Q135 / 200 · Manipal (MET) 2014 · Mathematics
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
- Correct answerA$\frac{x-2}{6}=\frac{y-4}{3}=\frac{z+1}{2}$
- B$\frac{x+2}{6}=\frac{y-4}{3}=\frac{z+1}{2}$
- C$\frac{x+2}{-6}=\frac{y-4}{3}=\frac{z+1}{2}$
- D$\frac{x+2}{6}=\frac{y+4}{3}=\frac{z+1}{2}$
Solution
Obtain the foot of the perpendicular $N(-4,1,-3)$.
Find the equations of the line passing through
$\begin{aligned}
& P(2,4,-1) \text { and } N(-4,1,-3) \\
& \qquad \frac{x-2}{6}=\frac{y-4}{3}=\frac{z+1}{2}
\end{aligned}$
Hence, the correct answer is option: 'A'.