Question

Equation of a line in the plane $\pi :2x-y+z-4=0$ which is perpendicular to the line $l$ whose equation is $\frac{x-2}{1}=\frac{y-2}{-1}=\frac{z-3}{-2}$ and which passes through the point of intersection of $l$ and $\pi$ is

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

Answer 1

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

Let direction ratios of the line be $(a,b,c)$; then
$2a-b+c=0$ and $a-b-2c=0$, i.e., $\frac{a}{3}=\frac{b}{5}=\frac{c}{-1}$

Therefore, direction ratios of the line are $(3,5,-1).$
Any point on the given line is $(2+\lambda ,2-\lambda ,3-2\lambda )$
It lies on the given plane $\pi$ if
$2(2+\lambda )-(2-\lambda )+(3-2\lambda )=4$
$\Rightarrow 4+2\lambda -2+\lambda +3-2\lambda =4$
$\Rightarrow \lambda =-1$

Therefore, the point of intersection of the line and the plane is $(1,3,5).$
Therefore, equation of the required line is
$\frac{x-1}{3}=\frac{y-3}{5}=\frac{z-5}{-1}$