Question

Find the equation of the plane passing through the line of intersection of the planes $\overrightarrow{r}.(\hat{i}+\hat{j}+\hat{k})=1$ and $\overrightarrow{r}.(2\hat{i}+3\hat{j}-\hat{k})+4=0$ and parallel to x-axis.

Answer 1

The given planes are
$\overrightarrow{r}.(\hat{i}+\hat{j}+\hat{k})=1$
$\Rightarrow$ $\overrightarrow{r}.(\hat{i}+\hat{j}+\hat{k})-1=0$

& $\overrightarrow{r}.(2\hat{i}+3\hat{j}-\hat{k})+4=0$

The equation of any plane passing through the line of intersection of these planes is

$[\overrightarrow{r}.(\hat{i}+\hat{j}+\hat{k})-7]+\lambda [\overrightarrow{r}.(2\hat{i}+3\hat{j}-\hat{k})+4]=0$

$\overrightarrow{r}.\left[\right(2\lambda +1)\hat{i}+(3\lambda +1)\hat{j}+(1-\lambda \left)\hat{k}\right]+(4\lambda +1)......\left(1\right)$

Its direction ratios are $(2\lambda +1),(3\lambda +1)$ and $(1-\lambda )$

The required plane is parallel to $x$-axis. Therefore, its normal is perpendicular to $x$-axis.

The direction ratios of $x$-axis are $1,0$ and $0$.

$\therefore$ $1.(2\lambda +1)+0(3\lambda +1)+0(1-\lambda )=0$

$\Rightarrow$ $2\lambda +1=0\Rightarrow \lambda =-\frac{1}{2}$

Substituting $\lambda =-\frac{1}{2}$ in equation (1), we obtain

$\Rightarrow$ $\overrightarrow{r}.[-\frac{1}{2}\hat{j}+\frac{3}{2}\hat{k}]+(-3)=0$

$\Rightarrow$ $\overrightarrow{r}\cdot (\hat{j}-3\hat{k})+6=0$

Therefore, its cartesian equation is $y-3z+6=0$