Question

Equation of the line in the plane $P:x+3y-z=9,$ which is perpendicular to the line $L:\overrightarrow{r}=\hat{i}+\hat{j}+\hat{k}+\lambda (2\hat{i}+\hat{j}-\hat{k})$ and passing through a point where the plane $P$ meets the given line $L,$ is

• A. $\frac{x-3}{2}=\frac{y-2}{1}=\frac{z}{5}$
• B. $\frac{x-3}{-2}=\frac{y-2}{-1}=\frac{z}{5}$
• C. $\frac{x-3}{-5}=\frac{y-2}{1}=\frac{z}{2}$
• D. $\frac{x-3}{1}=\frac{y-2}{1}=\frac{z}{2}$

Answer 1

The correct option is A $\frac{x-3}{2}=\frac{y-2}{1}=\frac{z}{5}$

Vector perpendicular to $2\hat{i}+\hat{j}-\hat{k}$ and $\hat{i}+3\hat{j}-\hat{k}$ is
$\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 2& 1& -1\\ 1& 3& -1\end{array}\right|=2\hat{i}+\hat{j}+5\hat{k}$

Any general point on the line $L$ is $(1+2\lambda ,1+\lambda ,1-\lambda )$
Since $L$ passes through the plane $x+3y-z=9,$
$\therefore (1+2\lambda )+3(1+\lambda )-(1-\lambda )=9$
$\Rightarrow \lambda =1$
$\therefore$ Point of intersection is $(3,2,0).$
Hence, required line is $\frac{x-3}{2}=\frac{y-2}{1}=\frac{z}{5}$