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 the point A in the line BC.

#### Solution

Let P be the foot of the perpendicular drawn from point A on the line joiningpoints B and C.

Let P’ (a, b, c) be the coordinates of image of point A.

Equation of line BC is given by,

`(x-x_1)/(x_2-x_1)=(y-y_1)/(y_2-y_1)=(z-z_1)/(z_2-z_1)`

`(x-0)/2=(y+1)/(-2)=(z-3)/(-4)=lambda`

General coordinates of P is (2λ,-2λ-1,-4λ+3)

Direction ratios of AP (2λ+1,-2λ-9,-4λ-1)

AP⊥BC

2(2λ+1)-2(2λ-9)-4(-4λ-1)=0

4λ+2+4λ+18+16λ+4=0

24+24λ=0

λ=-1

P(-2,1,7)

Coordinates of foot of perpendicular is (-2,1,7)

Coordinates of image of A is P' (a, b, c) is

`(a-1)/2=-2, a=-3`

`(b+8)/2=1, b= -6`

`(c+4)/2=7, c=10`

P'(-3,-6,10)