The distance of the point $(7,10,11)$ from the line $\frac{x-4}{1}=\frac{y-4}{0}=\frac{z-2}{3}$ along the line $\frac{x-9}{2}=\frac{y-13}{3}=\frac{z-17}{6}$ is

Question

Aicharya · Accepted Answer

$\because$ line $P Q$ is parallel to line $\frac{x-9}{2}=\frac{y-3}{3}=\frac{z-17}{6}$
$\begin{aligned}
& \therefore \frac{\lambda-3}{2}=\frac{-6}{3}=\frac{3 \lambda-9}{6} \Rightarrow \lambda=-1 \\ & Q=(3,4,-1) \\ & \therefore P Q=\sqrt{16+36+144}=14
\end{aligned}$ ^

The distance of the point $(7,10,11)$ from the line $\frac{x-4}{1}=\frac{y-4}{0}=\frac{z-2}{3}$ along the…

Solution