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} $