Finding a unit vector perpendicular to another vector For example we have the vector $8i + 4j - 6k$, how can we find a unit vector perpendicular to this vector?
Let $\vec{v}=x\vec{i}+y\vec{j}+z\vec{k}$, a perpendicular vector to yours. Their inner product (the dot product - $\vec{u}.\vec{v}$ ) should be equal to 0, therefore: $$8x+4y-6z=0 \tag{1}$$ Choose for example x,y and find z from equation 1. In order to make its length equal to 1, calculate $\|\vec{v}\|=\sqrt{x^2+y^2+z^2}$ and divide $\vec{v}$ with it. Your unit vector would be: $$\vec{u}=\frac{\vec{v}}{\|\vec{v}\|}$$
You need to find a*i + b*j + c* k so that the dot product of it with 8i +4j -6k is 0.

It means
8a + 4b - 6 c = 0. 

You need to choose a,b,c satisfy above. For example, you can choose a = 1, b = 1, c = 2.