Chapter 9, Problem 36RE

To determine

To find: The vector $v$ in $xy-$ plane.

Answer to Problem 36RE

The vector $v$ in $xy-$ plane is $\frac{3}{2}i+\frac{3\sqrt{3}}{2}j$ .

Explanation of Solution

Given information:

The magnitude of the vector is $\Vert v\Vert =3$ .

The direction angle of $v$ is $\alpha =60°$ .

Calculation:

The vector with magnitude and angle is given by,

  $v=\Vert v\Vert \left(\cos \alpha i+\sin \alpha j\right)$

Substitute the values in the above expression.

  $\begin{array}{c}v=3\left(\cos 60°i+\sin 60°j\right)\\ =3\left(\frac{1}{2}i+\frac{\sqrt{3}}{2}j\right)\\ =\frac{3}{2}i+\frac{3\sqrt{3}}{2}j\end{array}$

Therefore, the vector $v$ in $xy-$ plane is $\frac{3}{2}i+\frac{3\sqrt{3}}{2}j$ .