Question

Find the coordinates of foot of the perpendicular drawn from the point $\left(2,3,-1\right)$ to line $\frac{x-3}{3}=\frac{y-2}{1}=\frac{z}{2}$

Answer 1

Let the given point be $P(2,3,-1)$

And

$\frac{x-3}{3}=\frac{y-2}{1}=\frac{z}{2}=r$

$x=3r+3,y=r+2,z=2r$ ---- $\left(1\right)$

General point on the line is

$(3r+3,r+2,2r)$

Let the foot of perpendicular be $Q(3r+3,r+2,2r)$

Direction ratios of the line $PQ$ are

$3r+3-2,r+2-3,2r+1$

which is

$3r+1,r-1,2r+1$

Since the line $PQ$ is perpendicular to given line

So,

$3(3r+1)+1(r-1)+2(2r+1)=0$

$r=\frac{-2}{7}$

from $\left(1\right)$

Hence the coordinates of foot of perpendicular are

$(\frac{15}{7},\frac{12}{7},\frac{-4}{7})$