Given:
There are given that the value of u to find the magnitude and direction:
[tex]u=30\langle\cos 60^{\circ},\sin 60^{\circ}\rangle[/tex]
Explanation:
From the concept of the magnitude:
The magnitude of any vector a and b are:
[tex]\langle a,b\rangle=\sqrt[]{a^2+b^2}[/tex]
Then,
From the given vector:
[tex]\begin{gathered} \lvert-4u\rvert=\sqrt[]{(-4\times30\times\frac{1}{2})^2+(-4\times30\times\frac{\sqrt[]{3}}{2})^2} \\ \lvert-4u\rvert=\sqrt[]{(-60)^2+(-60\sqrt[]{3})^2} \\ \lvert-4u\rvert=\sqrt[]{3600+10800} \end{gathered}[/tex]
Then,
[tex]\begin{gathered} \lvert-4u\rvert=\sqrt[]{3600+10800} \\ \lvert-4u\rvert=\sqrt[]{14400} \\ \lvert-4u\rvert=120 \end{gathered}[/tex]
Now,
For finding the direction of the given vector:
From the formula to find the direction vector:
[tex]\theta=\tan ^{-1}(\frac{y}{x})[/tex]
Then,
From the given vector:
[tex]\begin{gathered} u=30\langle\cos 60^{\circ},\sin 60^{\circ}\rangle \\ -4u=\langle-120\cos 60^{\circ},-120\sin 60^{\circ}\rangle \end{gathered}[/tex]
Then,
From the formula:
[tex]\begin{gathered} \theta=\tan ^{-1}(\frac{y}{x}) \\ \theta=\tan ^{-1}(\frac{-120\sin ^{}60^{\circ}}{-120\cos 60^{\circ}}) \\ \theta=\tan ^{-1}(\frac{\frac{\sqrt[]{3}}{2}^{}}{\frac{1}{2}}) \\ \theta=\tan (\sqrt[]{3}) \\ \theta=59.9 \\ \theta=60^{\circ} \end{gathered}[/tex]
Final answer:
So, the magnitude of the given vector is 120 and the direction of the given vector is 60 degrees.
Hence, the correct option is D.
