### Unit Vector

• To be able to find length of a vector as well as unit normal vector.

#### Length (Magnitude) of a Vector:

Imagine a box with length, width and height equal to the i,j,k components of a vector. Then the length of the vector will be diagonal of this box, and it can be calculated using Pythagorean theorem.

If $\vec{a} = a_{1}\bar{i} + a_{2}\bar{j} + a_{3}\bar{k}$ then length of the vector = $\left | \vec{a} \right | = \sqrt{a_{1}^{2} + a_{2}^{2} + a_{3}^{2}}$

e.g. if $\bar{a}$ = 2i + 3j + 4k then, $\left | \bar{a} \right | = \sqrt{2^{2} + 3^{2} + 4^{2}} = \sqrt{29}$

## Calculating Unit Vector:

Unit vector has a magnitude (length) = 1 and is given by: $\hat{a} = \pm \dfrac{\bar{a}}{\left | \bar{a} \right |}$ e.g. if $\bar{a}$ = 2i + 3j + 4k then, $\left | \bar{a} \right | = \sqrt{2^{2} + 3^{2} + 4^{2}} = \sqrt{29}$ $\hat{a} = \pm \dfrac{2i + 3j + 4k}{\sqrt{29}}$ The $\pm$ sign is because there can be two unit vectors opposite to each other.

### Solved Example:

#### 4-4-01

Determine the magnitude of the force vector $\overline {F}$ = 20$\overline {i}$ + 60$\overline {j}$ - 90$\overline {k}$ (N).

Solution:
$\overline {F} =20\overline {i}+60\overline {j}-90\overline {k}$ \begin{align*} \left| \overline {F}\right|&=\sqrt {20^{2}+60^{2}+\left( -90\right) ^{2}}\\ &=\sqrt {400+3600+8100}\\ &=\sqrt {12100}\\ &= 110 \ N \end{align*}

### Solved Example:

#### 4-4-02

For the spherical surface $x^2+ y^2+ z^2 = 1$, the unit outward normal vector at the point $(\dfrac {1}{\sqrt {2}}, \dfrac {1}{\sqrt {2}}, 0)$ is given by:

Solution:

Given : $x^2+ y^2+ z^2 = 1$

This is a equation of sphere with radius r = 1 \begin{align*} \overline {OA} &=\left( \dfrac {1}{\sqrt {2}}-0\right) \overline {i}+\left( \dfrac {1}{\sqrt {2}}-0\right) \overline {j}+\left( 0-0\right) \overline {k}\\ &=\dfrac {1}{\sqrt {2}}\overline {i}+\dfrac {1}{\sqrt {2}}\overline {j} \end{align*}