Perpendicular Distance of a Point From a Plane

The distance of a point from a plane is the shortest distance between a point and a plane. We can draw an infinite number of lines from a point but we have to draw the shortest line. The perpendicular distance between a point and a plane is the shortest distance between them. In real life, we use the distance of a point from a plane to find the shortest distance to reach a road or a place, in navigation and computational geometry.

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This Story also Contains

1. What is the Distance of a Point From a Plane?
2. Perpendicular Distance in Vector Form: Formula
3. Derivation of the formula of Perpendicular Distance in Vector Form
4. Perpendicular Distance in Cartesian Form: Formula
5. Distance Between The Parallel Planes: Formula
6. Derivation of Distance Between The Parallel Planes
7. Solved Examples Based on the Distance of a Point From a Plane

In this article, we will cover the concept of the Distance of a Point From a Plane. 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 twenty-three questions have been asked on this topic in JEE Main from 2013 to 2023 including one in 2015, one in 2016, one in 2017, two in 2019, four in 2020, five in 2021, and seven in 2023.

What is the Distance of a Point From a Plane?

The distance of a point from a plane is the shortest distance between them. Generally, the perpendicular distance between a point and a plane is the shortest distance between them.

Perpendicular Distance in Vector Form: Formula

The perpendicular distance (D) from a point having a position vector $\overrightarrow{\mathbf{a}}$ to the plane $\overrightarrow{\mathbf{r}} \cdot \overrightarrow{\mathbf{n}}=d$ is given by

$
\mathbf{D}=\frac{|\overrightarrow{\mathbf{a}} \cdot \overrightarrow{\mathbf{n}}-d|}{|\overrightarrow{\mathbf{n}}|}
$

Derivation of the formula of Perpendicular Distance in Vector Form

Consider a point $P$ with a position vector $\overrightarrow{\mathbf{a}}$ and a plane $\pi_1$ whose equation is $\overrightarrow{\mathbf{r}} \cdot \overrightarrow{\mathbf{n}}=d$

Let R be a point in the plane such that $\overrightarrow{P R}$ is orthogonal to the plane $\pi_1$. since line PR passes through $\mathrm{P}(\mathrm{a})$ and is parallel to the vector $\overrightarrow{\mathbf{n}}$ which is normal to the plane $\pi_1$. So, the vector equation of line $P R$ is

$
\overrightarrow{\mathbf{r}}=\overrightarrow{\mathbf{a}}+\lambda \overrightarrow{\mathbf{n}}
$ . . . . . (i)

Point $R$ is the intersection of Eq. (i) and the given plane $\pi_1$.

$
\begin{array}{lc}
\therefore & (\overrightarrow{\mathbf{a}}+\lambda \overrightarrow{\mathbf{n}}) \cdot \overrightarrow{\mathbf{n}}=d \\
\Rightarrow & \overrightarrow{\mathbf{a}} \cdot \overrightarrow{\mathbf{n}}+\lambda \overrightarrow{\mathbf{n}} \cdot \overrightarrow{\mathbf{n}}=d \\
\Rightarrow & \lambda=\frac{d-\overrightarrow{\mathbf{a}} \cdot \overrightarrow{\mathbf{n}}}{|\overrightarrow{\mathbf{n}}|^2}
\end{array}
$

On putting the value of $\lambda$ in Eq. (i), we obtain the position vector of $R$ given by

$\begin{aligned} \overrightarrow{\mathbf{r}} & =\overrightarrow{\mathbf{a}}+\left(\frac{d-\overrightarrow{\mathbf{a}} \cdot \overrightarrow{\mathbf{n}}}{|\overrightarrow{\mathbf{n}}|^2}\right) \overrightarrow{\mathbf{n}} \\ \overrightarrow{\mathbf{P R}} & =\text { Position vector of } R-\text { Position vector of } P \\ & =\overrightarrow{\mathbf{a}}+\left(\frac{d-\overrightarrow{\mathbf{a}} \cdot \mathbf{n}}{|\overrightarrow{\mathbf{n}}|^2}\right) \overrightarrow{\mathbf{n}}-\overrightarrow{\mathbf{a}} \\ \overrightarrow{\mathbf{P R}} & =\left(\frac{d-\overrightarrow{\mathbf{a}} \cdot \overrightarrow{\mathbf{n}}}{|\overrightarrow{\mathbf{n}}|^2}\right) \overrightarrow{\mathbf{n}} \\ \Rightarrow \quad|\overrightarrow{\mathbf{P R}}| & =\left|\left(\frac{d-\overrightarrow{\mathbf{a}} \cdot \overrightarrow{\mathbf{n}}}{|\overrightarrow{\mathbf{n}}|^2}\right) \overrightarrow{\mathbf{n}}\right| \\ \Rightarrow \quad|\overrightarrow{\mathbf{P R}}| & =\frac{|d-(\overrightarrow{\mathbf{a}} \cdot \overrightarrow{\mathbf{n}})|}{|\overrightarrow{\mathbf{n}}|} \\ \text { or } \quad D & =\frac{|(\overrightarrow{\mathbf{a}} \cdot \overrightarrow{\mathbf{n}})-d|}{|\overrightarrow{\mathbf{n}}|}\end{aligned}$

Perpendicular Distance in Cartesian Form: Formula

Let $P\left(x_1, y_1, z_1\right)$ be the given point with position vector $\overrightarrow{\mathbf{a}}$ and $a x+b y+c z+d=0$ be the Cartesian equation of the given plane. Then

$
\begin{aligned}
\vec{a} & =x_1 \hat{i}+y_1 \hat{j}+z_1 \hat{k} \\
\vec{n} & =\mathrm{a} \hat{i}+\mathrm{b} j+\mathrm{c} \hat{k}
\end{aligned}
$

Hence, from Vector form of the perpendicular from $P$ to the plane is

$
\left|\frac{\left(x_1 \hat{i}+y_1 \hat{j}+z_1 \hat{k}\right) \cdot(a \hat{i}+b \hat{j}+c \hat{k})-(-d)}{\sqrt{a^2+b^2+c^2}}\right|=\left|\frac{a x_1+b y_1+c z_1+d}{\sqrt{a^2+b^2+c^2}}\right|
$

Distance Between The Parallel Planes: Formula

The distance between the two parallel planes $\mathrm{ax}+\mathrm{by}+\mathrm{cz}+\mathrm{d}_1=0$ and ax + by $+c z+d_2=0$ is given by

$
D=\left|\frac{\left(d_2-d_1\right)}{\sqrt{a^2+b^2+c^2}}\right|
$

Derivation of Distance Between The Parallel Planes

Let $\mathrm{P}\left(\mathrm{x}_1, \mathrm{y}_1, \mathrm{z}_1\right)$ be any point in the plane $\mathrm{ax}+$ by $+\mathrm{cz}+\mathrm{d}_1=0$.
Then the distance of the point $P$ from plane $a x+b y+c z+d_2=0$ is

$
D=\left|\frac{a x_1+b y_1+c z_1+d_2}{\sqrt{a^2+b^2+c^2}}\right|
$

Also,

$
\begin{array}{ll}
& a x_1+b y 1+c z_1+d_1=0 \\
\Rightarrow & \mathbf{D}=\left|\frac{\left(\mathbf{d}_2-\mathbf{d}_1\right)}{\sqrt{\mathbf{a}^2+\mathbf{b}^2+\mathbf{c}^2}}\right|
\end{array}
$

Recommended Video Based on the Distance of the Point from a Plane

Solved Examples Based on the Distance of a Point From a Plane

Example 1: Let the plane $\mathrm{P}: 8 \mathrm{x}+\alpha_1 \mathrm{y}+\alpha_2 \mathrm{z}+12=0$ be parallel to the line $\mathrm{L}: \frac{\mathrm{x}+2}{2}=\frac{\mathrm{y}-3}{3}=\frac{\mathrm{z}+4}{5}$. If the intercept of P on the $\mathrm{y}-\mathrm{axis}$ is 1 , then the distance between P and L is:
[JEE MAINS 2023]

Solution:
$\mathrm{y}-$ intercept $=\frac{-12}{\alpha_1}=1$

$
\begin{aligned}
& \alpha_1=-12 \& \overrightarrow{\mathrm{n}}=\left(8, \alpha_1, \alpha_2\right) \\
& \vec{\ell}=(2,3,5)
\end{aligned}
$

$\overrightarrow{\mathrm{n}} \vec{\ell}=0(\therefore$ plane $\mathrm{P} \&$ line $L$ are parallel $)$

$
\begin{aligned}
& 16+3 \alpha_1+5 \alpha_2=0 \\
& 16-36+5 \alpha_2=0 \\
& 5 \alpha_2=20 \\
& \alpha_2=4
\end{aligned}
$
$
\begin{aligned}
& 8 x-12 y+4 z+12=0 \\
& \Rightarrow 2 x-3 y+z+3=0
\end{aligned}
$

$(-2,3,-4)$ is a point on the line L distance between the point $(-2,3,-4)$ and the plane P is:-

$
d=\frac{|-4-9-4+3|}{\sqrt{2^2+3^2+1^2}}
$
$
=\frac{14}{\sqrt{14}}=\sqrt{14}
$
Hence, the answer is $\sqrt{14}$

Example 2: Let the plane P pass through the intersection of the planes $2 x+3 y-z=2$ and $x+2 y+3 z=6$ and be perpendicular to the plane $2 x+y-z=0$. If d is the distance of P from the point $(-7,1,1$,$) then d^2 is equal to :
[JEE MAINS 2023]

Solution:
Plane P is passing through the intersection of the two planes, so,

$
\begin{aligned}
& 2 x+3 y-z-2+\lambda(x+2 y+3 z-6)=0 \\
& x(2+\lambda)+y(3+2 \lambda)+z(3 \lambda-1)-2-6 \lambda=0
\end{aligned}
$
It is perpendicular to the plane, $2 x+y-2+1=0$
So, $(2+\lambda) 2+(3+2 \lambda) 1+(3 \lambda-1)(-1)=0$

$
\lambda=-8
$
So, plane $p_1-6 x-13 y-25 z+46=0$
distance of plane p from the point $(-7,1,1)$

$
\begin{aligned}
& d=\frac{|+42-13-25+46|}{\sqrt{36+169+625}}=\frac{50}{\sqrt{30}}= \\
& d^2=\frac{2500}{830}=\frac{250}{83}
\end{aligned}
$
Hence, the answer is $\frac{250}{83}$

Example 3: Let the plane containing the line of intersection of the planes $P 1: x+(\lambda+4) y+z=1$ and $P 2: 2 x+y+z=2$ pass through the points ( $0,1,0$ ) and $(1,0,1)$. Then the distance of the point $(2 \lambda, \lambda,-\lambda)$ from the plane $P 2$ is
[JEE MAINS 2023]

Solution
$
\begin{aligned}
& {[x+(\lambda+4) \mathrm{y}+\mathrm{z}-1]+\mu[2 \mathrm{x}+\mathrm{y}+\mathrm{z}-2]=0} \\
& (0,1,0)
\end{aligned}
$
$
\begin{aligned}
& \text { (i) }(\lambda+4-1)+\mu[-1]=0 \\
& \quad \lambda-\mu=-3 \\
& (1,0,1)(\text { ii }) 1+\mu[1]=0 \Rightarrow \mu=-1, \lambda=-4 \\
& \therefore \text { point }(-8,-4,4) ; 2 x+y+z-2=0 \\
& \quad d=\left|\frac{-16-4+4-2}{\sqrt{6}}\right|=\frac{18}{\sqrt{6}}=3 \sqrt{6}
\end{aligned}
$
Hence, the answer is $3 \sqrt{6}$

Example 4: Let the line $\frac{x}{1}=\frac{6-y}{2}=\frac{z+8}{5}$ intersect the lines $\frac{x-5}{4}=\frac{y-7}{3}=\frac{z+2}{1}$ and $\frac{x+3}{6}=\frac{3-y}{3}=\frac{z-6}{1}$ at the points $A$ and $B$ respectively. Then the distance of the mid-point of the line segment $A B$ from the plane $2 \mathrm{x}-2 \mathrm{y}+\mathrm{z}=14$ is:
[JEE MAINS 2023]

Solution
$
\begin{aligned}
& \frac{x}{1}=\frac{y-6}{-2}=\frac{z+8}{5}=\lambda ...... (1) \\
& \frac{x-5}{4}=\frac{y-7}{3}=\frac{z+2}{1}=\mu ...... (2) \\
& \frac{x+3}{6}=\frac{y-3}{-3}=\frac{z-6}{1}=\gamma ...... (3)
\end{aligned}
$
Intersection of (1) \& (2) "A"

$
\begin{aligned}
& (\lambda,-2 \lambda+6,5 \lambda-8) \&(4 \mu+5,3 \mu+7, \mu-2) \\
& \lambda=1, \mu=-1 \\
& A(1,4,-3)
\end{aligned}
$
Intersection (1) \& (3) "B"

$
\begin{aligned}
& (\lambda,-2 \lambda+6,5 \lambda-8) \&(6 \gamma-3,-3 \gamma+3, \gamma+6) \\
& \lambda=3 \\
& \gamma=1
\end{aligned}
$
$
B(3,0,7)
$
Mod point of $\mathrm{A} \& \mathrm{~B} \Rightarrow(2,2,2)$
The perpendicular distance from the plane

$
\begin{aligned}
& 2 x-2 y+z=14 \\
& \left|\frac{2(2)-2(2)+2-14}{\sqrt{4+4+1}}\right|=4
\end{aligned}
$
Hence, the answer is 4

Example 5: Let the equation of the plane P containing the line $\mathrm{x}+10=\frac{8-\mathrm{y}}{2}=\mathrm{z}$ be ax $+\mathrm{by}+3 \mathrm{z}=2(\mathrm{a}+\mathrm{b})$ and the distance of the plane P from the point $(1,27,7)$ be c . Then $a^2+b^2+c^2$ is equal to
[JEE MAINS 2023]

Solution:
The given equation of the plane is $a x+b y+3 z=2(a+b)$ It contains the line $\frac{x-(-10)}{1}=\frac{y-8}{-2}=\frac{z-0}{1}$
$\therefore$ plane (1) must pass through $(-10,8,0)$ and parallel to $1,-2,1$ Hence,

$
\begin{aligned}
& a(-10)+8 b=2 a+2 b \\
& \Rightarrow \quad 12 \mathrm{a}-6 \mathrm{~b}=0 ............. (2)
\end{aligned}
$

and $\quad a-2 b+3=0$ ...........(3)
on solving (2) and (3), we get

$
b=2, a=1
$

on solving (2) and (3), we get

$
\mathrm{b}=2, \mathrm{a}=1
$

$\therefore$ equation of the plane is

$
x+2 y+3 z=6
$

c is the perpendicular distance from $(1,27,7)$ the plane (4)

$
\Rightarrow \mathrm{c}=\left|\frac{1+2 \times 27+3 \times 7-6}{\sqrt{1^2+2^2+3^2}}\right|=\left|\frac{70}{\sqrt{14}}\right|=\frac{10 \sqrt{7}}{\sqrt{2}}
$

Now, $a^2+b^2+c^2=1+4+\frac{700}{2}=\frac{710}{2}=355$

Hence, the answer is the 355.

Summary
Understanding and calculating the distance from a point to a plane provides essential information and helps in solving practical problems. It exemplifies the intersection of geometric principles with real-world applications. In engineering and physics, this distance is used for determining clearances, collision detection, and ensuring objects are properly aligned relative to planes or surfaces.