Question

The perpendicular distance from $(4,-3,2)$ to the line $\frac{x-2}{3}=\frac{y-3}{-2}=\frac{z-5}{6}$ is

• A. $7\sqrt{2}$
• B. $14$
• C. $7$
• D. $49$

Answer 1

The correct option is C $7$

D.R of line are $3,-2,6$
Now in general the coordinates of any point on line can be represented in the form
$(3\lambda +2,-2\lambda +3,6\lambda +5)$
DR of line perpendicular to the given line $(3\lambda -2,-2\lambda +6,6\lambda +3)$
Now,
$3(3\lambda -2)-2(-2\lambda +6)+6(6\lambda +3)=0$
$\lambda =0$
Hence, the coordinates are $(2,3,5)$
Distance between points $(2,3,5)$ and $(4,-3,2)=\sqrt{2^2+6^2+3^2}=7$
Hence, option C is correct.