Question

Find the length of foot of the perpendicular from the point $(5,7,3)$ to the line $\frac{x-15}{3}=\frac{y-29}{8}=\frac{5-z}{5}$

Answer 1

The given point $P(5,7,3)$ and the line is
$\frac{x-15}{3}=\frac{y-29}{8}=\frac{z-5}{-5}$ ....(1)
Any point on it is $N(15+3r,29,+8r,5-5r)$ ...(2)
Therefore the d.r's of line $PN$ are
$(15+3r)-5,(29+8r)-7,(5-5r)-3\Rightarrow 10+3r,22+8r,2-5r$ ...(3)
Let $N$ be the foot of the perpendicular from $P$ to (1), then $PN$ is perpendicular to (1), and so we have
$3(10+3r)+8(22+8r)-5(2-5r)=0\Rightarrow r=-2$
Therefore from (2) the foot $N$ of the perpendicular $PN$ is
$(15-6,29-16,5+10)\Rightarrow (9,13,15)$
Therefore $PN=$ distance between $P(5,7,3)$ and $N(9,13,15)$
$=\sqrt{{(9-5)}^2+{(13-7)}^2{(15-3)}^2}=\sqrt{16+36+144}=14$