Question

Find the perpendicular distance of the point $(1,0,0)$ from the line $\frac{x-1}{2}=\frac{y+1}{-3}=\frac{z+0}{8}$. Also, find the coordinates of the foot of perpendicular.

Answer 1

$\begin{array}{c}\frac{x-1}{2}=\frac{y+1}{-3}=\frac{z+10}{8}=\lambda \\ D{R}^{\prime }atPQ:\left(2\lambda ,-3\lambda -1,8\lambda -10\right)\\ 2\times 2\lambda +3\left(3\lambda +1\right)+8\left(8\lambda -10\right)=0\\ 77\lambda -77=0\\ \lambda =1\\ Q:\left(3,-4,-2\right)\\ PQ=\sqrt{4+16+4}\\ =\sqrt{24}unit.\end{array}$