Question

The distance of the point $(-2,4,-5)$ from the line $\frac{x+3}{3}=\frac{y-4}{5}=\frac{z+8}{6}$ is

• A. $\frac{\sqrt{37}}{10}$
• B. $\sqrt{\frac{37}{10}}$
• C. $\frac{37}{\sqrt{10}}$
• D. $\frac{37}{10}$

Answer 1

The correct option is B $\sqrt{\frac{37}{10}}$

Given the line
$\frac{x+3}{3}=\frac{y-4}{5}=\frac{z+8}{6}=\lambda \left(let\right)$
Therefore the line passes through $(-3,4,-8)$
Coordinates of the foot of the perpendicular of the point on the line is $(3\lambda -3,5\lambda +4,6\lambda -8)$
DR's of the perpendicular to the line is $(3\lambda -3+2,5\lambda +4-4,6\lambda -8+5)=(3\lambda -1,5\lambda ,6\lambda -3)$

$\therefore 3(3\lambda -1)+5\left(5\lambda \right)+6(6\lambda -3)=0\lambda =\frac{3}{10}$
Coordinates of foot of perpendicular is $(\frac{-21}{10},\frac{65}{10},\frac{-62}{10})$

$\therefore d=\sqrt{\frac{1}{100}+\frac{225}{100}+\frac{144}{100}}=\sqrt{\frac{37}{10}}$