Question

Find the perpendicular distance of $(3,-1,1)$ from the line $\frac{x}{2}=\frac{y-2}{-3}=\frac{z-3}{4}$

Answer 1

L:- $\frac{x}{2}=\frac{y-2}{-3}=\frac{z-3}{4}$

$\therefore$ Direction ratio of the line $OA=(2,-3,4)$

Direction ratio of the line $PA=(3,-3,-2)$

$\therefore \cos \theta =\frac{2\times 3+9-8}{\sqrt{4+9+16}\sqrt{9+9+4}}$

$=\frac{11}{\sqrt{29}\sqrt{22}}$

$=\frac{\sqrt{11}}{\sqrt{58}}$

$\therefore \sin \theta =\frac{\sqrt{47}}{\sqrt{58}}$

$\therefore PQ=PA\sin \theta =\sqrt{3^2+{(-3)}^2+{\left(2\right)}^2}\frac{\sqrt{47}}{58}$

$=\sqrt{22}\times \frac{\sqrt{47}}{58}$ Ans