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

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?

• 1. Construct the line through (1, 2, 3) perpendicular to x+ z= 2. 2, Find the point, P, where that line intersects the plane. 3. Find the distance from (1, 2, 3) to P. 4. Find the point on that line the same distance from P on the other side of the plane. – user247327 Apr 1 '17 at 13:51

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 $$A'=A+2t\,\vec n.$$

• Let me just ask you this and you correct me if I'm wrong: What we're doing here is using a direction vector $\vec n$ (plane's normal) and asking ourselves this: "We've got a direction vector which is perpendicular to the plane. Now, what we want to find out is how much we have to scale it in order to get from point A to plane $\Pi$ (and we call our real number $t$). Then we say: "OK, since the points are symetric, then we have to move just as much in the other direction." Hence the second equation. Am I right? – NumberSymphony Apr 2 '17 at 8:27
• It's as simple as that. In more technical terms, the first step consists in writing the parametric equation of the line through $A$, with directing vector $\vec n$. – Bernard Apr 2 '17 at 10:27

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: