Find the distance from $U$ to the $span(v)$. Let $U$ and $V$ be the points in $\mathbb{R}^4$  with position vectors $u=(1, 2, -1, -3)$ and $v=(1, 0, -2, 3)$.
Find the distance from $U$ to the $span(v)$.
I'm not quite sure if the definition $span(v)$ that I know was right so I decided to use what I know which is $span(v)=t(1, 0, -2, 3)$ wherein to find the distance, I find the norm of $span(v)$-$U$. However, I am not quite sure if I'm going the right way with this. 
 A: 
so I decided to have 

If you don't know, then you should not decide it for yourself, but check your notes!
Nevertheless, your assuption was right. For a set $A$ we define
$$
span(A)=\left\{\sum_{k=1}^n \alpha_ka_k~:~n\in\mathbb N, \alpha_k\in\mathbb R,~a_k\in A\right\}.
$$
In your case, the set contains just the vector $v$, so you can simplify it to
$$
span(v)=\{tv~:~t\in\mathbb R\}.
$$
The minimal distance between $u$ and $span(v)$ is the minimum of $f(t)=d(u,tv)$ where $d$ is your metric. If you use the euclidean metric, then you get $f(t)=\|u-tv\|_2$. But to simplify the computation, you can consider 
\begin{align}
g(t)&=(f(t))^2=\|u-tv\|_2^2=\|(1-t,2,-1-2t,-3-3t)\|_2^2\\
&=(1-t)^2+2^2+(-1-2t)^2+(-3-3t)^2
\end{align}
So, $g$ is a quadratic polynomial. The square root of the minimum of $g$ is your minimal distance, you are looking for. The rest is up to you.
A: $\vec u$ can be broken to two vectors: its image on $\vec v$ and the normal on $\vec v$. For finding the image on $\vec v$ we have$$\vec {u_{image}}=\dfrac{\vec u\cdot\vec v}{|\vec v|^2}\vec v=\dfrac{-6}{14}\vec v=(-\dfrac{3}{7},0,\dfrac{6}{7},-\dfrac{9}{7})$$also$$\vec{u_{normal}}=\vec u-\vec {u_{image}}=(\dfrac{10}{7},2,-\dfrac{13}{7},-\dfrac{12}{7})$$and the distance is$$|\vec{u_{normal}}|=\dfrac{\sqrt{609}}{7}$$
