# How do you find the magnitude and direction angle of the vector v=3(cos60i+sin60j)?

Oct 26, 2017

$3$
${60}^{o}$ or $\frac{\pi}{3}$ radians

#### Explanation:

To find the magnitude we use a similar idea as used to find the distance between two points.

Magnitude is:

For vector $\underline{c}$ with component vectors $a i$ and $b j$

|c|= sqrt((ai)^2+(bj)^2

$3 \cos \left(60\right) = \frac{3}{2}$

$3 \sin \left(60\right) = \frac{3 \sqrt{3}}{2}$

$| v | = \sqrt{{\left(\frac{3}{2}\right)}^{2} + {\left(\frac{3 \sqrt{3}}{2}\right)}^{2}} = \sqrt{\frac{9}{4} + \frac{27}{4}} = \pm \sqrt{9} = 3 , - 3$
( negative root not applicable)

So:

$| v | = 3$

The direction is the angle formed between the vector and the x axis.

So we find the tangent ratio, which is:

$\tan \left(\theta\right) = \frac{3 \sin \left(60\right)}{3 \cos \left(60\right)} = \frac{\sin \left(60\right)}{\cos \left(60\right)} = \frac{\frac{\sqrt{3}}{2}}{\frac{1}{2}} = \sqrt{3}$

$\arctan \left(\sqrt{3}\right) = {60}^{o}$ or $\frac{\pi}{3}$ radians