JCECE 2025 - Date, Registration, Seat Allotment, Choice Filling, Merit List, Counselling, Cutoff

Equation of a plane passing through three non collinear point - Practice Questions & MCQ

Edited By admin | Updated on Sep 18, 2023 18:34 AM | #JEE Main

Syllabus

Quick Facts

18 Questions around this concept.

Solve by difficulty

MEDIUM

Let $P$ be the plane passing through the points (5, 3, 0), (13, 3, -2), and (1, 6, 2). For $\alpha \in N$ , if the distances of the points $\mathrm{A}(3,4, \alpha)\; and \; \mathrm{B}(2, \alpha,a)$ from the plane $\mathrm{P}$ are 2 and 3 respectively, then the positive value of a is:

View Solution

Concepts Covered - 1

Equation of a plane passing through three non collinear point

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

The vectors, $\overrightarrow{AB}=\vec{\mathbf b}-\vec{\mathbf a}$ and $\overrightarrow{AC}=\vec{\mathbf c}-\vec{\mathbf a}$ are in the given plane. Therefore, the vector $\overrightarrow{AB}\times \overrightarrow{AC}$ is perpendicular to the plane containing points A, B and C.

Let P be any point in the plane with position vector $\vec{\mathbf r}$ .

Therefore, the equation of the plane passing through OP and perpendicular to the vector $\overrightarrow{AB}\times \overrightarrow{AC}$ is

$\\\mathrm{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;}(\vec{r}-\vec{a}) \cdot(\overrightarrow{\mathrm{AB}} \times \overrightarrow{\mathrm{AC}})=0\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\left ( \because \overrightarrow{AR}=(\vec{r}-\vec{a}) \right )\\\text{or}\;\;\;\;\;\;\;\;\mathbf{(\vec{r}-\vec{a}) \cdot[(\vec{b}-\vec{a}) \times(\vec{c}-\vec{a})]=0}$

This is the equation of the plane in vector form passing through three noncollinear points.

Cartesian Form

Let (x₁, y₁, z₁), (x₂, y₂, z₂) and (x₃, y₃, z₃) be the coordinates of points A, B and C respectively.

Let P(x, y, z) be any point on the plane.

Then, the vectors $\overrightarrow{PA},\;\overrightarrow{BA}$ and $\overrightarrow{CA}$ are coplanar.
$\\\mathrm{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;}\left [ \overrightarrow{\mathbf{PA}}\;\;\overrightarrow{\mathbf{BA}}\;\;\overrightarrow{\mathbf{CA}} \right ]=0\\\\\mathrm{\;\;\;}{\begin{vmatrix} \mathbf{x-x_1 }&\mathbf{y-y_1} &\mathbf{z-z_1} \\ \mathbf{x_2-x_1} &\mathbf{y_2-y_1} &\mathbf{z_2-z_1} \\ \mathbf{x_3-x_1} &\mathbf{y_3-y_1} &\mathbf{z_3-z_1} \end{vmatrix}=0}\\\\\text{which is required equation of the plane .}$

Intercept form of the equation of a plane

Cartesian Form

The equation of a plane having intercepting lengths a, b and c with X-axis, Y-axis and Z-axis, respectively is

$\mathbf{\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1}$

Let the plane meets X, Y and Z-axes at (a, 0, 0), (0, b, 0), (0, 0, c) respectively and P(x, y, z) be any point on the plane.

Since these three points are non-collinear points.

Then, the vectors $\overrightarrow{PA},\;\overrightarrow{BA}$ and $\overrightarrow{CA}$ are coplanar. where P be any point in the plane ABC.

$\\\mathrm{\;\;\;\;\;\;\;\;\;\;\;}\left [ \overrightarrow{\mathbf{PA}}\;\;\overrightarrow{\mathbf{BA}}\;\;\overrightarrow{\mathbf{CA}} \right ]=0\\\\\mathrm{\;\;\;}{\begin{vmatrix} \mathbf{x-a }&\mathbf{y-0} &\mathbf{z-0} \\ \mathbf{0-a} &\mathbf{b-0} &\mathbf{0-0} \\ \mathbf{0-a} &\mathbf{0-0} &\mathbf{c-0} \end{vmatrix}=0}\\\\\Rightarrow\mathbf{(x-a)bc-y(-ac)+z(ba)=0}$