Question

The perpendicular distance of $P(1,2,3)$ from the line $\frac{x-6}{3}=\frac{y-7}{2}=\frac{z-7}{-2}$ is

• A. 7
• B. 5
• C. 0
• D. 6

Answer 1

Answer: (A)

The point $A(6,7,7)$ is on the line .
Let the perpendicular from $P$ meet the line in $L$.
Then $A P^{2}=(6-1)^{2}+(7-2)^{2}+(7-3)^{2}=66$
Also $A L=$ projection of $A P$ on line

$\left(\right.$ actual $\left. d.c's \frac{3}{\sqrt{17}}, \frac{2}{\sqrt{17}}, \frac{-2}{\sqrt{17}}\right)$
$\Rightarrow (6-1) \cdot \frac{3}{\sqrt{17}}+(7-2) . \frac{2}{\sqrt{17}}+(7-3) \frac{-2}{\sqrt{17}}$
$=\sqrt{17}$
$\therefore \perp$ distance $d$ of $P$ from the line is given by
$d^{2}=A P^{2}-A L^{2}=66-17=49$ so that $d =7$