### 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 $\mathbf{v},$ I need to use the formula $\text{proj}_{\mathbf{v}} \mathbf{u} = \frac{\mathbf{u} \cdot \mathbf{v}}{\mathbf{v} \cdot \mathbf{v}} \mathbf{v}.$
Since the projection of both $\begin{pmatrix} 4 \\ 1 \end{pmatrix}$ and $\begin{pmatrix} -1 \\ 3 \end{pmatrix}$ onto $\mathbf{v}$ is the same vector $\mathbf{p},$ I can set the formulas for the projections equal to each other and try to find $\mathbf{v}.$
So, I have $\frac{\begin{pmatrix} 4 \\ 1 \end{pmatrix} \cdot \mathbf{v}}{\mathbf{v} \cdot \mathbf{v}} \mathbf{v} = \frac{\begin{pmatrix} -1 \\ 3 \end{pmatrix} \cdot \mathbf{v}}{\mathbf{v} \cdot \mathbf{v}} \mathbf{v}.$
Simplifying, I get $\begin{pmatrix} 4 \\ 1 \end{pmatrix} \cdot \mathbf{v} = \begin{pmatrix} -1 \\ 3 \end{pmatrix} \cdot \mathbf{v}.$
I can write $\mathbf{v}$ as $\begin{pmatrix} x \\ y \end{pmatrix}$ and expand the dot products to get $4x + y = -x + 3y.$
Rearranging, I get $5x - 2y = 0.$
This is a linear equation in $x$ and $y,$ so I can find infinitely many solutions for $\mathbf{v}.$
One possible solution is $\mathbf{v} = \begin{pmatrix} 2 \\ 5 \end{pmatrix}.$
To check if this works, I can plug it into the projection formula and see if I get the same $\mathbf{p}$ for both vectors.
For $\begin{pmatrix} 4 \\ 1 \end{pmatrix},$ I get $\text{proj}_{\begin{pmatrix} 2 \\ 5 \end{pmatrix}} \begin{pmatrix} 4 \\ 1 \end{pmatrix} = \frac{\begin{pmatrix} 4 \\ 1 \end{pmatrix} \cdot \begin{pmatrix} 2 \\ 5 \end{pmatrix}}{\begin{pmatrix} 2 \\ 5 \end{pmatrix} \cdot \begin{pmatrix} 2 \\ 5 \end{pmatrix}} \begin{pmatrix} 2 \\ 5 \end{pmatrix} = \frac{13}{29} \begin{pmatrix} 2 \\ 5 \end{pmatrix}.$
For $\begin{pmatrix} -1 \\ 3 \end{pmatrix},$ I get $\text{proj}_{\begin{pmatrix} 2 \\ 5 \end{pmatrix}} \begin{pmatrix} -1 \\ 3 \end{pmatrix} = \frac{\begin{pmatrix} -1 \\ 3 \end{pmatrix} \cdot \begin{pmatrix} 2 \\ 5 \end{pmatrix}}{\begin{pmatrix} 2 \\ 5 \end{pmatrix} \cdot \begin{pmatrix} 2 \\ 5 \end{pmatrix}} \begin{pmatrix} 2 \\ 5 \end{pmatrix} = \frac{13}{29} \begin{pmatrix} 2 \\ 5 \end{pmatrix}.$
So, indeed, the projection is the same for both vectors, and it is $\mathbf{p} = \frac{13}{29} \begin{pmatrix} 2 \\ 5 \end{pmatrix}.$