Find the magnitude and direction angle of $\mathbf { v } = 6 \left( \cos 80 ^ { \circ } \mathbf { i } + \sin 80 ^ { \circ } \mathbf { j } \right)$ .
A) $\| \mathbf { v } \| = \sqrt { 6 } ; \theta = 80 ^ { \circ }$
B) $\| \mathbf { v } \| = 6 ; \theta = 100 ^ { \circ }$
C) $\| \mathbf { v } \| = 6 ; \theta = 80 ^ { \circ }$
D) $\| \mathbf { v } \| = 6 ; \theta = 10 ^ { \circ }$
E) $\| \mathbf { v } \| = \sqrt { 6 } ; \theta = 100 ^ { \circ }$

Question

Find the magnitude and direction angle of $\mathbf { v } = 6 \left( \cos 80 ^ { \circ } \mathbf { i } + \sin 80 ^ { \circ } \mathbf { j } \right)$ .
A)  $\| \mathbf { v } \| = \sqrt { 6 } ; \theta = 80 ^ { \circ }$
B)  $\| \mathbf { v } \| = 6 ; \theta = 100 ^ { \circ }$
C)  $\| \mathbf { v } \| = 6 ; \theta = 80 ^ { \circ }$
D)  $\| \mathbf { v } \| = 6 ; \theta = 10 ^ { \circ }$
E)  $\| \mathbf { v } \| = \sqrt { 6 } ; \theta = 100 ^ { \circ }$

Quizplus · Accepted Answer

The magnitude of $\mathbf { v }$ is found by multiplying 6 by the magnitude of the vector $\cos 80 ^ { \circ } \mathbf { i } + \sin 80 ^ { \circ } \mathbf { j }$, which is 1 since $\cos^2 \theta + \sin^2 \theta = 1$. Therefore, $\| \mathbf { v } \| = 6$. The direction angle $\theta$ is given directly as $80^\circ$.

Find the Magnitude and Direction Angle Of v=6(cos⁡80∘i+sin⁡80∘j)\mathbf { v } = 6 \left( \cos 80 ^ { \circ } \mathbf { i } + \sin 80 ^ { \circ } \mathbf { j } \right)v=6(cos80∘i+sin80∘j)

Find the Magnitude and Direction Angle Of $\mathbf { v } = 6 \left( \cos 80 ^ { \circ } \mathbf { i } + \sin 80 ^ { \circ } \mathbf { j } \right)$