Question

A perpendicular is drawn from the point $P(2,4,-1)$ to the line $\frac{x+5}{1}=\frac{y+3}{4}=\frac{z-6}{-9}$. The equation of the perpendicular from $P$ to the given line is

• A. $\frac{x-2}{6}=\frac{y-4}{3}=\frac{z+1}{2}$
• B. $\frac{x+2}{6}=\frac{y-4}{3}=\frac{z+1}{2}$
• C. $\frac{x-2}{-2}=\frac{y-4}{5}=\frac{z+1}{2}$
• D. $\frac{x+2}{6}=\frac{y+4}{3}=\frac{z+1}{2}$

Answer 1

Answer: (A)

Let $Q$ be foot of perpendicular from $P$ i.e. $(\lambda -5,4\lambda -3,-9\lambda +6)$

$\therefore \overrightarrow{PQ}=(\lambda -7)\hat{i}+(4\lambda -7)\hat{j}+(-9\lambda +7)\hat{k}$
$\Theta \overrightarrow{PQ}\cdot (\hat{i}+4\hat{j}-9\hat{k})=0$
$\Rightarrow (\lambda -7)+4(4\lambda -7)-9(-9\lambda +7)=0$
$98\lambda -98=0\Rightarrow \lambda =1$
$\therefore Q=(-4,1,-3)$
$\therefore \text{ Equation of }PQ\text{ is }\frac{x-2}{6}=\frac{y-4}{3}=\frac{z+1}{2}$