How to find reflection point over the line?

Question

The $P$ point is $(2,-1,3)$ and the line is $$l: \begin{cases} x-y-z=0,\\ x+y-2x=0 \end{cases}.$$ How to compute the reflection point $P$ over the line $l$? What is the general formula for the line going through $P$ being perpendicular to $l$?

Answer 1

A generic way of dealing with such problems is to specify the line in a parametric form (some origin point $A$ and a normalized vector $\vec{t}$ that specifies a direction). Your line is specified as an intersection of two planes through the origin, so one of the points on the line is simply the origin, and its direction is the cross product of plane normals, in that case, $(1,-1,-1)\times(1-2,1,0)$ and then you normalize it (that's assuming that $-2x$ is not a typo).

Once you have this, you can project the point onto the line:

$$N=A+((P-A)\cdot\vec{t})\vec{t}$$

where $N$ is now a perpendicular projection of $P$ onto the line ($A=0$ in your case, but I'm including it for better understanding). The dot product gives you the projection length of $P-A$, and then you just move from $A$ in the $\vec{t}$ direction by that amount.

The reflected point is then just going from $P$ to $N$, but twice too far:

$$P'=P+2(N-P)$$

You can plug the above result into this formula to get the final result.

Answer 2

The reflection of a point in a line in $\mathbb{R}^3$ is the same as the rotation by $\pi$ about the line.

So in this case, assuming your axis is $\mathbf{n}=(\frac1{\sqrt2},\frac1{\sqrt2},0)$ as you claimed in the comments, then the point $\mathbf{r}=(2,1,3)$ gives $$ -\mathbf{r}+2(\mathbf{r}\cdot\mathbf{n})\mathbf{n}=(1,2,-3). $$