Image of a Point in a Plane

Let $A$ (position vector $\vec u$) be a given point and $\vec r\cdot\vec n=p$ a given plane. The mirror image (reflection) $A'$ of $A$ in the plane is the point with position vector $\vec v$ such that the plane is the perpendicular bisector of segment $AA'$.

Derivation. Since $\vec{AA'}$ is perpendicular to the plane, it is parallel to $\vec n$: $\vec{AA'}=\lambda\vec n$, i.e. $\vec v-\vec u=\lambda\vec n$, so $\vec v=\vec u+\lambda\vec n$ for some scalar $\lambda$ … (1). Let $M$ be the midpoint of $AA'$; its position vector is $\dfrac{\vec u+\vec v}{2}$, and — crucially — $M$ lies on the plane (it is in fact the foot of the perpendicular from $A$ to the plane), so $\dfrac{\vec u+\vec v}{2}\cdot\vec n=p$ … (2). Substituting (1) into (2): $\dfrac{2\vec u+\lambda\vec n}{2}\cdot\vec n=p\ \Rightarrow\ \lambda=\dfrac{2\big(p-\vec u\cdot\vec n\big)}{|\vec n|^2}$.

Substituting back into (1) gives the position vector of the image:

$\boxed{\vec v=\vec u+\frac{2\big(p-\vec u\cdot\vec n\big)}{|\vec n|^2}\,\vec n.}$ …