Given:
There are given the vector to find the magnitude:
[tex]w=-168i-160j[/tex]
Explanation:
To find the magnitude, first, we need to find the vector for -4w.
Then,
From the given vector:
[tex]\begin{gathered} w=-168\imaginaryI-160j \\ -4w=-4(-168\mathrm{i}-160j) \\ -4w=672i+640j \end{gathered}[/tex]
Then,
The magnitude of the given vector is:
[tex]\begin{gathered} |-4w|=\sqrt{(672)^2+(640)^2} \\ =\sqrt{451584+409600} \\ =\sqrt{861184} \\ =928 \end{gathered}[/tex]
Now,
For the direction of the vector:
[tex]\theta=tan^{-1}(\frac{y}{x})[/tex]
Then,
[tex]\begin{gathered} \theta=tan^{-1}(\frac{y}{x}) \\ \theta=tan^{-1}(\frac{640}{672}) \end{gathered}[/tex]
Then,
[tex]\begin{gathered} \theta=tan^{-1}(\frac{640}{672}) \\ \theta=44^{\circ} \end{gathered}[/tex]
Final answer:
The magnitude and direction of the given vector is shown below:
[tex]\begin{gathered} magnitude:928 \\ direction:44^{\circ} \end{gathered}[/tex]
Hence, the correct option is D.
