Find the components of the vector v with given point P and terminal point Q, find te unit vector in the direction of v.
1. p: (3.2.0) Q: (5, -2,0)

Given vectors P=<3,2,0>, \;\ Q=<5,-2,0>

If \overline{PQ} is a vector in 3-space with initial point P(x_{1},y_{1},z_{1}) and terminal Q(x_{2},y_{2},z_{2}), then

\overline{PQ}=<x_{2}-x_{1}, \;\ y_{2}-y_{1}, \;\ z_{2}-z_{1}>

So, from the given vectors, we have <5-3, \;\ -2-2, \;\ 0-0>=<2,-4,0>

When we find a unit vector, we normalize.

Find the norm, which is just Pythagoras. A norm is a distance or length. The length of the vector.

||PQ||=\sqrt{2^{2}+(-4)^{2}}=\sqrt{20}=4\sqrt{2}

To normalize, we take the reciprocal and get \frac{1}{4\sqrt{2}}=\frac{\sqrt{2}}{8}

A unit vector is \frac{1}{||v||}\cdot v=\frac{\sqrt{2}}{8}\cdot <2,4,0>