Find the angle between the given vectors to the nearest tenth of a degree.
-$\mathbf { u } = \mathbf { i } + \sqrt { 7 } \mathbf { j } , \mathbf { v } = - \mathbf { i } - 9 \mathbf { j }$
A) $178.2 ^ { \circ }$
B) $165.6 ^ { \circ }$
C) $153.0 ^ { \circ }$
D) $113.1 ^ { \circ }$

Question

Quizplus · Accepted Answer

To find the angle between two vectors, use the dot product formula: $\mathbf{u} \cdot \mathbf{v} = |\mathbf{u}| |\mathbf{v}| \cos \theta$. Calculate $\mathbf{u} \cdot \mathbf{v} = (-1) + (-9\sqrt{7})$. Find magnitudes: $|\mathbf{u}| = \sqrt{1^2 + (\sqrt{7})^2} = \sqrt{8}$, $|\mathbf{v}| = \sqrt{(-1)^2 + (-9)^2} = \sqrt{82}$. Solve for $\theta$: $\cos \theta = \frac{-1 - 9\sqrt{7}}{\sqrt{8} \cdot \sqrt{82}}$. Calculate $\theta$ to get approximately $165.6^\circ$.

Find the Angle Between the Given Vectors to the Nearest u=i+7j,v=−i−9j\mathbf { u } = \mathbf { i } + \sqrt { 7 } \mathbf { j } , \mathbf { v } = - \mathbf { i } - 9 \mathbf { j }u=i+7​j,v=−i−9j

Find the Angle Between the Given Vectors to the Nearest $\mathbf { u } = \mathbf { i } + \sqrt { 7 } \mathbf { j } , \mathbf { v } = - \mathbf { i } - 9 \mathbf { j }$