Equation of Plane Passing Through 3 Non Collinear Points

A plane is also determined by a line and any point that does not lie on the line. Non-collinear points are the points which do not lie on the same line. 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 three non-collinear points. 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 fifteen questions have been asked on this topic in JEE Main from 2013 to 2023 including one in 2019, four in 2021, one in 2022, and eight in 2023.

This Story also Contains

1. What is a Plane?
2. What is a collinear Point?
3. Equation of a plane passing through three non-collinear points in Vector Form
4. Equation of a plane passing through three non-collinear points in Cartesian Form
5. Intercept form of the equation of a plane in Cartesian Form
6. Solved Examples Based on the equation of a plane passing through three non-collinear points

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.

What is a collinear Point?

If the points do not lie on a straight line it is called non-collinear points. We can not draw a straight line joining these points.

Equation of a plane passing through three non-collinear points in Vector Form

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

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

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

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

$
\begin{aligned}
(\vec{r}-\vec{a}) \cdot(\overrightarrow{\mathrm{AB}} \times \overrightarrow{\mathrm{AC}}) & =0 \\
\text { or } \quad(\tilde{\mathbf{r}}-\tilde{\mathbf{a}}) \cdot[(\tilde{\mathbf{b}}-\tilde{\mathbf{a}}) \times(\tilde{\mathbf{c}}-\tilde{\mathbf{a}})] & =0
\end{aligned} \quad(\because \overrightarrow{A R}=(\vec{r}-\vec{a}))
$

This is the equation of the plane in vector form passing through three non-collinear points.

Equation of a plane passing through three non-collinear points in Cartesian Form

Let $\left(x_1, y_1, z_1\right),\left(x_2, y_2, z_2\right)$ and $\left(x_3, y_3, z_3\right)$ be the coordinates of points $A$ , $B$ and $C$ respectively.

Let $P(x, y, z)$ be any point on the plane.
Then, the vectors $\overrightarrow{P A}, \overrightarrow{B A}$ and $\overrightarrow{C A}$ are coplanar.
which is required equation of the plane.

$
\begin{aligned}
& {\left[\begin{array}{lll}
\overrightarrow{\mathrm{PA}} & \overrightarrow{\mathrm{BA}} & \overrightarrow{\mathrm{CA}}
\end{array}\right]=0} \\
& \left|\begin{array}{ccc}
\mathrm{x}-\mathrm{x}_1 & \mathrm{y}-\mathrm{y}_1 & \mathrm{z}-\mathrm{z}_1 \\
\mathrm{x}_2-\mathrm{x}_1 & \mathrm{y}_2-\mathrm{y}_1 & \mathrm{z}_2-\mathrm{z}_1 \\
\mathrm{x}_3-\mathrm{x}_1 & \mathrm{y}_3-\mathrm{y}_1 & \mathrm{z}_3-\mathrm{z}_1
\end{array}\right|=0
\end{aligned}
$

which is required equation of the plane.

Intercept form of the equation of a plane in Cartesian Form

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

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

Derivation of Intercept form of the equation of a plane

Let the plane meet $X, Y$, 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 $\overrightarrow{P A}, \overrightarrow{B A}$ and $\overrightarrow{C A}$ are coplanar. where $P$ is any point in the plane $A B C$.

$
\begin{gathered}
{\left[\begin{array}{lll}
\overrightarrow{\mathrm{PA}} & \overrightarrow{\mathrm{BA}} & \overrightarrow{\mathrm{CA}}
\end{array}\right]=0} \\
\left|\begin{array}{lll}
\mathrm{x}-\mathbf{a} & \mathbf{y}-\mathbf{0} & \mathrm{z}-0 \\
\mathbf{0}-\mathbf{a} & \mathbf{b}-0 & 0-0 \\
0-\mathbf{a} & \mathbf{0}-\mathbf{0} & \mathbf{c}-0
\end{array}\right|=0 \\
\Rightarrow(\mathrm{x}-\mathbf{a}) \mathbf{b c}-\mathbf{y}(-\mathbf{a c})+\mathrm{z}(\mathbf{b a})=0
\end{gathered}
$

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

which is required equation of the plane.

This is the equation of the plane in cartesian form when the plane makes Intercepts a, b, and c on the coordinate axes.

Recommended Video Based on the Equation of a Plane Passing Through Three Non Collinear Points

Solved Examples Based on the equation of a plane passing through three non-collinear points

Example 1: Let the foot perpendicular of the point $\mathrm{P}(3,-2,-9)$ on the plane passing through the points $(-1,-2,-3),(9,3,4),(9,-2,1)$ be $Q(\alpha, \beta, \gamma)$. Then the distance of Q from the origin is
[JEE MAINS 2023]

Solution

Equation of plane through $\mathrm{A}, \mathrm{B}, \mathrm{C}$.

$
\begin{aligned}
& \left|\begin{array}{ccc}
x+1 & y+2 & z+3 \\
10 & 5 & 7 \\
10 & 0 & 4
\end{array}\right|=0 \\
& 2 x+3 y-5 z-7=0
\end{aligned}
$

Foot of $\mathrm{I}^{\mathrm{r}}$ of $\mathrm{P}(3,-2,-9)$ is

$\begin{aligned}
& \frac{x-3}{2}=\frac{y+2}{3}=\frac{z+9}{-5}=-\frac{(6-\phi+45-7)}{4+9+25} \\
& \frac{x-3}{2}=\frac{y+2}{3}=\frac{z+9}{-5}=-1 \\
& \mathrm{Q}(1,-5,-4)=(\alpha, \beta, \gamma) \\
& \mathrm{OQ}=\sqrt{\alpha^2+\beta^2+\gamma^2}=\sqrt{42}
\end{aligned}$

Hence, the answer is $\sqrt{42}$

Example 2: Let the plane $P$ contain the line
$2 x+y-z-3=0=5 x-3 y+4 z+9$ and be parallel to the line $\frac{x+2}{2}=\frac{3-y}{-4}=\frac{z-7}{5}$ Then the distance of the point $A(8,-1,-19)$ from the plane $P$ measured parallel to the line $\frac{x}{-3}=\frac{y-5}{4}=\frac{2-z}{-12}$ is equal to
[JEE MAINS 2023]

Solution
$
\begin{aligned}
& \text { Plane }=P_1=\lambda \cdot P_2=0 \\
& (2 x+y-z-3)+\lambda(5 x-3 y)+4 z+9)=0 \\
& (5 \lambda+2) x+(1-3 \lambda) y+(4 \lambda-1) z+9 \lambda-3=0 \\
& \overrightarrow{\mathrm{n}} \cdot \vec{b}=0 \text { where } \vec{b}(2,4,5) \\
& 2(5 \lambda+2)+4(1-3 \lambda)+5(4 \lambda-1)=0 \\
& \lambda=-\frac{1}{6}
\end{aligned}
$

Plane $7 x+9 y-10 z-27=0$

Equation of line $A B$ is

$
\frac{x-8}{-3}=\frac{y+1}{4}=\frac{x+19}{12}=\lambda
$

Let $B=(8-3 \lambda,-1+4 \lambda,-19+12 \lambda)$ lies on plane $P$

$
\begin{aligned}
& \therefore 7(8-3 \lambda)+9(4 \lambda-1)-10(12 \lambda-19)=27 \\
& \lambda=2 \\
& \therefore \text { Point } B=(2,7,5) \\
& A B=\sqrt{6^2+8^2+24^2}=26
\end{aligned}
$

Hence, the answer is 26

Example 3: If $(1,5,35),(7,5,5),(1, \lambda, 7)$ and $(2 \lambda, 1,2)$ are coplanar, then the sum of all possible values of $\lambda$ is :
[JEE MAINS 2023]

Solution

$\mathrm{A}(1,5,35), \mathrm{B}(7,5,5), \mathrm{C}(1, \lambda, 7), \mathrm{D}(2 \lambda, 1,2)$

$\overrightarrow{\mathrm{AB}}=(7-1) \hat{i}+(5-5) \hat{j}+(5-35) \hat{k}=6 \hat{i} -30\hat{k}$

$\overrightarrow{\mathrm{BC}}=(1-7) \hat{i}+(\lambda-5) \hat{j}+(7-5) \hat{k}=-6 \hat{i}+(\lambda-5) \hat{j} +2\hat{k}$

$\overrightarrow{\mathrm{CD}}=(2 \lambda-1) \hat{i}+(1-\lambda) \hat{j}+(2-7) \hat{k}=(2 \lambda-1) \hat{\mathrm{i}}+(1-\lambda) \hat{\mathrm{j}} -5\hat{k}$

These points are collinear, then

$\begin{aligned} & \Rightarrow 0=\left|\begin{array}{ccc}6 & 0 & -30 \\ -6 & \lambda-5 & 2 \\ 2 \lambda-1 & 1-\lambda & -5\end{array}\right| \\ & =6(-5 \lambda+25-2+2 \lambda)-30\left(-6+6 \lambda-\left(2 \lambda^2-\lambda-10 \lambda+5\right)\right. \\ & =6(-3 \lambda+23)-30\left(-2 \lambda^2+11 \lambda-5-6+6 \lambda\right) \\ & =6(-3 \lambda+23)-30\left(-2 \lambda^2+17 \lambda-11\right) \\ & =6\left(-3 \lambda+23+10 \lambda^2-85 \lambda+55\right) \\ & =6\left(10 \lambda^2-88 \lambda+78\right)=12\left(5 \lambda^2-44 \lambda+39\right) \\ & \Rightarrow 0=12\left(5 \lambda^2-44 \lambda+39\right) \\ & \lambda_1+\lambda_2=\frac{44}{5}\end{aligned}$

Hence, the answer is $\frac{44}{5}$

Example 4: If a plane passes through the points $(-1, k, 0),(2, k,-1),(1,1,2)$ and is parallel to the line $\frac{x-1}{1}=\frac{2 \mathrm{y}+1}{2}=\frac{\mathrm{z}+1}{-1}$, then the value of $\frac{\mathrm{k}^2+1}{(\mathrm{k}-1)(\mathrm{k}-2)}$ is
[JEE MAINS 2023]

Solution:
Equation of plane

$
\begin{aligned}
& \left|\begin{array}{ccc}
x-2 & y-k & 3+1 \\
1-2 & 1-k & 2+1 \\
-1-2 & k-k & 0+1
\end{array}\right|=0 \\
& \Rightarrow\left|\begin{array}{ccc}
x-2 & y-k & 3+1 \\
-1 & 1-k & 3 \\
-3 & 0 & 1
\end{array}\right|=0
\end{aligned}
$

$
\begin{aligned}
& (x-2)(1-k-0)-(y-k)(-1+9)+(3+1)(0+3-3 k)=0 \\
& \Rightarrow(1-\mathrm{k}) \mathrm{x}-8 \mathrm{y}+(3-3 \mathrm{k}) \mathrm{z}-2+2 \mathrm{k}+8 \mathrm{k}+3-3 \mathrm{k}=0 \\
& (1-\mathrm{K}) \mathrm{x}-8 \mathrm{y}+(3-3 \mathrm{~K}) \mathrm{z}+7 \mathrm{~K}+1=0
\end{aligned}
$

The plane is parallel to the line L :

$
\begin{aligned}
& \therefore(1-\mathrm{k}) 1-8.1+(3-3 \mathrm{~K})(-1)=0 \\
& \Rightarrow 1-\mathrm{k}-8-3+3 \mathrm{k}=0 \\
& 2 \mathrm{k}=10 \\
& \mathrm{k}=5
\end{aligned}
$

Now,

$
\frac{k^2+1}{(k-1)(k-2)}=\frac{25+1}{(5-1)(5-2)}=\frac{26}{4 \times 3}=\frac{13}{6}
$

Hence, the answer is $\frac{13}{6}$

Example 5: If the points with position vectors $
\alpha \hat{i}+10 \hat{j}+13 \hat{k}, 6 \hat{i}+11 \hat{j}+11 \hat{k}, \frac{9}{2} \hat{i}+\beta \hat{j}-8 \hat{k}
$ are collinear,
then $(19 \alpha-6 \beta)^2$ is equal to:
[JEE MAINS 2023]

Solution
$
\begin{array}{ll}
(\alpha, 10,13) ;(6,11,11),\left(\frac{9}{2}, \beta,-8\right) & \\
\frac{\alpha-6}{3 / 2}=\frac{-1}{11-\beta}=\frac{2}{19} & -19=22-2 \beta \\
\alpha-6=\frac{3}{19} & 2 \beta=41 \\
\alpha=6+\frac{3}{19}=\frac{117}{19} & \\
\therefore(19 \alpha-6 \beta)^2=(117-123)^2=36 &
\end{array}
$
Hence, the answer is 36

Summary

The equation of a plane passing through three non-collinear points helps us in the representation of a plane in three-dimensional space. It helps us to describe surfaces, calculate distances, and solve problems. Understanding the equation of the plane helps us to analyze and solve problems related to it.