Question

Find the vector v in the xy-plane with magnitude 3 if the direction angle of v is 60°.

Short answer

The vector v in the xy-plane with magnitude 3 and direction angle 60° of v is $\frac{3}{2}i+\frac{3\sqrt{3}}{2}j$.

Step-by-step solution

Step 1. Given information

We have been given:

Magnitude$=3$

and direction angle of v$=60°$

We have to find the above vector.

Step 2. Find the vector v using formula v=∥v∥(icosθ+jsinθ).

$$\begin{gathered} \mathrm{v}=\parallel \mathrm{v}\parallel (\mathrm{icos}\theta +\mathrm{jsin}\theta ) \\ =3\left(\mathrm{icos}{60}^{\circ }+\mathrm{jsin}{60}^{\circ }\right) \\ =3\left(\frac{1}{2}\mathrm{i}+\frac{\sqrt{3}}{2}\mathrm{j}\right) \\ =\frac{3}{2}\mathrm{i}+\frac{3\sqrt{3}}{2}\mathrm{j} \end{gathered}$$