Question

Find the magnitude and direction of the vector F = (3,2).A. FL= V13,56.30 N of EB. FL= \13, 1° N of EC. F1 = 13,0.6°N of ED. F = /13, 33.7° N of E

Answer 1

The way you find a vector's magnitude and direction is:

$\begin{gathered} |r|=\sqrt[]{x^2+y^2} \\ \tan \theta=\frac{y}{x} \end{gathered}$

So, the given data is x=3; y=2. Then the magnitude is:

$\begin{gathered} |r|=\sqrt[]{3^2+2^2\text{ }} \\ |r|=\sqrt[]{9+4^{}\text{ }} \\ |r|=\sqrt[]{13\text{ }} \end{gathered}$

And the direction :

$\begin{gathered} \tan \theta=\frac{2}{3} \\ \theta=\tan ^{-1}(\frac{2}{3}) \\ \theta=33.7\text{ degress} \end{gathered}$

The answer is D.