Question

Find the magnitude and direction angle of the following vector. Write your angle in degrees rounded to four decimal places.

Answer 1

Recall that the magnitude and direction of a vector:

$\vec{v}=ai+bj$

are given by the following formulas:

$\begin{gathered} \vec{|v}|=\sqrt{a^2+b^2}, \\ \theta=tan^{-1}(\frac{b}{a}). \end{gathered}$

Substituting

$\begin{gathered} a=2, \\ b=8, \end{gathered}$

in the above formulas, we get:

$\begin{gathered} \vec{|v}|=\sqrt{2^2+8^2}=\sqrt{4+64}=\sqrt{68}. \\ \theta=\tan^{-1}(\frac{8}{2})=\tan^{-1}(4). \end{gathered}$

Finally, we get:

$\begin{gathered} \vec{|v}|=2\sqrt{17}, \\ \theta=75.9638^{\circ}. \end{gathered}$

Answer: (|v|=2√17, θ=75.9638°)

$\begin{gathered} \vec{\lvert\rvert v}\lvert\rvert=2\sqrt{17}, \\ \theta=75.9638^{\operatorname{\circ}}. \end{gathered}$