### Instruction: Find the reflection of $\begin{pmatrix} 0 \\ 4 \end{pmatrix}$ over the vector $\begin{pmatrix} 1 \\ 3 \end{pmatrix}.$.### Response: I want to find the point that is symmetric to $\begin{pmatrix} 0 \\ 4 \end{pmatrix}$ across the line that passes through the origin and has the same direction as $\begin{pmatrix} 1 \\ 3 \end{pmatrix}.$
To do that, I can use the formula for the projection of a vector onto another vector, which is $\text{proj}_{\vec{v}}(\vec{u}) = \frac{\vec{u} \cdot \vec{v}}{\vec{v} \cdot \vec{v}} \vec{v}.$
The projection of $\begin{pmatrix} 0 \\ 4 \end{pmatrix}$ onto $\begin{pmatrix} 1 \\ 3 \end{pmatrix}$ is $\frac{\begin{pmatrix} 0 \\ 4 \end{pmatrix} \cdot \begin{pmatrix} 1 \\ 3 \end{pmatrix}}{\begin{pmatrix} 1 \\ 3 \end{pmatrix} \cdot \begin{pmatrix} 1 \\ 3 \end{pmatrix}} \begin{pmatrix} 1 \\ 3 \end{pmatrix} = \frac{12}{10} \begin{pmatrix} 1 \\ 3 \end{pmatrix} = \begin{pmatrix} 1.2 \\ 3.6 \end{pmatrix}.$
The reflection of $\begin{pmatrix} 0 \\ 4 \end{pmatrix}$ over the line is the point that is twice as far from the projection as $\begin{pmatrix} 0 \\ 4 \end{pmatrix}$ is, in the opposite direction.
So I can subtract $\begin{pmatrix} 0 \\ 4 \end{pmatrix}$ from $\begin{pmatrix} 1.2 \\ 3.6 \end{pmatrix}$ and then add the result to $\begin{pmatrix} 1.2 \\ 3.6 \end{pmatrix}$ again, to get the reflection.
That is, $\begin{pmatrix} 1.2 \\ 3.6 \end{pmatrix} + (\begin{pmatrix} 1.2 \\ 3.6 \end{pmatrix} - \begin{pmatrix} 0 \\ 4 \end{pmatrix}) = \begin{pmatrix} 2.4 \\ 3.2 \end{pmatrix}.$