Find the angle between the pair of vectors to the nearest tenth of a degree.
-$$9 \mathbf { i } + 8 \mathbf { j } , - 3 \mathbf { i } + 6 \mathbf { j }$$
A) $74.9 ^ { \circ }$
B) $99.3 ^ { \circ }$
C) $158.2 ^ { \circ }$
D) $75.9 ^ { \circ }$

Question

Quizplus · Accepted Answer

The angle $\theta$ between two vectors $\mathbf{a}$ and $\mathbf{b}$ can be found using the dot product formula: $\mathbf{a} \cdot \mathbf{b} = |\mathbf{a}| |\mathbf{b}| \cos(\theta)$. For vectors $9\mathbf{i} + 8\mathbf{j}$ and $-3\mathbf{i} + 6\mathbf{j}$, the dot product is $(9)(-3) + (8)(6) = -27 + 48 = 21$. The magnitudes are $|\mathbf{a}| = \sqrt{9^2 + 8^2} = \sqrt{81 + 64} = \sqrt{145}$ and $|\mathbf{b}| = \sqrt{(-3)^2 + 6^2} = \sqrt{9 + 36} = \sqrt{45}$. Thus, $\cos(\theta) = \frac{21}{\sqrt{145} \sqrt{45}}$. Solving for $\theta$ gives $\theta \approx 74.9^\circ$.

Find the Angle Between the Pair of Vectors to the Nearest