Family of Planes

For any three points that do not all lie on the same line, there is a unique plane that passes through these points. 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. In real life, we use planes to measure the circumference, area, and volume.

In this article, we will cover the concept of the Family of 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-five questions have been asked on this topic in JEE Main from 2013 to 2023 including four in 2019, nine in 2021, eight in 2022, and two in 2023.

What is a plane parallel to a given plane?

Parallel planes have the same normal vectors, so the equation of plane parallel to $\overrightarrow{\mathbf{r}} \cdot \overrightarrow{\mathbf{n}}=d_1$ is of the form $\overrightarrow{\mathbf{r}} \cdot \overrightarrow{\mathbf{n}}=d_2$, where $d_2$ can be determined by using the given conditions.

In cartesian form, if $a x+b y+c z+d=0$ is the given plane, then the plane parallel to this plane is $a x+b y+c z+k=0$, where $k$ is any scalar.

The plane passes through the intersection of two given planes

Equation of plane parallel to a given plane in Vector Form

The equation of a plane passing through the intersection of planes $\overrightarrow{\mathbf{r}} \cdot \overrightarrow{\mathbf{n}}_1=d_1$ and $\overrightarrow{\mathbf{r}} \cdot \overrightarrow{\mathbf{n}}_2=d_2$ is

$
\overrightarrow{\mathbf{r}} \cdot\left(\overrightarrow{\mathbf{n}}_1+\lambda \overrightarrow{\mathbf{n}}_2\right)=d_1+\lambda d_2
$

Derivation of Equation of plane parallel to a given plane

Let $\pi_1$ and $\pi_2$ be two planes with equations $\overrightarrow{\mathbf{r}} \cdot \overrightarrow{\mathbf{n}}_1=d_1$ and $\overrightarrow{\mathbf{r}} \cdot \overrightarrow{\mathbf{n}}_2=d_2$ respectively. The position vector of any point on the line of intersection must satisfy both equations.

If $\overrightarrow{\mathbf{P}}$ is the position vector of a point $P$ on the line, then $\vec{p} \cdot \hat{n}_1=d_1$ and $\vec{p} \cdot \hat{n}_2=d_2$

Therefore, for all real values of $\lambda$, we have

$
\vec{p} \cdot\left(\hat{n}_1+\lambda \hat{n}_2\right)=d_1+\lambda d_2
$
Since $p$ is arbitrary, it satisfies any point on the line.
Hence, the equation $\vec{r} \cdot\left(\vec{n}_1+\lambda \vec{n}_2\right)=d_1+\lambda d_2$ represents a plane $\pi_3$ which is such that if any vector $\overrightarrow{\mathbf{r}}$ satisfies both the equations $\pi_1$ and $\pi_2$, it also satisfies the equation $\pi_3$ i.e. any plane passing through the intersection of the planes.

Equation of plane parallel to a given plane in Cartesian Form

In the Cartesian system, let

$
\begin{aligned}
\vec{n}_1 & =\mathrm{a}_1 \hat{i}+\mathrm{b}_2 \hat{j}+\mathrm{c}_1 \hat{k} \\
\vec{n}_2 & =\mathrm{a}_2 \hat{i}+\mathrm{b}_2 \hat{j}+\mathrm{c}_2 \hat{k} \\
\vec{r} & =x \hat{i}+y \hat{j}+z k
\end{aligned}
$

then the vector equation, $\vec{r} \cdot\left(\vec{n}_1+\lambda \vec{n}_2\right)=d_1+\lambda d_2$ become

$
\begin{gathered}
x\left(\mathrm{a}_1+\lambda \mathrm{a}_2\right)+y\left(\mathrm{~b}_1+\lambda \mathrm{b}_2\right)+z\left(\mathrm{c}_1+\lambda \mathrm{c}_2\right)=d_1+\lambda d_2 \\
\text { or }\left(\mathbf{a}_1 x+\mathbf{b}_1 y+\mathbf{c}_1 z-d_1\right)+\lambda\left(\mathbf{a}_2 x+\mathbf{b}_2 y+\mathbf{c}_2 z-d_2\right)=0
\end{gathered}
$

which is the required Cartesian form of the equation of the plane passing through the intersection of the given planes for each value of $\lambda$.

Solved Examples Based on Family of Plane

Example 1: If the equation of the plane passing through the line of intersection of the planes $2 x-y+z=3,4 x-3 y+5 z+9=0$ and parallel to the line $\frac{x+1}{-2}=\frac{y+3}{4}=\frac{z-2}{5}$ is $a x+b y+c z+6=0$, ,then $\mathrm{a}+\mathrm{b}+\mathrm{c}$ is equal to :
[JEE MAINS 2023]

Solution:
Using a family of plane

$
\begin{aligned}
& P: P_1+\lambda P_2=0 \Rightarrow P(2+4 \lambda) x-(1+3 \lambda) y+(1+5 \lambda) z=(3-9 \lambda) \\
& P \text { is } \| \text { to } \frac{x+1}{-2}=\frac{\mathrm{y}+3}{4}=\frac{\mathrm{z}-2}{5}
\end{aligned}
$

Then for $\lambda: \vec{n}_p \cdot \vec{v}_L=0$

$
\begin{aligned}
& -2(2+4 \lambda)-4(1+3 \lambda)+5(1+5 \lambda)=0 \\
& -3+5 \lambda=0 \Rightarrow \lambda=\frac{3}{5}
\end{aligned}
$

Hence : $\mathrm{P}: 22 \mathrm{x}-14 \mathrm{y}+20 \mathrm{z}=-12$

$
\begin{aligned}
& P: 11 x-7 y+10 z+6=0 \\
& \Rightarrow \mathrm{a}=11 \\
& \mathrm{~b}=-7 \\
& \mathrm{c}=10 \\
& \Rightarrow \mathrm{a}+\mathrm{b}+\mathrm{c}=14
\end{aligned}
$

Hence, the answer is 14

Example 2: Let the equation of the plane passing through the line of intersection of the planes $x+2 y+a z=2$ and $x-y+z=3$ be $5 \mathrm{x}-11 \mathrm{y}+\mathrm{bz}=6 \mathrm{a}-1$. For $\mathrm{c} \epsilon \mathrm{Z}$, if the distance of this plane from the point $(\mathrm{a},-\mathrm{c}, \mathrm{c})$ is $\frac{2}{\sqrt{a}}$, then $\frac{a+b}{c}$ is equal
[JEE MAINS 2023]

Solution:
$(x+2 y+a z-2)+\lambda(x-y+z-3)=0$

$
\begin{aligned}
& \frac{1+\lambda}{5}=\frac{2-\lambda}{-11}=\frac{a+\lambda}{b}=\frac{2+3 \lambda}{6 a-1} \\
& \lambda=-\frac{7}{2}, a=3, b=1 \\
& \frac{2}{\sqrt{a}}=\left|\frac{5 a+11 c+b c-6 a+1}{\sqrt{25+121+1}}\right| \\
& c=-1 \\
& \therefore \frac{a+b}{c}=\frac{3+1}{-1}=-4
\end{aligned}
$

Hence, the answer is -4

Example 3: Let $\mathrm{P}_1: \overrightarrow{\mathrm{r}} \cdot(2 \hat{i}+\hat{j}-3 \hat{k})=4$ be a plane. Let $\mathrm{P}_2$ be another plane that passes through the points $(2,-3,2),(2,-2,-3)$ and $(1,-4,2)$. If the direction ratios of the line of intersection of $\mathrm{P}_1$ and $\mathrm{P}_2$ be $16, \alpha, \beta$, then the value of $\alpha+\beta$ is equal to
[JEE MAINS 2022]

Solution:
Let $\mathrm{A}(2,-3,2), \mathrm{B}(2,-2,-3), \mathrm{C}(1,-4,2)$

$
\begin{aligned}
\overrightarrow{n_1} & =2 \mathrm{i}+\mathrm{j}-3 \mathrm{k} \\
\vec{n}_2 & =\text { vector perpendicular to } 2^{\text {nd }} \text { plane } \\
& =\overrightarrow{A B} \times \overrightarrow{B C} \\
& =\left|\begin{array}{ccc}
i & j & k \\
0 & 1 & -5 \\
-1 & -2 & 5
\end{array}\right| \\
& =(5-10) \mathrm{i}-(-5) \mathrm{j}+(1) \mathrm{k} \\
= & -5 \mathrm{i}+5 \mathrm{j}+\mathrm{k}
\end{aligned}
$
To find the Direction ratios of the line of intersection, we first find $\vec{n}_1 \times \vec{n}_2$

$
\begin{aligned}
\vec{n}_1 \times \vec{n}_2 & =\left|\begin{array}{ccc}
i & j & k \\
2 & 1 & -3 \\
-5 & 5 & 1
\end{array}\right| \\
& =(16) \mathrm{i}-(-13) \mathrm{j}+(15) \mathrm{k} \\
& =16 \mathrm{i}+13 \mathrm{j}+15 \mathrm{k}
\end{aligned}
$

$\begin{aligned} & \therefore \quad \alpha=13, \beta=15 \\ & \Rightarrow \alpha+\beta=28\end{aligned}$

Hence, the answer is 28

Example 4: The acute angle between the planes $\mathrm{P}_1$ and $\mathrm{P}_2$, when $\mathrm{P}_1$ and $\mathrm{P}_2$ are the planes pass through the intersection of the planes $5 \mathrm{x}+8 \mathrm{y}+13 \mathrm{z}-29=0$ and $8 \mathrm{x}-7 \mathrm{y}+\mathrm{z}-20=0$ and the points $(2,1,3)$ and $(0,1,2)$, respectively, is
[JEE MAINS 2022]

Solution:
Let $\mathrm{P}_1:(5 \mathrm{x}+2 \mathrm{y}+13 \mathrm{z}-29)+\lambda(8 \mathrm{x}-7 \mathrm{y}+\mathrm{z}-20)=0$
Passes through $(2,1,3)$

$
\begin{aligned}
& \Rightarrow(10+8+39-29)+\lambda(16-7+3-20)=0 \Rightarrow \lambda=\frac{7}{2} \\
& \Rightarrow P_1:(10 \mathrm{x}+16 \mathrm{y}+26 \mathrm{z}-58)+(56 \mathrm{x}-49 \mathrm{y}+73-140)=0 \\
& \Rightarrow 66 \mathrm{x}-33 \mathrm{y}+33 \mathrm{z}-198=0 \Rightarrow 2 \mathrm{x}-\mathrm{y}+\mathrm{z}-6=0---(1)
\end{aligned}
$
Let $\mathrm{P}_2:(5 \mathrm{x}+8 \mathrm{y}+13 \mathrm{z}-29)+\mu(8 \mathrm{x}-7 \mathrm{y}+\mathrm{z}-20)=0$
Passes through $(0,1,2)$

$
\begin{aligned}
& \Rightarrow(0+8+26-29)+\mu(0-7+2-20)=0 \Rightarrow \mu=1 / 5 \\
& \Rightarrow \mathrm{P}_2:(25 \mathrm{x}+40 \mathrm{y}+65 \mathrm{z}-145)+8 \mathrm{x}-7 \mathrm{y}+\mathrm{z}-20=0 \Rightarrow 33 \mathrm{x}+33 \mathrm{y}+663-165=0 \\
& \Rightarrow \mathrm{x}+\mathrm{y}+2 \mathrm{z}-5=0---(2) \\
& \cos \theta=\left|\frac{(2,-1,1) \cdot(1,1,2)}{\sqrt{2^2+1^2+1^2} \sqrt{1^2+1^2+2^2}}\right|=\frac{3}{6}=\frac{1}{2} \Rightarrow \theta=\pi / 3
\end{aligned}
$

Hence, the answer is $\frac{\pi}{3}$

Example 5: The vector equation of the plane passing through the intersection of the planes $\vec{r} \cdot(\hat{i}+\hat{j}+\hat{k})=1$ and $\vec{r} \cdot(\hat{i}-2 \hat{j})=-2$, and the point $(1,0,2)$ is:
[JEE MAINS 2021]

Solution

$\begin{aligned} & \overrightarrow{\mathrm{r}} \cdot(\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}})=1 \\ & \overrightarrow{\mathrm{r}} \cdot(\hat{\mathrm{i}}-2 \hat{\mathrm{j}})=-2 \\ & \text { point }(1,0,2) \\ \end{aligned}$

Equation of Plane

$
\begin{aligned}
&\overrightarrow{\mathrm{r}} \cdot (\hat{\mathrm{i}} + \hat{\mathrm{j}} + \hat{\mathrm{k}}) - 1 + \lambda \left\{ \overrightarrow{\mathrm{r}} \cdot (\hat{\mathrm{i}} - 2 \hat{\mathrm{j}}) + 2 \right\} = 0 \\ & \overrightarrow{\mathrm{r}} \cdot \left\{ \hat{\mathrm{i}}(1 + \lambda) + \hat{\mathrm{j}}(1 - 2\lambda) + \hat{\mathrm{k}} \right\} - 1 + 2\lambda = 0 \\ & \text{Point: } \hat{\mathrm{i}} + 0\hat{\mathrm{j}} + 2\hat{\mathrm{k}} = \overrightarrow{\mathrm{r}} \\ & \therefore (\hat{\mathrm{i}} + 2\hat{\mathrm{k}}) \cdot \left\{ \hat{\mathrm{i}}(1 + \lambda) + \hat{\mathrm{j}}(1 - 2\lambda) + \hat{\mathrm{k}} \right\} - 1 + 2\lambda = 0 \\ & 1 + \lambda + 2 - 1 + 2\lambda = 0 \\ & \lambda = -\frac{2}{3} \\ & \therefore \quad \overrightarrow{\mathrm{r}} \cdot \left[ \hat{\mathrm{i}}\left(\frac{1}{3}\right) + \hat{\mathrm{j}}\left(\frac{7}{3}\right) + \hat{\mathrm{k}} \right] = \frac{7}{3} \\ & \overrightarrow{\mathrm{r}} \cdot \left[ \hat{\mathrm{i}} + 7\hat{\mathrm{j}} + 3\hat{\mathrm{k}} \right] = 7 \\ & \text{Hence, the answer is } \overrightarrow{\mathrm{r}} \cdot (\hat{\mathrm{i}} + 7\hat{\mathrm{j}} + 3\hat{\mathrm{k}}) = 7
\end{aligned}
$

Summary

Parallel planes have the same normal vectors, so the equation of plane parallel to $\overrightarrow{\mathbf{r}} \cdot \overrightarrow{\mathbf{n}}=d_1$ is of the form $\overrightarrow{\mathbf{r}} \cdot \overrightarrow{\mathbf{n}}=d_2$, where $d_2$ can be determined by using the given conditions.

In cartesian form, if $a x+b y+c z+d=0$ is the given plane, then the plane parallel to this plane is $a x+b y+c z+k=0$, where $k$ is any scalar.