Let A 1 1 2 and 5 6 12 12 Find SAb Here SA is the reflecti

Let A = [-1, 1, 2] and 5 = [6, -12, 12]. Find S_A(b). Here S_A is the reflection about the plane perpendicular to A such that (0, 0, 0) belongs to the plane.

Solution

plane prpendiculat to AB and passing through A

n = (6, -12, 12) -(-1,1,2) = (7,-13, 10)

plane satisfying the above condition will be

7x -13y +10z +d =0

since passing through (0,0,0)

7*0 -13*0 +0 +d =0

d =0

hence equation of plane is

7x -13y +10z =0 eqn 1

let Sab be reflection of point B

equastion of line perpendicular to B

x-6/7 = y+12/-13 = z-12/10 equation number 2

solving 1 and 2 we get x , y , z

x=-2, y=2, z=2

now

a+6/2 = -2

a =-4-6 =-10

(b-12)/2 = 2

b -12 =4

b =16

c-12/2 = 2

c =4+12 = 16

hence reflection point is -10, 16,16