Image of a point in a plane

The image of a point in a plane is the reflection of the point in that plane. It lies on the normal through the point, at the same distance on the other side, and the foot of the perpendicular is the mid-point of the point and its image.

Let P(2, −1, 3) and the plane be 3x − 2y − z − 9 = 0. Its normal direction is (3, −2, −1).

Step 1: Normal line through P

Any point of the normal through P is (2 + 3t, −1 − 2t, 3 − t).

Step 2: Foot of the perpendicular

Put it in the plane:

3(2+3t) − 2(−1−2t) − (3−t) − 9 = 0
6 + 9t + 2 + 4t − 3 + t − 9 = 0
14t − 4 = 0   →   t = 2/7

Foot F = (2 + 6/7, −1 − 4/7, 3 − 2/7) = (20/7, −11/7, 19/7).

Check: 3(20/7) − 2(−11/7) − 19/7 = (60 + 22 − 19)/7 = 9, so F is on the plane.

Step 3: Image

F is the mid-point of P and its image Q, so the image is given by t = 4/7 on the normal:

Q = (2 + 12/7, −1 − 8/7, 3 − 4/7) = (26/7, −15/7, 17/7)

Check: the mid-point of P and Q is ((2+26/7)/2, (−1−15/7)/2, (3+17/7)/2) = (20/7, −11/7, 19/7) = F.

Answer: the image is (26/7, −15/7, 17/7).