Question

Find the foot of the perpendicular from the point $(0,2,3)$ on the line $\frac{x+3}{5}=\frac{y-1}{2}=\frac{z+4}{3}$ Also, find the lenght of the perpendicular.

Answer 1

Let $L$ be the foot of perpendicular drawn from the point to the given line.

And the given line is

$\frac{x+3}{5}=\frac{y-1}{2}=\frac{z+4}{3}=t$ Consider

Therefore, $x=5t-3,y=2t+1,z=3t-4$

which is direction ratios of this line

and also the point $(0,2,3)$

$0\left(5t-3\right)+2\left(2t+1\right)+3\left(3t-4\right)=0$

Since they are perpendicular.

$4t+2+9t-12=0$

$t=\frac{10}{13}$

So, coordinates of $L$ are $x=5\left(\frac{10}{13}\right)-3,y=2\left(\frac{10}{13}\right)+1,z=3\left(\frac{10}{13}\right)-4;x=\frac{11}{13},y=\frac{33}{13},z=\frac{-22}{13}$

Foot of perpendicular from the point $(0,2,3)$ is $\left(\frac{11}{13},\frac{33}{13},\frac{-22}{13}\right)$

And the length

$=\sqrt{(\frac{11}{13}-0{)}^2+(\frac{33}{13}-2{)}^2+(\frac{-22}{13}-3{)}^2}$

$=\sqrt{{\left(\frac{11}{13}\right)}^2+{\left(\frac{7}{13}\right)}^2+{\left(\frac{-61}{13}\right)}^2}$

$=4.8$ sq. units