### 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: I know that the projection of a vector $\mathbf{a}$ onto another vector $\mathbf{b}$ is given by the formula $\frac{\mathbf{a} \cdot \mathbf{b}}{\|\mathbf{b}\|^2} \mathbf{b},$ where $\mathbf{a} \cdot \mathbf{b}$ is the dot product and $\|\mathbf{b}\|$ is the magnitude of $\mathbf{b}.$
So, to find $\mathbf{p},$ I need to find $\mathbf{v}$ first, and then apply the projection formula to either $\begin{pmatrix} 4 \\ 1 \end{pmatrix}$ or $\begin{pmatrix} -1 \\ 3 \end{pmatrix}.$
Since both vectors have the same projection onto $\mathbf{v},$ it must be that their difference is orthogonal to $\mathbf{v}.$
That is, $(\begin{pmatrix} 4 \\ 1 \end{pmatrix} - \begin{pmatrix} -1 \\ 3 \end{pmatrix}) \cdot \mathbf{v} = 0.$
Simplifying, I get $(\begin{pmatrix} 5 \\ -2 \end{pmatrix}) \cdot \mathbf{v} = 0.$
Let $\mathbf{v} = \begin{pmatrix} x \\ y \end{pmatrix}.$
Then, I have $5x - 2y = 0,$ or $y = \frac{5}{2} x.$
This means that $\mathbf{v}$ is a scalar multiple of $\begin{pmatrix} 2 \\ 5 \end{pmatrix}.$
That is, $\mathbf{v} = k \begin{pmatrix} 2 \\ 5 \end{pmatrix}$ for some nonzero constant $k.$
The value of $k$ does not affect the projection, so I can choose any convenient value.
For simplicity, I will take $k = 1.$
So, $\mathbf{v} = \begin{pmatrix} 2 \\ 5 \end{pmatrix}.$
Now, to find $\mathbf{p},$ I can use the projection formula with either $\begin{pmatrix} 4 \\ 1 \end{pmatrix}$ or $\begin{pmatrix} -1 \\ 3 \end{pmatrix}.$
I will use the former.
I have $\mathbf{p} = \frac{\begin{pmatrix} 4 \\ 1 \end{pmatrix} \cdot \begin{pmatrix} 2 \\ 5 \end{pmatrix}}{\|\begin{pmatrix} 2 \\ 5 \end{pmatrix}\|^2} \begin{pmatrix} 2 \\ 5 \end{pmatrix}.$
Evaluating the dot product, I get $\begin{pmatrix} 4 \\ 1 \end{pmatrix} \cdot \begin{pmatrix} 2 \\ 5 \end{pmatrix} = 4 \times 2 + 1 \times 5 = 13.$
Evaluating the magnitude, I get $\|\begin{pmatrix} 2 \\ 5 \end{pmatrix}\| = \sqrt{2^2 + 5^2} = \sqrt{29}.$
Simplifying the fraction, I get $\frac{13}{\sqrt{29}^2} = \frac{13}{29}.$