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 $\mathbf{p}.$
Note that the vector $\mathbf{p}$ must lie on the line passing through $\begin{pmatrix} 4 \\ 1 \end{pmatrix}$ and $\begin{pmatrix} -1 \\ 3 \end{pmatrix}.$  This line can be parameterized by
\[\begin{pmatrix} 4 \\ 1 \end{pmatrix} + t \left( \begin{pmatrix} -1 \\ 3 \end{pmatrix} - \begin{pmatrix} 4 \\ 1 \end{pmatrix} \right) = \begin{pmatrix} 4 \\ 1 \end{pmatrix} + t \begin{pmatrix} -5 \\ 2 \end{pmatrix} = \begin{pmatrix} -5t + 4 \\ 2t + 1 \end{pmatrix}.\][asy]
usepackage("amsmath");

unitsize(1 cm);

pair A, B, O, P;

A = (4,1);
B = (-1,3);
O = (0,0);
P = (O + reflect(A,B)*(O))/2;

draw((-2,0)--(5,0));
draw((0,-1)--(0,4));
draw(O--A,Arrow(6));
draw(O--B,Arrow(6));
draw(O--P,Arrow(6));
draw(interp(A,B,-0.1)--interp(A,B,1.1),dashed);

label("$\begin{pmatrix} 4 \\ 1 \end{pmatrix}$", A, N);
label("$\begin{pmatrix} -1 \\ 3 \end{pmatrix}$", B, N);
label("$\mathbf{p}$", P, N);
[/asy]

The vector $\mathbf{p}$ itself will be orthogonal to the direction vector $\begin{pmatrix} -5 \\ 2 \end{pmatrix},$ so
\[\begin{pmatrix} -5t + 4 \\ 2t + 1 \end{pmatrix} \cdot \begin{pmatrix} -5 \\ 2 \end{pmatrix} = 0.\]Hence, $(-5t + 4)(-5) + (2t + 1)(2) = 0.$  Solving, we find $t = \frac{18}{29}.$  Hence, $\mathbf{p} = \boxed{\begin{pmatrix} 26/29 \\ 65/29 \end{pmatrix}}.$
\begin{pmatrix}26/29\65/29\end{pmatrix}
Level 4
Precalculus
Problem: 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 $\mathbf{p}.$
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To find the projection of the vectors onto a common vector $\mathbf{v}$, we start by recalling the formula for the projection of a vector $\mathbf{a}$ onto another vector $\mathbf{b}$, which is given by:

\[
\text{proj}_{\mathbf{b}} \mathbf{a} = \frac{\mathbf{a} \cdot \mathbf{b}}{\mathbf{b} \cdot \mathbf{b}} \mathbf{b}
\]

Let $\mathbf{a_1} = \begin{pmatrix} 4 \\ 1 \end{pmatrix}$ and $\mathbf{a_2} = \begin{pmatrix} -1 \\ 3 \end{pmatrix}$. We denote the common projection result as $\mathbf{p}$. 

Since both vectors project to the same vector $\mathbf{p}$, we can express this as:

\[
\text{proj}_{\mathbf{v}} \mathbf{a_1} = \text{proj}_{\mathbf{v}} \mathbf{a_2} = \mathbf{p}
\]

This leads to the equations:

\[
\frac{\mathbf{a_1} \cdot \mathbf{v}}{\mathbf{v} \cdot \mathbf{v}} \mathbf{v} = \mathbf{p}
\]
\[
\frac{\mathbf{a_2} \cdot \mathbf{v}}{\mathbf{v} \cdot \mathbf{v}} \mathbf{v} = \mathbf{p}
\]

Setting these two expressions for $\mathbf{p}$ equal gives:

\[
\frac{\mathbf{a_1} \cdot \mathbf{v}}{\mathbf{v} \cdot \mathbf{v}} \mathbf{v} = \frac{\mathbf{a_2} \cdot \mathbf{v}}{\mathbf{v} \cdot \mathbf{v}} \mathbf{v}
\]

Assuming $\mathbf{v} \neq \mathbf{0}$, we can cancel $\mathbf{v}$ from both sides, leading to:

\[
\mathbf{a_1} \cdot \mathbf{v} = \mathbf{a_2} \cdot \mathbf{v}
\]

This implies:

\[
\begin{pmatrix} 4 \\ 1 \end{pmatrix} \cdot \mathbf{v} = \begin{pmatrix} -1 \\ 3 \end{pmatrix} \cdot \mathbf{v}
\]

Let $\mathbf{v} = \begin{pmatrix} v_1 \\ v_2 \end{pmatrix}$. Then we have:

\[
4v_1 + 1v_2 = -1v_1 + 3v_2
\]

Rearranging gives:

\[
5v_1 - 2v_2 = 0 \quad \Rightarrow \quad v_2 = \frac{5}{2}v_1
\]

Now, we can express $\mathbf{v}$ in terms of a scalar $k$:

\[
\mathbf{v} = \begin{pmatrix} k \\ \frac{5}{2}k \end{pmatrix}
\]

Next, we can substitute $\mathbf{v}$ back into the projection formula to find $\mathbf{p}$. We can choose $k = 2$ for simplicity, giving:

\[
\mathbf{v} = \begin{pmatrix} 2 \\ 5 \end{pmatrix}
\]

Now we compute the projections:

1. For $\mathbf{a_1}$:

\[
\mathbf{a_1} \cdot \mathbf{v} = 4 \cdot 2 + 1 \cdot 5 = 8 + 5 = 13
\]
\[
\mathbf{v} \cdot \mathbf{v} = 2^2 + 5^2 = 4 + 25 = 29
\]
\[
\text{proj}_{\mathbf{v}} \mathbf{a_1} = \frac{13}{29} \begin{pmatrix} 2 \\ 5 \end{pmatrix} = \begin{pmatrix} \frac{26}{29} \\ \frac{65}{29} \end{pmatrix}
\]

2. For $\mathbf{a_2}$:

\[
\mathbf{a_2} \cdot \mathbf{v} = -1 \cdot 2 + 3 \cdot 5 = -2 + 15 = 13
\]
\[
\text{proj}_{\mathbf{v}} \mathbf{a_2} = \frac{13}{29} \begin{pmatrix} 2 \\ 5 \end{pmatrix} = \begin{pmatrix} \frac{26}{29} \\ \frac{65}{29} \end{pmatrix}
\]

Both projections yield the same result, confirming that $\mathbf{p} = \begin{pmatrix} \frac{26}{29} \\ \frac{65}{29} \end{pmatrix}$.<|END_OF_LATENT|> The final answer is $\begin{pmatrix}26/29\65/29\end{pmatrix}$.