### Instruction: When the vectors $\begin{pmatrix} 4 \\ 1 \end{pmatrix}$ and $\begin{pmatrix} -1 \\ 3 \end{pmatrix}$ are both projected onto the same vector $\mathbf{v},$ the result is $\mathbf{p}$ in both cases.  Find $\mathbf{p}.$.### Response: To project a vector $\mathbf{u}$ onto another vector $\mathbf{v},$ we use the formula $\frac{\mathbf{u} \cdot \mathbf{v}}{\mathbf{v} \cdot \mathbf{v}} \mathbf{v}.$
So, to find $\mathbf{p},$ we need to find $\mathbf{v}$ such that $\frac{\begin{pmatrix} 4 \\ 1 \end{pmatrix} \cdot \mathbf{v}}{\mathbf{v} \cdot \mathbf{v}} \mathbf{v} = \frac{\begin{pmatrix} -1 \\ 3 \end{pmatrix} \cdot \mathbf{v}}{\mathbf{v} \cdot \mathbf{v}} \mathbf{v}.$
This implies that $\begin{pmatrix} 4 \\ 1 \end{pmatrix} \cdot \mathbf{v} = \begin{pmatrix} -1 \\ 3 \end{pmatrix} \cdot \mathbf{v}.$
Writing $\mathbf{v} = \begin{pmatrix} x \\ y \end{pmatrix},$ this becomes $4x + y = -x + 3y.$
Simplifying, we get $5x = 2y.$
We can choose any nonzero vector $\mathbf{v}$ that satisfies this equation, for example $\mathbf{v} = \begin{pmatrix} 2 \\ 5 \end{pmatrix}.$
Then, using the projection formula, we get $\mathbf{p} = \frac{\begin{pmatrix} 4 \\ 1 \end{pmatrix} \cdot \begin{pmatrix} 2 \\ 5 \end{pmatrix}}{\begin{pmatrix} 2 \\ 5 \end{pmatrix} \cdot \begin{pmatrix} 2 \\ 5 \end{pmatrix}} \begin{pmatrix} 2 \\ 5 \end{pmatrix} = \frac{13}{29} \begin{pmatrix} 2 \\ 5 \end{pmatrix}.$
We can check that this also works for the other vector, as $\frac{\begin{pmatrix} -1 \\ 3 \end{pmatrix} \cdot \begin{pmatrix} 2 \\ 5 \end{pmatrix}}{\begin{pmatrix} 2 \\ 5 \end{pmatrix} \cdot \begin{pmatrix} 2 \\ 5 \end{pmatrix}} \begin{pmatrix} 2 \\ 5 \end{pmatrix} = \frac{13}{29} \begin{pmatrix} 2 \\ 5 \end{pmatrix}.$