Image of a Point in the Plane

A point in three dimensions is denoted by three coordinates dimension ( x , $y, z)$. The image of a point in the given plane is the reflection of the point over the given plane. We use the image of the point to find the reflection of the point which makes our calculations easy.

In this article, we will cover the concept of the Image of a Point in the 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 2019, three in 2021, eight in 2022, and seven in 2023.

What is the Plane?

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. A plane is also determined by a line and any point that does not lie on the line.

Equation of a plane in Vector Form

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

Equation of a plane in Cartesian form

The equation $\overrightarrow{\mathbf{r}} \cdot \hat{\mathbf{n}}=d$ gives

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

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

What is the Image of a Point in the Plane?

The image of a point in the given plane is the reflection of the point over the given plane. If the point lies on the plane, its image remains unchanged. If the point is above or below the plane, its image is found by reflecting it across the plane.

Equation of the Image of a Point in the Plane

The image of the point $\mathrm{P}\left(\mathrm{x}_1, \mathrm{y}_1, \mathrm{z}_1\right)$ in the plane $a x+b y+c z+d=0$ is $Q\left(x_3, y_3, z_3\right)$ then coordinates of point $Q$ is given by

$
\frac{\mathrm{x}_3-\mathrm{x}_1}{\mathrm{a}}=\frac{\mathrm{y}_3-\mathrm{y}_1}{\mathrm{~b}}=\frac{\mathrm{z}_3-\mathrm{z}_1}{\mathrm{c}}=-\frac{2\left(\mathrm{ax}_1+\mathrm{by}_1+\mathrm{cz}_1+\mathrm{d}\right)}{\mathrm{a}^2+\mathrm{b}^2+\mathrm{c}^2}
$

Derivation of the Image of a Point in the Plane

Let point $Q\left(x_3, y_3, z_3\right)$ is the image of the point $P\left(x_1, y_1, z_1\right)$ in the plane $\pi$ $: a x+b y+c z+d=0$

Let line $P Q$ meet the plane $a x+b y+c z+d=0$ at point $R$, the direction ratio of normal to plane $\pi$ are ( $a, b, c)$, since, PQ is perpendicular to plane $\pi$.

So direction ratio of $P Q$ is $\mathrm{a}, \mathrm{b}, \mathrm{c}$
$\Rightarrow$ equation of line PQ is

$
\frac{x-x_1}{a}=\frac{y-y_1}{b}=\frac{z-z_1}{c}=r
$

any point on line $P Q$ may be taken as

$
\left(a r+x_1, b r+y_1, c r+z_1\right)
$
Let $Q \equiv\left({a r+x_1}, {b r+y_1}, {c r+z_1}\right)$
since, $R$ is the middle point of $P Q$

$
\begin{aligned}
& R \equiv\left(\frac{x_1+a r+x_1}{2}, \frac{y_1+b r+y_1}{2}, \frac{z_1+c r+z_1}{2}\right) \\
& \Rightarrow \quad R \equiv\left(x_1+\frac{a r}{2}, y_1+\frac{b r}{2}, z_1+\frac{c r}{2}\right)
\end{aligned}
$

Since point $R$ lies in the plane $\pi$, we get

$
\begin{array}{cc}
\Rightarrow & R \equiv\left(x_1+\frac{a r}{2}, y_1+\frac{b r}{2}, z_1+\frac{c r}{2}\right) \\
\Rightarrow & a\left(x_1+\frac{a r}{2}\right)+b\left(y_1+\frac{b r}{2}\right)+c\left(z_1+\frac{c r}{2}\right)+d=0 \\
\Rightarrow & \left(a^2+b^2+c^2\right) \frac{r}{2}=-\left(a x_1+b y_1+c z_1+d\right) \\
\Rightarrow & r=\frac{-2\left(a x_1+b y_1+c z_1+d\right)}{a^2+b^2+c^2}
\end{array}
$

In a similar method, we can also find the coordinates of point R.

Steps To Determine the Image of a Point in a Plane

Consider two points $P$ and $Q$. Let's assume a plane $\pi$ such that:

• A perpendicular line $P Q$ to the plane $\pi$ exists.
• The midpoint of PQ lies on the plane $\pi$. In such a case, the image of the point is either of the points to one another in the plane $\pi$.

The steps to find the image of a point in a given plane:
Step 1: The equations of the normal to the given plane and the line passing through the point $P$ are written as

$
\frac{x-x_0}{a}=\frac{y-y_0}{b}=\frac{z-z_0}{c}
$
Step 2: The coordinates of the image Q are ( $\mathrm{x}_1+\mathrm{ar}, \mathrm{y}_1+\mathrm{br}, \mathrm{z}_1+\mathrm{cr}$ ).
Step 3: The coordinates of the mid-point R of PQ are then determined.
Step 4: The value of $r$ is obtained by substituting the coordinates of $R$ in the plane equation.

Step 5: Finally, the $r$-value is substituted in the coordinates of $Q$ to get the final result.

Solved Examples on Image if a Point in a Plane

Example 1: Let the image of the point $P(2,-1,3)$ in the plane $x+2 y-z=0$ be $Q$. Then the distance of the plane $3 x+2 y+z+29=0$ from the point $Q$ is
$[$ JEE MAINS 2023]
Solution:
Let $\mathrm{Q}(\alpha, \beta, \gamma)$ is image of $\mathrm{P}(2,-1,3)$ in the plane $\mathrm{x}+2 \mathrm{y}-\mathrm{z}=0$

$
\begin{aligned}
& \frac{\alpha-2}{1}=\frac{\beta+1}{2}=\frac{\gamma-3}{-1}=\frac{-2(2-2-3)}{1^2+2^2+(-1)^2}=1 \\
& \alpha=3, \beta=1, \gamma=2
\end{aligned}
$

Distance of $\mathrm{Q}(3,1,2)$ from

$
\begin{aligned}
& 3 x+2 y+z+29=0 \\
& \mathrm{D}=\frac{|3(3)+2(1)+2+29|}{\sqrt{3^2+2^2+1^2}} \\
& =\frac{42}{\sqrt{14}}=3 \sqrt{14}
\end{aligned}
$
Hence, the answer is $3 \sqrt{14}$

Example 2: Let the food of perpendicular to the point $\mathrm{A}(4,3,1)$ on the plane $\mathrm{P}: \mathrm{x}-\mathrm{y}+2 \mathrm{z}+3=0$ be N . If $\mathrm{B}(5, \alpha, \beta), \alpha, \beta \in \mathrm{Z}{\text {is a }}$ point on a plane P such that the area of the triangle $A B N$ is $3 \sqrt{2}$, then $\alpha^2+\beta^2+\alpha \beta$ is equal to _____________. [JEE MAINS 2023]

Solution

$
\begin{aligned}
& \mathrm{AN}=\sqrt{6} \\
& 5-\alpha+2 \beta+3=0 \\
& \Rightarrow \alpha=8+2 \beta
\end{aligned}
$
N is given by

$
\begin{aligned}
& \frac{x-4}{1}=\frac{y-3}{-1}=\frac{z-1}{2}=\frac{-(4-3+2+3)}{1+1+4} \\
& \mathrm{x}=3, \mathrm{y}=4, \mathrm{z}=-1 \\
& \mathrm{~N} \\
& (3,4,-1) \\
& \mathrm{BN}=\sqrt{4+(\alpha-4)^2+(\beta+1)^2} \\
& =\sqrt{4+(2 \beta+4)^2+(\beta+1)^2} \\
& \text { Area of } \triangle \mathrm{ABN}=\frac{1}{2} \mathrm{AN} \times \mathrm{BN}=3 \sqrt{2}
\end{aligned}
$
$
\begin{aligned}
& \frac{1}{2} \times \sqrt{6} \times \mathrm{BN}=3 \sqrt{2} \\
& \mathrm{BN}=2 \sqrt{3} \\
& 4+(2 \beta+4)^2+(\beta+1)^2=12 \\
& (2 \beta+4)^2+(\beta+1)^2-8=0 \\
& 5 \beta^2+18 \beta+9=0 \\
& (5 \beta+3)(\beta+3)=0 \\
& \beta=-3 \\
& \alpha=2 \\
& \alpha^2+\beta^2+\alpha \beta=9+4-6=7
\end{aligned}
$
Hence, the answer is 7

Example 3: Let $(\alpha, \beta, \gamma)$ be the image of the point $\mathrm{P}(2,3,5)$ in the plane $2 x+y-3 z=6$ then $\alpha+\beta+\gamma$ is equal to
[JEE MAINS 2023]

Solution

$\begin{array}{llc}\frac{\alpha-2}{2}=\frac{\beta-3}{1}=\frac{\gamma-5}{-3}=-2\left(\frac{2 \times 2+3-3 \times 5-6}{2^2+1^2+1-3^2}\right)=2 & & \\ \frac{\alpha-2}{2}=2 & \beta-3=2 & \gamma-5=-6 \\ \alpha=6 & \beta=5 & \gamma=-1\end{array}$

Hence, the answer is 10

Example 4: Let two vertices of a triangle $\operatorname{ABC}$ be $(2,4,6)$ and $(0,-2,-5)$ and its centroid be $(2,1,-1)$. If the image of the third vertex in the plane $x+2 y$ $+4 z=11$ is $(\alpha, \beta, \gamma)$ then $\alpha \beta+\beta \gamma+\gamma \alpha$ is equal to
[JEE MAINS 2023]

Solution

Given the Two vertices of the Triangle
$\mathrm{A}(2,4,6)$ and $\mathrm{B}(0,-2,-5)$ and if centroid $\mathrm{G}(2,1,-1)$. Let Third vertices be $(x, y, z)$
Now $\frac{2+0+x}{3}=2, \frac{4-2+y}{3}=1, \frac{6-z+5}{3}=-1$

$
x=4, y=1, z=-1
$

Third vertices $C(4,1,-4)$ Now, Image of vertices $C(4,1,-4)$ in the given plane is $D=(\alpha, \beta, \gamma)$

Now

$
\begin{aligned}
& \frac{\alpha-4}{1}=\frac{\beta-1}{2}=\frac{\gamma+4}{4}=-2 \frac{(4+2-16-11)}{1+4+16} \\
& \frac{\alpha-4}{1}=\frac{\beta-1}{2}=\frac{\gamma+4}{4}=\frac{42}{21} \Rightarrow 2 \\
& \alpha=6, \beta=5, \gamma=4 \\
& \text { Then } \alpha \beta+\beta \gamma+\gamma \alpha \\
& \quad=(6 \times 5)+(5 \times 4)+(4 \times 6) \\
& =30+20+24 \\
& =74
\end{aligned}
$
Hence, the answer is 74

Example 5 : If the foot of the perpendicular from the point $\mathrm{A}(-1,4,3)$ on the plane $\mathrm{P}: 2 \mathrm{x}+\mathrm{my}+\mathrm{nz}=4$, is $\left(-2, \frac{7}{2}, \frac{3}{2}\right)$, then the distance of the point A from the plane P , measured parallel to a line with direction ratios $3,-1,-4$, is equal to :
[JEE MAINS 2022]

Solution:
Let B be foot of $\perp$ coordinates of $\mathrm{B}=\left(-2, \frac{7}{2}, \frac{3}{2}\right)$
Direction ratio of line AB is $\langle 2,1,3\rangle$ So

$
\mathrm{m}=1, \mathrm{n}=3
$
So equation of AC is $\frac{\mathrm{x}+1}{3}=\frac{\mathrm{y}-4}{-1}=\frac{\mathrm{z}-3}{-4}=\lambda$
So point c is $(3 \lambda-1,-1+9,-4, \lambda+3)$.but c lies on the plane, so

$
\begin{aligned}
& 6 \lambda-2-\lambda-1-12 \lambda+9=4 \\
& \Rightarrow \lambda=1 \Rightarrow \mathrm{C}(2,3,-1) \\
& \Rightarrow \quad \mathrm{AC}=\sqrt{26}
\end{aligned}
$
Hence, the answer is $\sqrt{26}$

Summary

The image of a point in a given plane is determined by reflecting the point across the plane. If the point is above or below the plane, its image is found by reflecting it across the plane, resulting in a transformation of its coordinates relative to the plane. This reflection process is fundamental in geometry and helps determine the relationship between points and planes in a coordinate plane.