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Answer 1 · 2018-01-10T07:58:45

The Unit Vector $\to u = - i$ $color(blue)(vec u = -i)$ is in the direction of the given vector $\to v = - 3 i$ $color(red)(vec v = -3i)$

Given:

$\to v = - 3 i = < - 3, 0 >$ $color(red)(vec v = -3i = <-3,0>)$

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

$\to u = \frac{\to v}{∣ ∣ ∣ ∣ \to v ∣ ∣ ∣ ∣}$ $color(green)(vec u = (vec v)/|| vec v||$ , where

$∣ ∣ ∣ ∣ \to v ∣ ∣ ∣ ∣$ $|| vec v||$ is the Magnitude of Vector $\to v$ $vec v$

$\to u = \frac{1}{∣ ∣ ∣ ∣ \to v ∣ ∣ ∣ ∣} \cdot \to v$ $vec u = 1/||vec v||*vec v$

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

$\to u = \frac{1}{\sqrt{9}} \cdot < - 3, 0 >$ $vec u = 1/sqrt(9)*<-3, 0>$

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

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

Hence, we get

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

$\Rightarrow < - 1, 0 >$ $color(blue)(rArr <-1, 0>)$

We can also write this as $\to u = - i$ $color(blue)(vec u =-i)$

$\to u = - i$ $color(blue)(vec u =-i)$ is our Unit Vector in the direction of the given vector $\to v = - 3 i$ $vec v = -3i$

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

The Magnitude of the vector must be equal to one

Let us now find

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

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

$\Rightarrow \sqrt{(1) + (0)}$ $rArr sqrt((1) + (0))$

$\Rightarrow \sqrt{1}$ $rArr sqrt(1)$

$\Rightarrow 1$ $rArr 1$

Hence, verified.

Hope you find this solution useful.