Family of Plane - Practice Questions & MCQ

Equation of 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 given plane, then the plane parallel to this plane is $a x+b y+c z+k=0$, where $k$ is any scalar.

Plane passing through the intersection of two given planes
Vector Form
Equation of 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
$

Proof:
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, $\overrightarrow{\mathbf{p}}$ is arbitrary, it satisfies for 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.

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 \hat{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 λ.