# Question 2ad9d

Jan 10, 2018

The Unit Vector $\textcolor{b l u e}{\vec{u} = - i}$ is in the direction of the given vector $\textcolor{red}{\vec{v} = - 3 i}$

#### Explanation:

Given:

$\textcolor{red}{\vec{v} = - 3 i = < - 3 , 0 >}$

The Formula to find the Unit Vector $\vec{u}$ is

color(green)(vec u = (vec v)/|| vec v||, where

$| | \vec{v} | |$ is the Magnitude of Vector $\vec{v}$

$\vec{u} = \frac{1}{|} | \vec{v} | | \cdot \vec{v}$

$\vec{u} = \frac{1}{\sqrt{{\left(- 3\right)}^{2} + {\left(0\right)}^{2}}} \cdot < - 3 , 0 >$

$\vec{u} = \frac{1}{\sqrt{9}} \cdot < - 3 , 0 >$

$\vec{u} = \frac{1}{3} \cdot < - 3 , 0 >$ This is Scalar Multiplication

We have multiplied the Scalar Value $\frac{1}{3}$ by each of the components of Vector $\vec{v}$

Hence, we get

$< - \frac{3}{3} , \frac{0}{3} >$

$\textcolor{b l u e}{\Rightarrow < - 1 , 0 >}$

We can also write this as $\textcolor{b l u e}{\vec{u} = - i}$

$\textcolor{b l u e}{\vec{u} = - i}$ is our Unit Vector in the direction of the given vector $\vec{v} = - 3 i$

We can also verify this solution by finding the Magnitude of the vector color(blue)(vec u =<-1,0>#

The Magnitude of the vector must be equal to one

Let us now find

$| | < - 1 , 0 > | |$

$\Rightarrow \sqrt{{\left(- 1\right)}^{2} + {\left(0\right)}^{2}}$

$\Rightarrow \sqrt{\left(1\right) + \left(0\right)}$

$\Rightarrow \sqrt{1}$

$\Rightarrow 1$

Hence, verified.

Hope you find this solution useful.