Question

Find the magnitude and direction angle (to the nearest tenth) for each vector. Give the measure of the direction angle as an angle in [0, 360°].

Answer 1

Answer:

magnitude = 8.5

direction angle = 45°

To get the magnitude of a vector , we will use the following formula:

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

Substituting a = 6 and b =6:

$\begin{gathered} ||\vec{u}||=\sqrt{a^2+b^2} \\ ||\vec{u}||=\sqrt{6^2+6^2} \\ ||\vec{u}||=\sqrt{36+36} \\ ||\vec{u}||=\sqrt{72} \\ ||\vec{u}||=8.485\approx8.5 \end{gathered}$

For the direction angle, we will use the following formula:

$\begin{gathered} \tan\theta=\frac{b}{a} \\ \theta=\tan^{-1}\frac{6}{6} \\ \theta=45 \end{gathered}$

The magnitude is 8.5 while the direction angle is 45°