Question

Find the coordinates of the foot of perpendicular and length of the perpendicular drawn from the point P(5,4,2) to the line $\frac{x+1}{2}=\frac{y-3}{3}=\frac{z-1}{-1}$ also find the image of the point

Answer 1

$\frac{x+1}{2}=\frac{y-3}{3}=\frac{z-1}{-1}$ has a point $(2r-1,3r+3,-r+1)$ on it if this point is foot of perpendicular from $(5,4,2)$ then,

$(2r-6,3r-1,-r-1)$ is $\perp$ to $(2,3,-1)$

$\therefore (2r-6)2+(3r-1)3+(-r-1)(-1)=0\Rightarrow r=1$

Foot of $\perp =(1,6,0)$

Length of $\perp =\sqrt{{(5-1)}^2+{(4-6)}^2+{(2-0)}^2}=2\sqrt{6}$

image of the point be $I$ and foot of perpendicular be $F(1,6,0)$.

$\therefore$ $F$ divides $PI$ in $1:1$.

$\therefore I=(-3,8,-2)$