Need help with second part please!

Question

 

Where for a) x - 3y + 4z -26 = 0

Answer 1

b)i) The plane equation is x - 3y + 4z - 26 = 0

Point (13, -1, -4)

The image point is p'.

Let p' = (x , y, z)

Find the second point which has same distance to plane.

The line is t.

Parametic equation p' = (13,-1, -4) + t(1, -3, 4)

p' = (13 + t, -1 - 3t, -4 + 4t)

Substitute the point in the plane.

(13 + t) - 3(- 1- 3t) + 4(-4 + 4t) - 26 = 0

13 + t  + 3 + 9t - 16 + 16t - 26 = 0

26t - 26 = 0

26t = 26

t = 26/26

t = 1

Therefore if you go twice as far you will get the mirror image.

P' = (13,-1, -4) + 2(1)(1, -3, 4)

P' = (13,- 1, -4) + 2(1,-3,4)

p' = (15, -7, 4)

Image point is (15, -7, 4).