, \mathrm{~B}=\frac{7}{28}, \mathrm{C}=\frac{11}{28} \\ & 140(\mathrm{C}-\mathrm{B}+\mathrm{A})=140\left(\frac{11}{28}-\frac{7}{28}-\frac{1}{28}\right) \\ & =\frac{140 \times 3}{28}=15\end{aligned}$

Hence, the answer is 15.