The way you find a vector's magnitude and direction is:
[tex]\begin{gathered} |r|=\sqrt[]{x^2+y^2} \\ \tan \theta=\frac{y}{x} \end{gathered}[/tex]
So, the given data is x=3; y=2. Then the magnitude is:
[tex]\begin{gathered} |r|=\sqrt[]{3^2+2^2\text{ }} \\ |r|=\sqrt[]{9+4^{}\text{ }} \\ |r|=\sqrt[]{13\text{ }} \end{gathered}[/tex]
And the direction :
[tex]\begin{gathered} \tan \theta=\frac{2}{3} \\ \theta=\tan ^{-1}(\frac{2}{3}) \\ \theta=33.7\text{ degress} \end{gathered}[/tex]
The answer is D.
