Equation of a Plane Passing Through a Given Point and Parallel to Two Given Vectors

A plane is determined by a line and any point that does not lie on the line and vector is represented by a directed line segment. An arrow from the initial point to the terminal point indicates the direction of the vector. In real life, we use planes to measure the circumference, area, and volume.

In this article, we will cover the concept of the Equation of a Plane Passing Through a Given Point and Parallel to Two Given Vectors. 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 ten questions have been asked on this topic in JEE Main from 2013 to 2023 including one in 2019, two in 2020, one in 2021, two in 2022, and three in 2023.

JEE Main 2025: Sample Papers | Mock Tests | PYQs | Study Plan 100 Days

JEE Main 2025: Maths Formulas  | Study Materials

JEE Main 2025: Syllabus | Preparation Guide | High Scoring Topics 

This Story also Contains

1. What is a Plane?
2. Equation of a Plane Passing Through a Given Point and Parallel to Two Given Vectors in Vector Form
3. Equation of a Plane Passing Through a Given Point and Parallel to Two Given Vectors in Cartesian Form
4. Solved Examples on the equation of a Plane Passing Through a Given Point and Parallel to Two Given Vectors
5. Example 1: If the system of equations

What is a Position Vector?

A vector is represented by a directed line segment (an arrow). An arrow from the initial point to the terminal point indicates the direction of the vector.

Let $P$ be any point in space, having coordinates $(x, y, z)$ with respect to the origin $\mathrm{O}(0,0,0)$.

Then, the vector $\overrightarrow{O P}$ having $O$ and $P$ as its initial and terminal points, respectively, is called the position vector of the point $P$ with respect to $O$ .

$OP$ vector can also be expressed as $\overrightarrow{O P}=\vec{r}=x \hat{\mathbf{i}}+y \hat{\mathbf{j}}+z \hat{\mathbf{k}}$
Using the distance formula, the magnitude of $\overrightarrow{O P}$ or $\vec{r}$ is given by

$
|\overrightarrow{\mathrm{OP}}|=\sqrt{x^2+y^2+z^2}
$

Where $\hat{\mathbf{i}}, \hat{\mathbf{j}}$ and $\hat{\mathbf{k}}$ are unit vectors parallel to the positive $X$-axis, $Y$-axis, and $Z$ -axis respectively.

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.

Equation of a Plane Passing Through a Given Point and Parallel to Two Given Vectors in Vector Form

Let a plane pass through point A with a position vector $\overrightarrow{\mathbf{a}}$ and parallel to two vectors $\overrightarrow{\mathbf{b}}$ and $\overrightarrow{\mathbf{c}}$.

Let $\overrightarrow{\mathbf{r}}$ be the position vector of any point $P$ on the plane.

$
\overrightarrow{A P}=\overrightarrow{O P}-\overrightarrow{O A}=\vec{r}-\vec{a}
$

Since $AP$ lies in the plane, hence, $\overrightarrow{\mathbf{r}}-\overrightarrow{\mathbf{a}}, \overrightarrow{\mathbf{b}}$ and $\overrightarrow{\mathbf{c}}$ are coplanar.
We have,
$
\begin{aligned}
(\overrightarrow{\mathbf{r}}-\overrightarrow{\mathbf{a}}) \cdot(\vec{b} \times \overrightarrow{\mathbf{c}}) & =0 \\
(\overrightarrow{\mathbf{r}}) \cdot(\vec{b} \times \overrightarrow{\mathbf{c}}) & =(\overrightarrow{\mathbf{a}}) \cdot(\overrightarrow{\mathbf{b}} \times \overrightarrow{\mathbf{c}}) \\
{\left[\begin{array}{lll}
\overrightarrow{\mathbf{r}} & \vec{b} & \overrightarrow{\mathbf{c}}
\end{array}\right] } & =\left[\begin{array}{lll}
\overrightarrow{\mathbf{a}} & \vec{b} & \overrightarrow{\mathbf{c}}
\end{array}\right]
\end{aligned}
$

Which is the required equation of plane.

Equation of a Plane Passing Through a Given Point and Parallel to Two Given Vectors in Cartesian Form

$
\begin{aligned}
& \text { From }(\overrightarrow{\mathbf{r}}-\overrightarrow{\mathbf{a}}) \cdot(\overrightarrow{\mathbf{b}} \times \overrightarrow{\mathbf{c}})=0 \text { we have, }[\overrightarrow{\mathbf{r}}-\overrightarrow{\mathbf{a}} \overrightarrow{\mathbf{b } } \overrightarrow{\mathbf{c}}] \\
& \Rightarrow \quad\left|\begin{array}{ccc}
x-x_1 & y-y_1 & z-z_1 \\
x_2 & y_2 & z_2 \\
x_3 & y_3 & z_3
\end{array}\right|=0
\end{aligned}
$

Which is the required equation of a plane in the cartesian form where

$
\overrightarrow{\mathbf{b}}=x_2 \hat{\mathbf{i}}+y_2 \hat{\mathbf{j}}+z_2 \hat{\mathbf{k}} \text { and } \overrightarrow{\mathbf{c}}=x_3 \hat{\mathbf{i}}+y_3 \hat{\mathbf{j}}+z_3 \hat{\mathbf{k}}
$

Recommended Videos Based on Equation of a Plane Passing Through a Given Point and Parallel to Two Given Vectors

Solved Examples on the equation of a Plane Passing Through a Given Point and Parallel to Two Given Vectors

Example 1: If the system of equations

$
\begin{aligned}
& x+2 y+3 z=3 \\
& 4 x+3 y-4 z=4
\end{aligned}
$

$8 x+4 y-\lambda z=9+\mu$ has infinitely many solutions, then the ordered pair $(\lambda, \mu)$ is equal to :
[JEE MAINS 2023]

Solution:
Planes are not parallel
$
\begin{aligned}
& \therefore(x+2 y+3 z-3)+a(4 x+3 y-4 z-4) \\
& =8 x+4 y-\lambda z-9-\mu=0 \\
& \frac{1+4 a}{8}=\frac{2+3 a}{4}=\frac{3-4 a}{-\lambda}=\frac{-3-4 a}{-9-\mu}
\end{aligned}
$

$
\begin{aligned}
& (i) 1+4 a=4+6 a \\
& a=\frac{-3}{2}
\end{aligned}
$

$
\begin{aligned}
& \text { (ii) } \frac{2-\frac{9}{2}}{4}=\frac{3+6}{-\lambda} \\
& -\lambda=\frac{36}{-5} \times 2 \\
& \lambda=\frac{72}{5}
\end{aligned}
$
$
\begin{aligned}
& \text { (iii) } \frac{-5}{8}=\frac{-3-4 a}{-9-\mu} \\
& \frac{5}{8}=\frac{3-6}{-9-\mu} \\
& -9-\mu=\frac{-24}{5} \\
& \mu=\frac{-45+24}{5} \\
& \mu=\frac{-21}{5}
\end{aligned}
$

Hence, the answer is $\left(\frac{72}{5},-\frac{21}{5}\right)$

Example 2: Let $P$ be the point of intersection of the line $\frac{x+3}{3}=\frac{y+2}{1}=\frac{1-z}{2}$ and the plane $\mathrm{x}+\mathrm{y}+\mathrm{z}=2$. If the distance of the point $P$ from the plane $3 x-4 y+12 z=32$ is $q$, then $q$ and $2 q$ are the roots of the equation :
[JEE MAINS 2023]

Solution
$
\begin{aligned}
& \frac{x+3}{3}=\frac{y+2}{1}=\frac{1-\mathrm{z}}{2}=\lambda \\
& \mathrm{x}=3 \lambda-3, \mathrm{y}=\lambda-2, \mathrm{z}=1-2 \lambda \\
& \mathrm{P}(3 \lambda-3, \lambda-2,1-2 \lambda) \text { will satisfy the equation of plane } \mathrm{x}+\mathrm{y}+\mathrm{z} \\
& =2 \\
& 3 \lambda-3+\lambda-2+1-2 \lambda=2 \\
& 2 \lambda-4=2 \\
& \lambda=3 \\
& \mathrm{P}(6,1,-5)
\end{aligned}
$

Perpendicular distance of $P$ from plane $3 x-4 y+12 z-32=0$ is

$
\begin{aligned}
& q=\left|\frac{3(6)-4(1)+12(-5)-32}{\sqrt{9+16+144}}\right| \\
& q=6 \\
& 2 q=12
\end{aligned}
$

Sum of roots $=6+12=18$
Product of roots $=6(12)=72$
$\therefore$ A quadratic equation having $q$ and $2 q$ as roots is $x^2-18+72$
Hence, the answer is $x^2+18 x+72=0$

Example 3 : A plane $P$ contains the line of intersection of the plane $\vec{r} \cdot(\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}})=6$ and $\vec{r} \cdot(2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}+4 \hat{\mathrm{k}})=-5$. If P passes through the point $(0,2,-2)$, then the square of distance of the point $(12,12,18)$ from the plane P is
[JEE MAINS 2023]

Solution:
Eqn of plane $\mathrm{P}_1+\lambda \mathrm{P}_2=0$

$
\begin{aligned}
& (x+y+z-6)+\lambda(2 x+3 y+4 z+5)=0 \\
& \text { pass th. }(0,2,-2) \\
& (-6)+\lambda(6-8+5)=0 \\
& (-6)+\lambda[3]=0 \\
\end{aligned}$

Equaiton of plane

$5 x+7 y+9 z+4=0 \\$
distance from
$\begin{aligned}(12,12,18) \\
d=\left|\frac{60+84+162+4}{\sqrt{25+49+81}}\right| \\
d=\frac{310}{\sqrt{155}} \\
d^2=\frac{310 \times 310}{155} \\
d^2=620
\end{aligned}
$

Hence, the answer is 620

Example 4: A vector $\vec{a}$ is parallel to the line of intersection of the plane determined by the vectors $\hat{i}, \hat{i}+\hat{j}$ and the plane determined by the vectors $\hat{\mathrm{i}}-\hat{\mathrm{j}}, \hat{\mathrm{i}}+\hat{\mathrm{k}}$. The obtuse angle between $\vec{a}$ and the vector $\tilde{b}=\hat{i}-2 \hat{j}+2 \hat{k}$ is
[JEE MAINS 2022]

Solution:
$\overrightarrow{n_1}=\mathrm{i} \times(\mathrm{i}+\mathrm{j})=\mathrm{i} \times \mathrm{i}+\mathrm{i} \times \mathrm{j}=0+\mathrm{k}=\mathrm{k}$

$
\begin{aligned}
& \overrightarrow{n_2}=(\mathrm{i}-\mathrm{j}) \times(\mathrm{i}+\mathrm{k}) \\
& =\mathrm{i} \times \mathrm{i}+\mathrm{i} \times \mathrm{k}-\mathrm{j} \times \mathrm{i}-\mathrm{j} \times \mathrm{k} \\
& =-\mathrm{j}+\mathrm{k}-\mathrm{i} \\
& =-\mathrm{i}-\mathrm{j}+\mathrm{k}
\end{aligned}
$

$\therefore \vec{a}$ is parallel to $\vec{n}_1 \times \vec{n}_2$

$
\begin{aligned}
& =\left|\begin{array}{rrr}
i & j & k \\
0 & 0 & 1 \\
-1 & -1 & 1
\end{array}\right| \\
& =\mathrm{i}-\mathrm{j}
\end{aligned}
$

Angle between $\vec{a}$ and $\vec{b}$
$
\begin{aligned}
\cos \theta & =\frac{\tilde{\mathrm{a}} \cdot \vec{b}}{|\vec{a}||\vec{b}|} \\
& =\frac{(\mathrm{i}-\mathrm{j}) \cdot(\mathrm{i}-2 \mathrm{j}+2 \mathrm{k})}{\sqrt{2} \cdot 3} \\
& =\frac{1}{\sqrt{2}}
\end{aligned}
$

$
\therefore \text { obtuse angle }=\frac{3 \pi}{4}
$

Hence, the answer is $\frac{3 \pi}{4}$

Example 5 : If the distance of the point $(1,-2,3)$ from the plane $x+2 y-3 z+10=0$ measured parallel to the line,
$\frac{x-1}{3}=\frac{2-y}{m}=\frac{z+3}{1}$ is $\sqrt{\frac{7}{2}}$, then the value of $|\mathrm{m}|$ is equal to :
[JEE MAINS 2021]

Solution

$\begin{aligned} & \text { DC of line } \equiv\left(\frac{3}{\sqrt{m^2+10}}, \frac{-m}{\sqrt{m^2+10}}, \frac{1}{\sqrt{m^2+10}}\right) \\ & Q \equiv\left(1+\frac{3 r}{\sqrt{m^2+10}},-2+\frac{-m r}{\sqrt{m^2+10}}, 3+\frac{r}{\sqrt{m^2+10}}\right) \\ & \text { Q lies on } x+2 y-3 z+10=0 \\ & 1+\frac{3 r}{\sqrt{m^2+10}}-4-\frac{2 m r}{\sqrt{m^2+10}}-9-\frac{3 r}{\sqrt{m^2+10}}+10=0 \\ & \Rightarrow \frac{r}{\sqrt{m^2+10}}(3-2 m-3)=2 \\ & \Rightarrow \frac{r}{\sqrt{m^2+10}}(-2 m)=2 \\ & r^2 m^2=m^2+10 \\ & \frac{7}{2} m^2=m^2+10 \Rightarrow \frac{5}{2} m^2=10 \Rightarrow m^2=4 \\ & |m|=2\end{aligned}$

Hence, the answer is 2

Summary
The equation of a plane passing through a given Point and Parallel to two given vectors defines the plane's orientation and positional relationship within a three-dimensional coordinate system. It is used in geometry, physics, engineering, and computer graphics for solving problems where precise alignment and positional accuracy are required. Knowledge of planes is necessary to analyze and solve real-life applications.