Find the angle between the given vectors to the nearest tenth of a degree.
-$$\mathbf { u } = \langle - 7,5 \rangle , \mathbf { v } = \langle - 6,3 \rangle$$
A) $4.5 ^ { \circ }$
B) $19.0 ^ { \circ }$
C) $9.0 ^ { \circ }$
D) $- 5.5 ^ { \circ }$

Question

Quizplus · Accepted Answer

The angle θ between two vectors u and v can be found using the dot product formula: $ \cos \theta = \frac{\mathbf{u} \cdot \mathbf{v}}{\|\mathbf{u}\| \|\mathbf{v}\|} $. Calculate the dot product $ \mathbf{u} \cdot \mathbf{v} = (-7)(-6) + (5)(3) = 42 + 15 = 57 $. The magnitudes are $ \|\mathbf{u}\| = \sqrt{(-7)^2 + 5^2} = \sqrt{49 + 25} = \sqrt{74} $ and $ \|\mathbf{v}\| = \sqrt{(-6)^2 + 3^2} = \sqrt{36 + 9} = \sqrt{45} $. Thus, $ \cos \theta = \frac{57}{\sqrt{74} \cdot \sqrt{45}} $. Solving for θ gives approximately $ 9.0^\circ $.

Find the Angle Between the Given Vectors to the Nearest u=⟨−7,5⟩,v=⟨−6,3⟩\mathbf { u } = \langle - 7,5 \rangle , \mathbf { v } = \langle - 6,3 \rangleu=⟨−7,5⟩,v=⟨−6,3⟩

Find the Angle Between the Given Vectors to the Nearest $\mathbf { u } = \langle - 7,5 \rangle , \mathbf { v } = \langle - 6,3 \rangle$