1. ## Reflection of light

I have the following problem that i have to solve:

A light ray emitted from Point A is reflected at a mirror and then passes through the Point B.
The plane of the mirror is given by the Plane E. Determine the point on the mirror at which the light ray is reflected.
Given:
Point A: (0,-2,1)
Point B: (4,3,-5)
Plane E: x+3y+2z-24=0

2. ## Re: Reflection of light

Take the desired point to be X= (x, y, z). The vector from point A to X is xi+ (y+2)j+ (z- 1)k.
The unit vector in that direction is $\frac{1}{\sqrt{x^2+ (y+2)^2+ (z- 1)^2}}(xi+ (y+2)j+ (z- 1)k)$

The vector from point B to x is (x- 4)k+ (y-3)j+ (z+ 5)k.
The unit vector in that direction is $\frac{1}{\sqrt{(x-4)^2+ (y-3)^2+ (z+5)^2}}((x- 4)i+ (y-3)j+ (z+5)k)$

The normal to the plane is given by the vector i+ 3j+ 2k.
The unit normal is $\frac{1}{\sqrt{14}}(i+ 3j+ 2k)$

The physical law here is that the light rays must make the same angle with the perpendicular. That that means that the dot products of each of those first two unit vectors with the unit normal must be the same.

3. ## Re: Reflection of light

Hello, timole!

I agree with HallsofIvy . . . but I've hit a wall.

Following his set-up: . $A(0,\text{-}2,1),\;B(4,3,\text{-}5),\;X(x,y,z)$

Let $\vec a$ = unit vector from A to X.
Let $\vec b$ = unit vector from B to X.
Let $\vec n$ = unit normal of the plane: $\tfrac{1}{14}\langle 1,3,2\rangle$

The angle between $\vec a$ and $\vec n$ must be equal to
. . the angle between $\vec b$ and $\vec n.$

This gives us one equation . . . in three variables.
. . We need two more equations.

I found one more equation.
. . The vectors $\vec a,\,\vec b ,\,\vec n$ must be coplanar.

Now what?

4. ## Re: Reflection of light

If an object does not emit its own light (which accounts for most objects in the world), it must reflect light in order to be seen.