# How do I find the magnitude and direction angle of the vector v =6i-6j?

May 6, 2018

The magnitude of $v$ is $| \vec{v} | = 6 \sqrt{2}$ and the direction angle $\theta$ is $\theta = - {45}^{\circ}$.

#### Explanation:

In this answer, we'll define the direction angle to be $\theta$ and the magnitude of the vector to be $| \vec{v} |$.

Now, for a more general case, how do we find the magnitude and direction angle of the vector $\vec{\nu} = a \vec{i} - b \vec{j}$? As you can see, the sides with lenghts $a$, $b$ and $| \vec{\nu} |$ form a right triangle. Thus,

$| \vec{\nu} {|}^{2} = {a}^{2} + {b}^{2}$

sin theta = a/|vec nu|"

$\cos \theta = \frac{b}{|} \vec{\nu} |$

Using these identities, we can solve for our particular case with $\vec{v} = 6 \vec{i} - 6 \vec{j}$.

$\therefore | \vec{v} {|}^{2} = {6}^{2} + {\left(- 6\right)}^{2} = 72 \implies | \vec{v} | = 6 \sqrt{2}$

$\sin \theta = \frac{b}{|} \vec{v} | = - \frac{6}{6 \sqrt{2}} = - \frac{1}{\sqrt{2}}$

$\cos \theta = \frac{a}{|} \vec{v} | = \frac{6}{6 \sqrt{2}} = \frac{1}{\sqrt{2}}$

Now, the sine function is negative in the fourth quadrant while the cosine is positive, so we must have $\theta = - {45}^{\circ}$.

Our vector has magnitude $6 \sqrt{2}$ and direction angle $- {45}^{\circ}$.