Linear Algebra - direction of ball after bounce off a plane In an orthonormal system, a ball is thrown from a point  =(2,6,5) towards a plane with
equation −=−2, in such a way that after it bounces off the plane, it passes through the point =(3,3,7). What is the ball's direction after the bounce.
This question has been posed here before but with a wrong answer.
Reflection of a line through a plane: ball bouncing
I had managed to solve the problem a few weeks ago obtaining the correct answer which is
that the ball's direction vector is  = (1,-1,0). But somehow I can't remember how I solved it
last time. Although, I remember being able to find the point of impact on the plane, which I
believe is the key to solving the problem.
At the moment, my line of thinking is:
The plane is normal vector is (1,0,-1). We could consider the reflection point ′ on the other side of the plane which maybe could be calculated as as a multiple of the plane's normal that extends to ′. The parameter equation of the line passing through  =(2,6,5) and normal to the plane is (x,y,z) = (2,6,5) + t(1,0,-1).  => x = 2 + t, y = 6, z = 5 - t.
For a certain t, this line will cross the plane at which point the equation of the plane must be satisfied, thus  (2+t) - (5-t) = -2  => t = t = 0.5. So it crosses the plane at
point M = (2.5,6,4.5), and for double the t amount we get the reflection point ′ = (3,6,4)
From this I thought maybe I can get the ball's direction vector by  = −′=(0,−3,3).
But I know is wrong. Last time constructing the direction vector  by
taking v = Q - 'point-of-impact' gave me the correct answer.
 A: I strongly believe that your answer is correct and the given answer is incorrect.
We can check the given solution as follows. If the direction vector is $(1,-1,0)$, then the trajectory after the bounce follows the line
$$
L(t) = Q + t(1,-1,0) = (3 + t,3-t,7).
$$
The point where the ball would bounce of the plane satisfies
$$
(3 + t) - 7 = -2 \implies t = 2 \implies L(t) = (5,1,7).
$$
From here, we already have an indication something is wrong: the solution $t \geq 0$ implies that we hit the plane after continuing along the direction of travel rather than by going backwards.
In any case, the direction from the point $A = (5,1,7)$ to $Q$ is $(-1,1,0)$, and the direction from $A$ to $P$ is $(2,6,5) - (5,1,7) = (-3,5,-2)$.
For this to make physical sense, the angle between $(-1,1,0)$ and the normal should be half the angle between $(-1,1,0)$ and $(-3,5,-2)$.  The first angle is given by
$$
\cos \theta = \frac{(-1,1,0) \cdot (1,0,-1)}{\sqrt{2}\cdot{\sqrt{2}}} = \frac{-1}{2}
$$
so $\theta  = 120^\circ$, which is already problematic. The other angle satisfies
$$
\cos \theta = \frac{(-1,1,0) \cdot (-3,5,-2)}{\sqrt{2}\cdot{\sqrt{38}}} = \frac{8}{2 \sqrt{17}},
$$
so $\theta \approx 14.04^\circ$. So, nothing about the solution makes sense.

An alternative way to find $P'$:
We note that $A = (0,0,2)$ is a point on the plane; any point from the plane will do; this on was computed by setting $x = y = 0$. To calculate the projected point $P'$, first calculate the projection of $P - A$ onto the plane's normal as follows. Let $n$ denote the normal vector $n = (1,0,-1)$.  The projection of $P - A$ onto the plane is given by
$$
v = \operatorname{proj}_n (P - A) = 
\frac{(P - A) \cdot n}{n \cdot n} n 
= \frac{(2,6,3) \cdot (1,0,-1)}{n \cdot n} (1,0,-1)
\\ = \frac{-1}{2}\cdot (1,0,-1) = (-1/2,0,1/2).
$$
Since $P - v$ is the projection of $P$ onto the plane, $P - 2v$ is the reflection of $P$ across the plane. That is,
$$
P' = P - 2v = (2,6,5) - 2(-1/2,0,1/2) = (3,6,4).
$$
