# Unit Vector

A unit vector is any vector which has a magnitude of one and is pointing in any direction. Since these vectors are in any direction so they do not have any dimension. For example velocity (m/s) and acceleration (m/sec^2).
The unit vector is obtained by
$u=\frac{v}{\left \| v \right \|}$
u is a scalar multiple of v The vector u has magnitude 1 and the same direction as v . The vector u is called the unit vector in the direction of v..

## Finding the unit vector

Example 1: Find the unit vector in the direction of v= $\left \langle -1,4 \right \rangle$ and verify the result that has a magnitude of 1.
Solution : The unit vector in the direction of v is
$\frac{v}{\left \| v \right \|}$ = $\frac{\left \langle -1,4 \right \rangle}{\sqrt{(-1)^{2}+4^2}}$

= $\frac{1}{\sqrt{17}}{\left \langle -1,4 \right \rangle}$

=${\left \langle \frac{-1}{\sqrt{17}},\frac{4}{\sqrt{17}} \right \rangle}$

This has a magnitude 1 because

$\sqrt{(\frac{-1}{\sqrt{17}}})^{2}+(\frac{4}{\sqrt{17}})^{2}$

=$\sqrt{(\frac{1}{17}})+(\frac{16}{17})$

=$\sqrt{\frac{17}{17}}$ = 1.

Example 2: Find the unit vector in the direction of v= $\left \langle 2,3 \right \rangle$ and verify the result that has a magnitude of 1.
Solution : The unit vector in the direction of v is
$\frac{v}{\left \| v \right \|}$ = $\frac{\left \langle 2,3 \right \rangle}{\sqrt{(2)^{2}+3^2}}$

= $\frac{1}{\sqrt{13}}{\left \langle 2,3 \right \rangle}$

=${\left \langle \frac{2}{\sqrt{13}},\frac{3}{\sqrt{13}} \right \rangle}$

This has a magnitude 1 because

$\sqrt{(\frac{2}{\sqrt{13}}})^{2}+(\frac{3}{\sqrt{13}})^{2}$

=$\sqrt{(\frac{4}{13}})+(\frac{9}{13})$

=$\sqrt{\frac{13}{13}}$ = 1.