Find $\text{proj}_{-2 {w}} (3 {v})$. Let ${v}$ and ${w}$ be vectors such that
$\text{proj}_{{w}} {v} = \begin{pmatrix} 4 \\ -7 \end{pmatrix}$.
Find $\text{proj}_{-2 {w}} (3 {v})$.
I don't understand where to start. I can't just find $w$ & $v$ so...
Help is appreciated!!
 A: Try to think of $\operatorname{proj}_w v$ as not just an operation you do with vectors $w$ and $v$ but as its own vector.  If you scale $v$, how does the projection change?  If you scale $w$, how does the projection change?
First, let's scale $v$ and see what happens.  Let $c$ be any scalar:
\begin{align*}
    \operatorname{proj}_w(cv) &= \frac{w\cdot(cv)}{w\cdot w} w \\
    &= \frac{c(w\cdot v)}{w\cdot w} w = c \frac{w\cdot v}{w\cdot w} w \\
    &= c \operatorname{proj}_w(v)
\end{align*}
Now, let's scale $w$:
\begin{align*}
    \operatorname{proj}_{cw}v &= \frac{(cw)\cdot v}{(cw)\cdot (cw)} (cw) \\
    &= \frac{c(w\cdot v)}{c^2 (w\cdot w)} (cw) = \frac{c^2}{c^2} \frac{w\cdot v}{w\cdot w} w \\
    &=  \operatorname{proj}_w(v)
\end{align*}
So scaling $v$ scales the projection by the same amount, while scaling $w$ doesn't change the projection at all.  This makes sense with the geometric description: we project onto a line rather than a vector, and scaling the direction vector of a line doesn't change the particular line it generates.
In your case, you need to scale both, but you can do one at a time:
$$
    \operatorname{proj}_{2w}(3v) = \operatorname{proj}_{w}(3v) 
    = 3\operatorname{proj}_{w}v
    = 3\begin{pmatrix} 4 \\ -7 \end{pmatrix}
    = \begin{pmatrix} 12 \\ -21 \end{pmatrix}
$$
A: Since $\text{proj}_{{w}} {v}=\langle 4,-7\rangle$, we have 
$$
\dfrac{v\cdot w}{w\cdot w} w= \langle 4,-7\rangle. 
$$
So 
\begin{align*}
\text{proj}_{{-2w}} {3v} 
&= \frac{3v \cdot (-2w)}{(-2w) \cdot (-2w)}(-2w) \\
&= \frac{-6(-2)}{(-2)^2}\frac{v \cdot w}{w\cdot w} w \\ 
&= 3 \frac{v \cdot w}{w\cdot w} w \\ 
&= 3 \langle 4,-7\rangle \\
&= \langle 12, -21 \rangle. 
\end{align*}
