How to find unit normal vector of a 2d line I have been given the line equation of $y-8=0$, and have to find the unit normal vector? How would I go about doing this? is there a specific equation I need to use?
Following this I have been given the question: Find the equation of a line passing through point $A(9, −6)$ and orthogonal to vector $v=[−4,0]^T$. Which i am also unsure how to complete.
Any help would be much appreciated, it's been quite a while since i have done this sort of maths.
Thanks
 A: If the line is put in the form
$$
ax+by=c\tag{1}
$$
that is equivalent to
$$
(a,b)\cdot(x,y)=c\tag{2}
$$
which says that the points on the line are perpendicular to the vector $(a,b)$. Thus, the unit normals would be
$$
\pm\frac{(a,b)}{|(a,b)|}\tag{3}
$$
In the case of $y-8=0$, you get $0x+1y=8$, so the unit normals would be
$$
\pm\frac{(0,1)}{|(0,1)|}=\pm(0,1)\tag{4}
$$

Reversing the process in $(1)$ and $(2)$, we get that
$$
-4x+0y=C\tag{5}
$$
is perpendicular to $\begin{bmatrix}-4&0\end{bmatrix}^T$. Plug $\begin{bmatrix}9&-6\end{bmatrix}^T$ into $(5)$ to get $C=-36$
$$
\begin{align}
-4x+0y&=-36\\
x&=9\tag{6}
\end{align}
$$
which would be the line orthogonal to $\begin{bmatrix}-4&0\end{bmatrix}^T$ and passing through $\begin{bmatrix}9&-6\end{bmatrix}^T$.
A: For the general case, to get a vector perpendicular to a given one, you want the dot product to be zero.  That means a vector perpendicular to $[a,b]$ is $[-b,a]$.  Then you have to scale it to make it a unit vector.
In your specific cases, there are zeros involved.  Have you plotted $y-8=0$?  You should be able to find a perpendicular vector by inspection.
For the second one, what is the direction of $v$?  What directions are perpendicular to it?
