Question

The reflection of the point $\overrightarrow{a}$ in the plane $\overrightarrow{r}\cdot \overrightarrow{n}=q$ is

• A. $\overrightarrow{a}+\frac{(\overrightarrow{q}-\overrightarrow{a}\cdot \overrightarrow{n})}{|\overrightarrow{n}|}$
• B. $\overrightarrow{a}+\frac{2(\overrightarrow{q}-\overrightarrow{a}\cdot \overrightarrow{n})}{|\overrightarrow{n}{|}^2}\overrightarrow{n}$
• C. $\overrightarrow{a}+\frac{2(\overrightarrow{q}-\overrightarrow{a}\cdot \overrightarrow{n})}{|\overrightarrow{n}|}\overrightarrow{n}$
• D. none of these

Answer 1

The correct option is B $\overrightarrow{a}+\frac{2(\overrightarrow{q}-\overrightarrow{a}\cdot \overrightarrow{n})}{|\overrightarrow{n}{|}^2}\overrightarrow{n}$

Plane is $\overrightarrow{r}.\overrightarrow{n}=q...\left(1\right)$
Let the image of $A\left(\overrightarrow{a}\right)$ in the plane be $B\left(\overrightarrow{b}\right)$.
Equation of $AC$ is $\overrightarrow{r}=\overrightarrow{a}+\lambda \overrightarrow{n}...\left(2\right)$
($\because AC$ is normal to the plane)

Solving $\left(1\right)$ and $\left(2\right),$ we get
$(\overrightarrow{a}+\lambda \overrightarrow{n}).\overrightarrow{n}=q$
$\Rightarrow \lambda =\frac{q-\overrightarrow{a}.\overrightarrow{n}}{|\overrightarrow{n}{|}^2}$
$\therefore \overrightarrow{AC}=\overrightarrow{a}+\frac{(q-\overrightarrow{a}.\overrightarrow{n})}{|\overrightarrow{n}{|}^2}.\overrightarrow{n}$
But $\overrightarrow{AC}=\frac{\overrightarrow{a}+\overrightarrow{b}}{2}$
$\therefore \overrightarrow{a}+\frac{(q-\overrightarrow{a}.\overrightarrow{n})\overrightarrow{n}}{|\overrightarrow{n}{|}^2}=\frac{\overrightarrow{a}+\overrightarrow{b}}{2}$
or $\overrightarrow{b}=\overrightarrow{a}+2\left(\frac{q-\overrightarrow{a}.\overrightarrow{n}}{|\overrightarrow{n}{|}^2}\right)\overrightarrow{n}$