Question

Find the coordinates of the foot of perpendicular drawn from the point A(-1, 8, 4) to the line joining the points B(0,-1,3) and C(2, -3, -1). Hence find the image of point A in line BC.

Answer 1

Equation of line BC: $\frac{x-0}{2-0}=\frac{y-(-1)}{-3-(-1)}=\frac{z-3}{-1-3}i.e.,\frac{x}{2}=\frac{y+1}{-2}=\frac{z-3}{-4}=\lambda$ say

$\therefore$ Coordintes of random point on this line is $P(2\lambda ,-2\lambda -1,-4\lambda +3)$.

The $AP\perp BC$ so, $2(2\lambda +1)-2(-2\lambda -9)-4(-4\lambda -1)=0$

$\Rightarrow \lambda =-1\therefore P(-2,1,7)$.

Let Q(h, p, s) be the image of A in the line BC,

So P must be the mid-point of AQ.

$\therefore \therefore P\left(\begin{array}{c}\frac{h-1}{2},\frac{p+8}{2},\frac{s+4}{2}\end{array}\right)=P(-2,1,7)$

On comparing the coordinates, we get h = -3, p=-6, s= 10,

Hence the image is Q(-3, -6, 10).