Find a 2d unit vector perpendicular to a given vector So I'm given the vector 5i-12j and I need to find a unit vector perpendicular to this line. I know I need to use the dot product in some way, shape, or form, but I just can't figure it out. 
Thank you for your help.
 A: given a non zero vector
$\vec{u}=(a,b)=a\vec{i}+b\vec{j}$
the vector  $\vec{v}=(b,\color{red}{-}a)$ is perpendicular to  $\vec{u}$.
the unit vector is  then obtained by dividing by its norm
$||\vec{v}||=\sqrt{a^2+b^2}$.
so, the vector you seek is
$$\left(\frac{b}{||\vec{v}||},\frac{\color{red}{-}a}{||\vec{v}||}\right)$$.
A: To figure it out do it that way:
$$
\vec{v} =  \left[\begin{matrix}
5 \\
-12
\end{matrix}\right] \\
\vec{w} = \left[\begin{matrix}
a \\
b
\end{matrix}\right]\\
\vec{u} = \frac{1}{\sqrt{a^2 + b^2}}\left[\begin{matrix}
a \\
b
\end{matrix}\right]
$$
what you need to do is find $a$ and $b$ to get $\vec{w}$ and then compute that unit vector $\vec{u}$.
By dot product definition, you know that $cos(\theta) = 0$ when $\theta = \frac{\pi}{2}$. Hence, by it's algebric deffition, you'll get that it needs to be 0 when $\theta = \frac{\pi}{2}$.
By dot product computation, you'll get:
$$
5a -12b = 0\\
a = 12, b = 5 \rightarrow 60 - 60 - 0
$$
So your $\vec{w}$ is:
$$
\vec{w} = \left[\begin{matrix}
12\\
5
\end{matrix}\right]
$$
Now all you need to do is normalize $\vec{w}$ to get $\vec{u}$, the unit vector perpendicular to $\vec{v}$:
$$
\sqrt{12^2 + 5^2} = \\
\sqrt{144+25} = 13\\
\vec{u} = \frac{1}{13}\vec{w}\\
\vec{u} = \left[\begin{matrix}
\frac{12}{13}\\
\frac{5}{13}
\end{matrix}\right]
$$
