Question

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

Answer 1

Given planes are $r.(\hat{r}+\hat{j}+\hat{k})=1andr.\left(2\widehat{i+3\hat{j}-\hat{k}}\right)+4=0$
The equation of any plane passing through the line of intersection of these planes is
$[r.(\hat{i}+\hat{j}+\hat{k})-1]+\lambda [r.(2\hat{i}+3\hat{j}-\hat{k})+4]=0$
$orr.\left[\right(2\lambda +1)\hat{i}+(3\lambda +1)\hat{j}+(1-\lambda )\hat{k}+(4\lambda -1\left)\right]=0$ . . . (i)
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.
$\therefore 1(2\lambda +1)+0(3\lambda +1)+0(1-\lambda )=0\Rightarrow 2\lambda +1=0\Rightarrow \lambda =-\frac{1}{2}$
Putting the value of $\lambda$ in Eq. (i) , we obtain
$r.(-\frac{1}{2}\hat{j}+\frac{3}{2}\hat{k})+(-3)=0\Rightarrow r.(\hat{j}-3\hat{k})+6=0$
Therefore, its Cartesian equation is y - 3z + 6 = 0 (Put r = $x\hat{i}+y\hat{j}+z\hat{k}$)
This is the equation of the required plane.