Solve the problem.
-For the vectors $\mathbf { u }$ and $\mathbf { v }$ with magnitudes $| \mathbf { u } | = 3$ and $| \mathbf { v } | = 6$, find the angle $\theta$ between $\mathbf { u }$ and $\mathbf { v }$ which makes $\mid$ proju $^ { \mathbf { v } } \mid = 2$
A) $70.53 ^ { \circ }$
B) $60.00 ^ { \circ }$
C) $48.19 ^ { \circ }$
D) $41.81 ^ { \circ }$

Question

Quizplus · Accepted Answer

The magnitude of the projection of $\mathbf{u}$ onto $\mathbf{v}$, denoted as $|\text{proj}_{\mathbf{v}}\mathbf{u}|$, is given by the formula $|\text{proj}_{\mathbf{v}}\mathbf{u}| = \frac{|\mathbf{u} \cdot \mathbf{v}|}{|\mathbf{v}|}$. Since $|\text{proj}_{\mathbf{v}}\mathbf{u}| = 2$, $|\mathbf{u}| = 3$, and $|\mathbf{v}| = 6$, we can use the formula for the dot product in terms of magnitudes and the cosine of the angle between them: $\mathbf{u} \cdot \mathbf{v} = |\mathbf{u}||\mathbf{v}|\cos(\theta)$. Substituting the given values, we get $2 = \frac{3 \cdot 6 \cdot \cos(\theta)}{6}$, simplifying to $2 = 3\cos(\theta)$, which gives $\cos(\theta) = \frac{2}{3}$. Solving for $\theta$ yields $\theta \approx 48.19^\circ$.

Solve the Problem u\mathbf { u }u And v\mathbf { v }v

Solve the Problem $\mathbf { u }$ And $\mathbf { v }$