Regarding solving for the coordinates of reflection points in 3D space

Question

In space, there exists a smooth plane, and additionally, there are two points, denoted as $A (x_a, y_a, z_a)$ and $B (x_b, y_b, z_b)$. A beam of light is emitted from point A, reflects off the mirror surface of the smooth plane at a reflection point P, and finally arrives at point B. Now, we can obtain the mirror image point of B with respect to the smooth plane, denoted as $B' = (x_b', y_b', z_b')$. How can I calculate the expression for point P in the most elegant and simple way?

Answer 1

Consider this figure.DP is perpendicular to plane FG and B' is the mirror of B about the plane. Fundamental optic theorem says for reflection we must have:

$\angle APD=\angle BPD$

To find coordinate of P we connect B' to A; P is the intersection of line AB' and plane FG. I show calculations in this example:

Let:

Equation of the plane: $Ax+By+Cz+D=0$

The coordinates of mirror of $B:(x_0, y_0, z_0) is:

$B': (x_0-2A\lambda, y_0-2B\lambda, Z_0-2C\lambda)$

$\lambda=\frac{Ax_0+By_0+Cz_0+D}{A^2+B^2+C^2}$

Example:

Plane: $2x+y+z+4=0$

$A:(1, -1, 2)$

$B: (0, 1, 1)$

Using the formula you get:

$\lambda=\frac{2\times 0+1\times 1+1\times 1 +4}{2^2+1^2+1^2}=1$

$B': (-4, -1, -1)$

Equation of line AB':

$\frac{x+4}5=\frac{y+1}{-2}=\frac{z+1}3=t\space\space\space\space\space\space\space\space(1)$

finding x, y and z in terms of t and putting in the equation of the plane we finally get:

$t=\frac 6{11}$

Putting t in (1) gives:

$P: ( \frac {-32}{11}, \frac {-23}{11}, \frac 7{11})$

Update: If you can obtain the coordinate of B' then you can find the equation of plan as follows:

1): the components of the direction vector of line BB' is :

$(l, m, n)=(x_b-x'_b, y_b-y'_b, z_b-z'_b)$

This is the normal of the plane. The point the plane passes is the mid point of BB'; lets denote it as M:

$(x_M, y_M, z_M)=(\frac{x_b+x'_b}2, \frac{y_b+y'_b}2, \frac{z_b+z'_b}2)$

Hence the equation of the plane will be:

$l(x-x_M)+m(y-y_M)+n(z-z_m)=0$

and the equation of line AB' is:

$\frac{x-x'_b}{x_a-x'_b}=\frac{y-y'_b}{y_a-y'_b}=\frac{z-z'_b}{z_a-z'_b}=t$

P will be the intersection of AB' and the plane. The rule to find the coordinates of P is as I said in the example.