Question

Find the foot of perpendicular from the point (2,3,-8) to $\frac{4-x}{2}=\frac{y}{6}=\frac{1-z}{3}$. Also, find the perpendicular distance from the given point to the line.

Answer 1

We have, equation of line as $\frac{4-x}{2}=\frac{y}{6}=\frac{1-z}{3}$
$\Rightarrow \frac{x-4}{-2}=\frac{y}{6}=\frac{z-1}{-3}\lambda$
$\Rightarrow x=-2\lambda +4,y=6\lambda andz=-3\lambda +1$
Let the coordinates of L be $(4-2\lambda ,6\lambda ,1-3\lambda )$ and direction ratios of PL are proportional to give line.
$\therefore -2(2-2\lambda )+6(6\lambda -3)-3(9-3\lambda )=0$
$\Rightarrow -4+4\lambda +36\lambda -18-27+9\lambda =0$
$\Rightarrow 49\lambda =49\Rightarrow \lambda =1$
So, the coordinates of L are $(4-2\lambda ,6\lambda ,1-3\lambda )$i.e., (2,6,-2).

Also, length of PL = $\sqrt{(2-2{)}^2+(6-3{)}^2+(-2+8{)}^2}$
$=\sqrt{0+9+36}=3\sqrt{5}$units.