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 $ax+by+cz+d=0$ is given plane, then the plane parallel to this plane is $ax+by+cz+k=0$, 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
$\overrightarrow{\mathbf{r}}\cdot ({\overrightarrow{\mathbf{n}}}_1+\lambda {\overrightarrow{\mathbf{n}}}_2)=d_1+\lambda d_2$

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 $\overrightarrow{p}\cdot {\hat{n}}_1=d_1$ and $\overrightarrow{p}\cdot {\hat{n}}_2=d_2$

Therefore, for all real values of $\lambda$, we have
$\overrightarrow{p}\cdot ({\hat{n}}_1+\lambda {\hat{n}}_2)=d_1+\lambda d_2$

Since, $\overrightarrow{\mathbf{p}}$ is arbitrary, it satisfies for any point on the line.
Hence, the equation $\overrightarrow{r}\cdot ({\overrightarrow{n}}_1+\lambda {\overrightarrow{n}}_2)=d_1+\lambda d_2$ 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
$\begin{array}{rl}{\overrightarrow{n}}_1& ={\mathrm{a}}_1\hat{i}+{\mathrm{b}}_2\hat{j}+{\mathrm{c}}_1\hat{k}\\ {\overrightarrow{n}}_2& ={\mathrm{a}}_2\hat{i}+{\mathrm{b}}_2\hat{j}+{\mathrm{c}}_2\hat{k}\\ \overrightarrow{r}& =x\hat{i}+y\hat{j}+z\hat{k}\end{array}$
then the vector equation, $\overrightarrow{r}\cdot ({\overrightarrow{n}}_1+\lambda {\overrightarrow{n}}_2)=d_1+\lambda d_2$ become
$\begin{array}{c}x({\mathrm{a}}_1+\lambda {\mathrm{a}}_2)+y({\mathrm{b}}_1+\lambda {\mathrm{b}}_2)+z({\mathrm{c}}_1+\lambda {\mathrm{c}}_2)=d_1+\lambda d_2\\ \text{ or }({\mathbf{a}}_{1}x+{\mathbf{b}}_{1}y+{\mathbf{c}}_{1}z-d_1)+\lambda ({\mathbf{a}}_{2}x+{\mathbf{b}}_{2}y+{\mathbf{c}}_{2}z-d_2)=0\end{array}$

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.