Two Sides of a Plane

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 Two Sides of 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 one questions have been asked on this topic in JEE Main from 2013 to 2023

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This Story also Contains

1. What is a Plane?
2. Equation of a plane in Vector Form
3. Equation of a plane in Cartesian form
4. Condition for two sides of a Plane
5. Proof of two sides of a Plane
6. Solved Examples on Two Sides of a Plane

What is a Plane?

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.

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.

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 Cartesian equation of the plane in the normal form is
$
\begin{aligned}
&\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}=\mathbf{d}
\end{aligned}
\end{aligned}
$

Condition for two sides of a Plane

Let $a x+b y+c z+d=0$ be the plane, then the points $\left(x_1, y_1, z_1\right)$ and $\left(x_2\right.$, $\mathrm{y}_2, \mathrm{z}_2$ ) lie on the same side or opposite side according to

$
\begin{array}{ll}
\frac{a x_1+b y_1+c z_1+d}{a x_2+b y_2+c z_2+d}>0 & \text { (same side) } \\
\frac{a x_1+b y_1+c z_1+d}{a x_2+b y_2+c z_2+d}<0 & \text { (opposite side) }
\end{array}
$

Proof of two sides of a Plane

Let the line segment joining the point $P\left(x_1, y_1, z_1\right)$ and $Q\left(x_2, y_2, z_2\right)$ be divided by a point $R$ (which lies in the plane) internally in the ratio m:n.

Point $R$ lies in the plane $a x+b y+c z+d=0$.

From the section formula

$
R \equiv\left(\frac{m x_2+n x_1}{m+n}, \frac{m y_2+n y_1}{m+n}, \frac{m z_2+n z_1}{m+n}\right)
$

Since point $R$ lies in the plane
Therefore,

$
\begin{array}{cc}
& a\left(\frac{m x_2+n x_1}{m+n}\right)+b\left(\frac{m y_2+n y_1}{m+n}\right)+c\left(\frac{m z_2+n z_1}{m+n}\right)+d=0 \\
\Rightarrow & a\left(m x_2+n x_1\right)+b\left(m y_2+n y_1\right)+c\left(m z_2+n z_1\right)+d(m+n)=0 \\
\Rightarrow & m\left(a x_2+b y_2+c z_2+d\right)+n\left(a x_1+b y_1+c z_1+d\right)=0 \\
\Rightarrow & \quad \frac{m}{n}=-\frac{\left(a x_1+b y_1+c z_1+d\right)}{\left(a x_2+b y_2+c z_2+d\right)}
\end{array}
$

Now, if ax ${ }_1+\mathrm{by}_1+\mathrm{cz}_1+\mathrm{d}$ and $\mathrm{ax}_2+\mathrm{by}_2+\mathrm{cz}_2+\mathrm{d}$ are of the same sign then $\mathrm{m} / \mathrm{n}<0$ (external division), and if the opposite sign then $\mathrm{m} / \mathrm{n}>0$ (internal division).

Therefore, if

$\begin{array}{ll}\frac{a x_1+b y_1+c z_1+d}{a x_2+b y_2+c z_2+d}>0 & \text { (same side) } \\ \frac{a x_1+b y_1+c z_1+d}{a x_2+b y_2+c z_2+d}<0 & \text { (opposite side) }\end{array}$

Recommended Video Based on Two Sides of a Plane

Solved Examples on Two Sides of a Plane

Example 1: The value of x for which $(x, x+1, x+2)$ and $(1-x, x+2, x)$ lie on the opposite sides of the plane $x+y+z-2=0$ is

Solution:
On putting points in L.H.S. of the equation of plane one by one, we get an expression

$
\begin{aligned}
& (x)+(x+1)+(x+2)-2=3 x+1 \\
& \text { and }(1-x)+(x+2)+(x-2)=x+1
\end{aligned}
$
For points on opposite sides of the plane,

$
\begin{aligned}
& (3 x+1)(x+1)<0 \\
& \Rightarrow x \in\left(-1, \frac{-1}{3}\right)
\end{aligned}
$
Hence, the answer is

$
x \in\left(-1, \frac{-1}{3}\right)
$

Example 2: 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 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: Let $P$ be the point of intersection of the line $\frac{x+3}{3}=\frac{y+2}{1}=\frac{1-z}{2}$ and the plane $\mathrm{x}+\mathrm{y}+\mathrm{z}=2$. If the distance of the point $P$ from the plane $3 x-4 y+12 z=32$ is $q$, then $q$ and $2 q$ are the roots of the equation :
[JEE MAINS 2023]

Solution
$
\begin{aligned}
& \frac{x+3}{3}=\frac{y+2}{1}=\frac{1-\mathrm{z}}{2}=\lambda \\
& \mathrm{x}=3 \lambda-3, \mathrm{y}=\lambda-2, \mathrm{z}=1-2 \lambda \\
& \mathrm{P}(3 \lambda-3, \lambda-2,1-2 \lambda) \text { will satisfy the equation of plane } \mathrm{x}+\mathrm{y}+\mathrm{z} \\
& =2 \\
& 3 \lambda-3+\lambda-2+1-2 \lambda=2 \\
& 2 \lambda-4=2 \\
& \lambda=3 \\
& \mathrm{P}(6,1,-5)
\end{aligned}
$

Perpendicular distance of $P$ from plane $3 x-4 y+12 z-32=0$ is

$
\begin{aligned}
& q=\left|\frac{3(6)-4(1)+12(-5)-32}{\sqrt{9+16+144}}\right| \\
& q=6 \\
& 2 q=12
\end{aligned}
$

Sum of roots $=6+12=18$
Product of roots $=6(12)=72$
$\therefore$ A quadratic equation having $q$ and $2 q$ as roots is $x^2-18+72$
Hence, the answer is $x^2+18 x+72=0$

Example 5: 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 $3 \sqrt{14}$

Summary

The two sides of the plane are 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.

Frequently Asked Questions (FAQs)

1. What is a Plane?
Answer:
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.

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

3. What is the equation of the plane in cartesian form?
Answer:
The Cartesian equation of the plane in the normal form is
$
\begin{aligned}
&\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}=\mathbf{d}
\end{aligned}
\end{aligned}
$

4. What are the conditions for the two sides of the plane?
Answer:
Let $a x+b y+c z+d=0$ be the plane, then the points $\left(x_1, y_1, z_1\right)$ and $\left(x_2\right.$, $\mathrm{y}_2, \mathrm{z}_2$ ) lie on the same side or opposite side according to

$
\begin{array}{ll}
\frac{a x_1+b y_1+c z_1+d}{a x_2+b y_2+c z_2+d}>0 & \text { (same side) } \\
\frac{a x_1+b y_1+c z_1+d}{a x_2+b y_2+c z_2+d}<0 & \text { (opposite side) }
\end{array}
$