## VECTORS

In three dimensions, the base vectors are
$$i=\begin{pmatrix} 1\\0 \\0 \end{pmatrix}$$, $$j=\begin{pmatrix} 0\\1 \\0 \end{pmatrix}$$, $$k=\begin{pmatrix} 0\\0 \\1 \end{pmatrix}$$
that are along the $$x, y$$ and $$z$$ coordinate directions, respectively, as shown in the figure.
For example, $$v=\begin{pmatrix} -4\\3 \\2 \end{pmatrix}$$ can be written as $$-4i+3j+2k$$.
The magnitude of $$v$$:

$$|v|=\sqrt{((-4)^2+3^2+2^2)}=\sqrt{29}$$
Given that if $$v=xi+yj+zk$$ then
$$|v|=\sqrt{x^2+y^2+z^2}$$

## UNIT VECTORS

If $$\overrightarrow{AB}$$ is a non-unit vector, then a unit vector in the direction of $$\overrightarrow{AB}$$ is

$${1\over |\overrightarrow{AB}|}\overrightarrow{AB}$$
LESSON 1
Determine the unit vector in the direction of the vector $$\overrightarrow{OA}=\begin{pmatrix} -4\\3 \\2 \end{pmatrix}$$.
SOLUTION
• Determine the length of the vector
$$|\overrightarrow{OA}|=\sqrt{(-4)^2+(3)^2+(2)^2}=\sqrt{29}$$
• Use the form for a unit vector
Unit Vector
$${1\over \sqrt{29}} \begin{pmatrix} -4\\3 \\2 \end{pmatrix}$$

TRY THIS

Complete Question 1 from worksheet.