[Community Question] Geometry: geometry - find the path a light ray must take to reach a destination with one bounce off a mirror

One of our user asked:

this seems like it should be simple, but i've run out of leads with a similar-triangles approach, and the algebraic approaches seem pretty daunting, so i'm asking for help.

i'm working on a simulation where a ray of light reaches a point by bouncing once off a planar mirror. i know the position of the source, the destination point, and a point & normal of the mirror. (or any other convenient means of describing the mirror) For simplicity i'm happy to assume the mirror is infinite in extent.

as i mentioned, a geometric solution via similar-triangles and so on has not revealed itself to me, but i'm good at missing those.

for the algebraic approach i've set up a system of equations constraining the on-mirror point to 1) lie on the line of the mirror and 2) the angle between the incoming ray and normal and the angle between the outgoing ray and the normal are equal, by asserting that those Dot-products are equal. I haven't carried that math through to its end because it looks pretty monstrous because the thing i'm trying to solve for appears in so many places including inside many radicals.

I'm not saying i can't solve this, but it looks like a lot of work and if there's a better approach i'd love some advice.

Here's my sketch of the situation. Apologies if i mis-use terminology! Speaking of terminology, is there a word for this point ? The point of reflection ? The point of incidence ?

Calculus: Intersection of hyperplanes in $\mathbb{R}^5$

Consider the $4$ hyperplanes in $\mathbb{R}^5$ given by the equations $$x_1-x_2+x_3+x_4+2x_5=0$$ $$2x_1-x_2+6x_3+2x_4+6x_5=0$$ $$3x_1-3x_2+3x_3+4x_4+7x_5=0$$ $$x_1+x_2+9x_3+x_4+6x_5=0$$ Let $V \leq \mathbb{R}^5$ be the intersection of these hyperplanes. Find a basis for $V$ .

[Community Question] Calculus: calculating 2 constants in a function

One of our user asked: $$M=\{f\in C[0,2\pi],\int_{0}^{2\pi}f(x)sinxdx=\pi,\int_{0}^{2\pi}f(x)sin2xdx=2\pi\} $$ $a,b\in \mathbb R, g\in M, g(x)=asinx+bsin2x,x\in [0,2\pi]$ I've read on the answers that $a=1,b=2$ and I don't know how to calculate them. Can somebody explain me,please? By the way, the problem is to determine $ \int_{0}^{2\pi}(g(x))^2 dx$ so you have to first get the constants $a$ and $b$ .

Calculus: A difficulty in understanding a step in a solution.

Here is the solution: But I could not understand how the last term in the fourth line came from the line before it, could anyone explain this for me please?

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