### Instruction: Consider two lines: line $l$ parametrized as
\begin{align*} 
x &= 1 + 4t,\\
y &=  4 + 3t
\end{align*}and the line $m$ parametrized as
\begin{align*} 
x &=-5  + 4s\\
y &=  6 + 3s.
\end{align*}Let $A$ be a point on line $l$, $B$ be a point on line $m$, and let $P$ be the foot of the perpendicular from $A$ to line $m$.

Then $\overrightarrow{PA}$ is the projection of $\overrightarrow{BA}$ onto some vector $\begin{pmatrix} v_1\\v_2\end{pmatrix}$ such that $v_1+v_2 = 2$. Find $\begin{pmatrix}v_1 \\ v_2 \end{pmatrix}$.### Response: I need to find a vector $\begin{pmatrix} v_1\\v_2\end{pmatrix}$ that is perpendicular to line $m$, since that is the direction of the projection of $\overrightarrow{BA}$ onto $\overrightarrow{PA}$.
A vector that is perpendicular to line $m$ is also parallel to the normal vector of line $m$, which is a vector that is orthogonal to the direction vector of line $m$.
The direction vector of line $m$ is $\begin{pmatrix} 4\\3 \end{pmatrix}$, so a normal vector of line $m$ is $\begin{pmatrix} -3\\4 \end{pmatrix}$, or any scalar multiple of it.
I can choose any such multiple, but I want one that satisfies $v_1+v_2 = 2$.
If I let $v_1 = -3k$ and $v_2 = 4k$, where $k$ is some scalar, then I have $v_1+v_2 = -3k+4k = k = 2$.