3D billiard with no gravity nor air drag

Question

If I have a projectile shot in a vacuum-box that has no gravity or wind resistance inside it (lets say the box is 1000 units tall and 1000 units wide), with the only rule being that the projectile bounces off the sides of the box how do I solve for the angles that will hit any target inside the box from a projectile launched at any point inside the box, in less than or equal to 1 side-bounces?

Of course, shooting at any angle would eventually hit the target because the projectile would bounce around forever until finally meeting the target, I'm looking for the angles that result in a single side-bounce only.

Answer 1

Pick the side you want to bounce your projectile off of.

Reflect the box, with target, in the side selected above.

Fire from your original location at the reflection of the target. Of course, the projectile won't enter the reflection, but it will bounce of the wall through the real target.

Expanded answer for OP:

Assume the box is a cube located with one corner at the origin, with edges running to $(1000,0,0)$, $(0, 1000, 0)$, and$((0, 0, 1000)$. Further, let the target be located inside the cube at $(a, b, c)$, and the gun is located at $(x, y, z)$, also inside the box.

Suppose you want to bounce your shot off the floor of the box. We reflect the box in the floor by leaving the $a$ and $b$ values the same, and reversing the value of $c$. So find the direction to fire the gun at a point $(a, b, -c)$ and you'll hit the target on the bounce.