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 $l$?
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2Are you sure about the $x+y-2x=0$? That is a strange way to write $x-y=0$. – user10354138 Jun 13 '19 at 14:44
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Yes, I'm sure, then $l={(x,x,0), ,x\in \mathbb{R}}$ I think – zorro47 Jun 13 '19 at 14:46
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Should it not be $$x+y-2z=0$$? – Dr. Sonnhard Graubner Jun 13 '19 at 14:49
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That's almost surely a typo in that second plane equation: there's no justification at that level to write $;-x+y=0;$ in such an absurd way. – DonAntonio Jun 13 '19 at 15:33
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There is an infinite number of lines through $P$ that are perpendicular to $l$. I suspect that you also want this line to intersect $l$. – amd Jun 13 '19 at 18:49
2 Answers
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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 $$ -\mathbf{r}+2(\mathbf{r}\cdot\mathbf{n})\mathbf{n}=(1,2,-3). $$
user10354138
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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.
orion
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