Equation of a Plane In the Normal Form

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. 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 In The Normal Form. 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, one in 2021, one in 2022, and one in 2023.

Equation of a plane in normal form

The equation of a plane in a cartesian coordinate system can be computed through different methods based on the available input values about the plane. One of the methods to represent the plane is normal form.

Vector form

The vector equation of a plane normal to unit vector $\hat{\mathbf{n}}$ and at a distance d from the origin is $\overrightarrow{\mathbf{r}} \cdot \hat{\mathbf{n}}=d$.

Derivation of Vector form

Let $O$ be the origin and let $(d \neq 0)$ be the length of the perpendicular from the origin $(O)$ to the given plane.

If $\overrightarrow{\mathrm{ON}}$ is normal from the origin to the plane, and $\hat{\mathbf{n}}$ is the unit normal vector along $\overrightarrow{O N}$. Then $\overrightarrow{O N}=d \hat{\mathbf{n}}$.

Let $P$ be any point on the plane with position vector $\overrightarrow{\mathbf{r}}$, so that $\overrightarrow{\mathrm{OP}}=\overrightarrow{\mathbf{r}}$.

Now, $\overrightarrow{\mathrm{NP}}$ is perpendicular to $\overrightarrow{\mathrm{ON}}$

$
\begin{array}{rrl}
\therefore & \overrightarrow{\mathrm{NP}} \cdot \overrightarrow{\mathrm{ON}}=0 & \\
\Rightarrow & (\overrightarrow{\mathrm{OP}}-\overrightarrow{\mathrm{ON}}) \cdot \overrightarrow{\mathrm{ON}}=0 & (\text { as } \overrightarrow{\mathrm{ON}}+\overrightarrow{\mathrm{NP}}=\overrightarrow{\mathrm{OP}}) \\ \Rightarrow & (\overrightarrow{\mathbf{r}}-d \hat{\mathbf{n}}) \cdot d \hat{\mathbf{n}}=0 \\
\Rightarrow & \overrightarrow{\mathbf{r}} \cdot d \hat{\mathbf{n}}-d^2 \hat{\mathbf{n}} \cdot \hat{\mathbf{n}}=0 & \\
\Rightarrow & d \overrightarrow{\mathbf{r}} \cdot \hat{\mathbf{n}}-d^2|\hat{\mathbf{n}}|^2=0 & (\because d \neq 0) \\
\Rightarrow & \overrightarrow{\mathbf{r}} \cdot \hat{\mathbf{n}}-d=0 & \left(\because|\hat{\mathbf{n}}|^2=1\right) \\
\Rightarrow & \overrightarrow{\mathbf{r}} \cdot \hat{\mathbf{n}}=d &
\end{array}
$

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

Cartesian form

If $\mathrm{P}(\mathrm{x}, \mathrm{y}, \mathrm{z})$ is any point in the plane and $\hat{\mathbf{n}}$ is the unit vector normal to the plane.

Let $\mathrm{l}, \mathrm{m}, \mathrm{n}$ be the direction cosines. Then Cartesian equation of the plane in the normal form is given by

$\begin{array}{r}(x \hat{i}+y \hat{j}+z \hat{k}) \cdot({l} \hat{i}+m \hat{j}+n \hat{k})=d \\ \mathbf{l x}+\mathbf{m y}+\mathbf{n z}=\mathbf{d}\end{array}$

Derivation of the Cartesian form

$\overrightarrow{\mathbf{r}} \cdot \hat{\mathbf{n}}=d$ is the vector form of the equation of a plane in the normal form where $\hat{\mathbf{n}}$ is the unit vector normal to the plane.

Let $\mathrm{P}(\mathrm{x}, \mathrm{y}, \mathrm{z})$ is any point in the plane.
Then, $\overrightarrow{O P}=\overrightarrow{\mathbf{r}}=x \hat{\mathbf{i}}+y \hat{\mathbf{j}}+z \hat{\mathbf{k}}$
Let $\mathrm{l}, \mathrm{m}, \mathrm{n}$ be the direction cosines of Then

$
\hat{\mathbf{r}}=l \hat{\mathbf{i}}+m \hat{\mathbf{j}}+z \hat{\mathbf{k}}
$

Therefore, the equation $\overrightarrow{\mathbf{r}} \cdot \hat{\mathbf{n}}=d$ gives

$
\begin{aligned}
& (x \hat{i}+y \hat{j}+z \hat{k}) \cdot(\hat{i}+m \hat{j}+n \hat{k})=d \\
& \mathbf{l x}+\mathbf{m y}+\mathbf{n z}=\mathrm{d}
\end{aligned}
$

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

NOTE:
If $\vec{r} \cdot(a \hat{i}+b \hat{j}+c \hat{k})=d$ is the vector equation of a plane, then ax + by $+c z=d$ is the Cartesian equation of the plane, where $a, b$, and $c$ are the direction ratios of the normal to the plane.

Solved Examples Based on the Equation of A Plane In The Normal Form

Example 1: Let $\alpha x+\beta y+y z=1$ be the equation of a plane through the point $(3,-2,5)$ and perpendicular to the line joining the points $(1,2,3)$ and $(-2,3,5)$. Then the value of $\alpha \beta y$ is equal to :

1) 6

2) 8

3) 11

4) 15

Solution:
The plane is perpendicular to the line joining the

$
(1,2,3) \&(-2,3,5)
$

So, the line will be along the normal of plane.

$
\vec{n}=(3,-1,-2) \text { or }(-3,1,2)
$

Compare it with eq. of the plane, $\alpha x+\beta y+r_2=1$

$
\alpha=3, \beta=-1, r=-2 \text { or } \alpha=-3, \beta=1, r=2
$
So, $\alpha \beta r=6$

Hence, the answer is 6 .

Example 2: Let $\frac{\mathrm{x}-2}{3}=\frac{\mathrm{y}+1}{-2}=\frac{\mathrm{z}+3}{-1}$ lie on the plane $\mathrm{px}-\mathrm{qy}+\mathrm{z}=5$ for some $\mathrm{p}, \mathrm{q} \in \mathbb{R}$. The shortest distance of the plane from the origin is:

1) $\sqrt{\frac{3}{109}}$
2) $\sqrt{\frac{5}{142}}$

3) $\frac{5}{\sqrt{71}}$

4) $\frac{1}{\sqrt{142}}$

Solution

For the line to lie on the plane, the point $\mathrm{A}(2,-1,-3)$ lying on the line should lie on the plane as well

$
\therefore \quad 2 p+q-3=5 \Rightarrow 2 p+q=8 \ldots (i)
$

Also vector parallel to the line $(3 \mathrm{i}-2 \mathrm{j}-\mathrm{k})$ should be perpendicular to a normal vector $\vec{n}$ of the plane $(\mathrm{pi}-\mathrm{qj}+\mathrm{k})$

$
\begin{aligned}
& \therefore(3 i-2 j-k) \cdot(p i-q j+k)=0 \\
& \Rightarrow 3 p+2 q-1=0 \\
& \Rightarrow 3 p+2 q=1 \quad \ldots (ii)
\end{aligned}
$

From (i) and (ii)

$
\mathrm{p}=15, \mathrm{q}=-22
$

$\therefore$ plane is $15 \mathrm{p}+22 \mathrm{y}+\mathrm{z}-5=0$
Distance from origin $(0,0,0)$

$
=\frac{|-5|}{\sqrt{15^2+22^2+1^2}}
$
$\begin{aligned} & =\frac{5}{\sqrt{710}} \\ & =\sqrt{\frac{5}{142}}\end{aligned}$

Hence, the answer is option 2

Example 3: Let the plane $a x+b y+c z+d=0$ bisect the line joining the points $(4,-3,1)$ and $(2,3,-5)$ at the right angles. If $a, b, c, d$ are integers, then the minimum value of $\left(a^2+b^2+c^2+d^2\right)$ is__________.

1) $16$

2) $12$

3) $32$

4) $28$

Solution

D.R's of line $\left(a_1, b_1, c_1\right)=(2-4,3+3,-5-1)=(-2,6,-6)$ and line $\perp$ to plane

$
\begin{aligned}
& \frac{a}{a_1}=\frac{b}{b_1}=\frac{c}{c_1}=k \\
& a=-2 k, b=6 k, c=-6 k
\end{aligned}
$

again midpoint of $(4,-3,1)$ and $(2,3,-5)$ is $(3,0,-2)$ lies on the plane

$
\begin{aligned}
& \therefore 3 a+0-2 c+d=0 \\
& \Rightarrow-6 \mathrm{k}+12 \mathrm{k}+\mathrm{d}=0 \\
& \Rightarrow \mathrm{d}=-6 \mathrm{k} \\
& a^2+b^2+c^2+d^2=112 k^2
\end{aligned}
$

$\left(a^2+b^2+c^2+d^2\right)_{\min }$ for $k=\frac{1}{2}(a, b, c, d$ are integer $)$

$
\left(a^2+b^2+c^2+d^2\right)_{\min }=28
$

Hence, the answer is option 4.

Example 4: The plane containing the line $\frac{x-3}{2}=\frac{y+2}{-1}=\frac{z-1}{3}$ and also containing its projection on the plane $2 x+3 y-z=5$, contains which one of the following points?

1) $(0,-2,2)$

2) $(-2,2,2)$

3) $(2,0,-2)$

4) $(2,2,0)$

Solution:
A plane passing through a point and a line (vector form) -

Let the plane pass through $A(\vec{a})$ and a line $\vec{r}=\vec{b}+\lambda \vec{c}$, then the plane is given by

$
\left[\begin{array}{rl}
r & b & c
\end{array}\right]+\left[\begin{array}{rcc}
r & c & a
\end{array}\right]=\left[\begin{array}{lll}
a & b & c
\end{array}\right]
$

- wherein

$
\begin{aligned}
& \vec{n}=(\vec{b}-\vec{a}) \times(\vec{c}) \\
& (\vec{r}-\vec{a}) \cdot(\vec{b}-\vec{a}) \times(\vec{c})=0
\end{aligned}
$

The normal vector of the plane is

$
(2 \hat{i}-\hat{j}+3 \hat{k}) \times(2 \hat{i}+3 \hat{j}-\hat{k})=8(-\hat{i}+\hat{j}+\hat{k})
$

Hence, the equation of the plane is $-(x-3)+(y+2)+(z-1)=0-x+3+y+2+z-1=0-x+y+z+4=0$

(2,0,-2) satisfies the plane

Hence, the answer is the option (3)

Example 5 : If the point $(2, \alpha, \beta)$ lies on the plane which passes through the points $(3,4,2)$ and $(7,0,6)$ and is perpendicular to the plane $2 x-5 y=15$, then $2 \alpha-3 \beta$ is equal to :

1) $17$

2) $5$

3) $7$

4) $12$

Solution:

Cartesian equation of plane passing through a given point and normal to a given vector -

$
\left(x-x_0\right) a+\left(y-y_0\right) b+\left(z-z_0\right) c=0
$

- wherein

$
\begin{aligned}
& \vec{r}=x \hat{i}+y \hat{j}+z \hat{k} \\
& \vec{a}=x_0 \hat{i}+y_0 \hat{j}+z_0 \hat{k} \\
& \vec{n}=a \hat{i}+b \hat{j}+c \hat{k}
\end{aligned}
$

Putting in

$
(\vec{r}-\vec{a}) \cdot \vec{n}=0
$

We get $\left(x-x_0\right) a+\left(y-y_0\right) b+\left(z-z_0\right) c=0$

Conversion of equation in normal form (vector form ) -
The equation $a x+b y+c z+D=0$
is converted in normal by
(i) $a x+b y+c z=-D$
(ii) Making RHS position

$
-a x-b y-c z=D
$

(iii) Dividing by $\sqrt{a^2+b^2+c^2}$
(iv) $\frac{-a x-b y-c z}{\sqrt{a^2+b^2+c^2}}=\frac{D}{\sqrt{a^2+b^2+c^2}}$

We get $l x+m y+n z=d$

Normal Vector of plane $=\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ 2 & -5 & 0 \\ 4 & -4 & 4\end{array}\right|=-4(5 \hat{i}+2 \hat{j}-3 \hat{k})$
Equation of plane is $5(x-7)+2 y-3(z-6)=0$

$
\begin{aligned}
& =>5 x+2 y-3 z=17 \\
& =>2 \alpha-3 \beta=17-(5 \times 2)=7
\end{aligned}
$

Hence, the answer is option (3).

Summary

The equation of plane is an important concept in three-dimensional geometry defining a plane as a flat, infinite surface extending in all directions. The equation of plane tells us about the algebraic and geometrical properties. Knowledge of planes is necessary to analyze and solve real-life applications.