How to determine the coordinates of a point symmetrical to a plane?

Question

I have a plane $\Pi ... x + z = 2$ and a point $A(1,2,3)$. I have to find a symetrical point $A'$ in regard to the plane $\Pi$.

I don't know how to solve this task. Can someone guide me, step-by-step?

Answer 1

Hint:

Let $\vec n$ be a vector normal to the plane (How do you get it from the equation of $\Pi$?). First, determine the parameter $t$ so that $A+t\,\vec n\in\Pi$. The symmetrical of $A$ is the point $$A'=A+2t\,\vec n.$$

Answer 2

The Plan:

One has a point $X$ of $\Pi$ which is closest to $A$.

This determines a line $L$ through $X$ and $A$.

One then chooses that point $B$ on the line which has the same distance $d = \lVert XA\rVert = \lVert XB\rVert$ but is different from $A$, thus on the other side on the plane.

Execution:

A direction vector of line $L$ is parallel to a normal vector of $\Pi$.

The equation for plane $\Pi$ can be written as $$ 2 = x + z = (1,0,1)^T \cdot (x,y,z)^T = n \cdot u $$ which is called normal form of the plane.

The line $L$ can be written as $$ u = t \, n + A \quad (t \in \mathbb{R}) $$ where $n$ is a direction vector of $L$, which we choose to be the normal vector from the equation for $\Pi$. $A$ is featuring both as point of the affine space and as vector $OA$ from the origin $O$ to the point $A$.

So $X \in \Pi$ fulfills $$ 2 = n \cdot X = n \cdot (t \, n + A) = t \lVert n \rVert^2 + n \cdot A \iff \\ t = \frac{2 - n \cdot A}{\lVert n \rVert^2} $$ which gives \begin{align} X &= \frac{2 - n \cdot A}{\lVert n \rVert^2} n + A \\ &= \frac{2 - (1,0,1)^T \cdot (1,2,3)^T}{\lVert (1,0,1)^T \rVert^2} (1,0,1) + (1,2,3)^T \\ &= -(1,0,1)^T + (1,2,3)^T \\ &= (0,2,2)^T \end{align} and $$ d = \lVert XA \rVert = \lVert X - A \rVert = \frac{\lvert 2 - n \cdot A\rvert}{\lVert n \rVert} = \sqrt{2} $$ We now write the line via a new parameter $s$ as $$ u = s \, r + X $$ where $r$ is a unit normal vector, which means we parameterize by the signed distance to $X$.

We want $$ A = d \, r + X = \frac{\lvert 2 - n \cdot A\rvert}{\lVert n \rVert} r + \frac{2 - n \cdot A}{\lVert n \rVert^2} n + A \iff \\ $$ \begin{align} r &= -\frac{2 - n \cdot A}{\lVert n \rVert^2} n \frac{\lVert n \rVert}{\lvert 2 - n \cdot A\rvert} \\ &= \frac{n \cdot A - 2}{\lvert n \cdot A - 2\rvert} \frac{n}{\lVert n \rVert} \\ &= \frac{1}{\sqrt{2}} (1,0,1)^T \end{align} So $$ \DeclareMathOperator{sgn}{sgn} u = s \sgn(n \cdot A - 2) \frac{n}{\lVert n \rVert} + X $$ and \begin{align} B &= -d \sgn(n \cdot A - 2) \frac{n}{\lVert n \rVert} + X \\ &= -\sqrt{2} \frac{1}{\sqrt{2}} (1,0,1)^T + (0,2,2)^T \\ &= (-1, 2, 1)^T \end{align}

Here is a visualization: