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 $AL=$ projection of $AP$ on line

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