Find the angle between the pair of vectors to the nearest tenth of a degree.
-$$\langle - 5 , - 7 \rangle , \langle 6 , - 4 \rangle$$
A) $45.9 ^ { \circ }$
B) $91.8 ^ { \circ }$
C) $101.8 ^ { \circ }$
D) $35.9 ^ { \circ }$

Question

Quizplus · Accepted Answer

The angle $\theta$ between two vectors $\vec{a}$ and $\vec{b}$ can be found using the dot product formula: $\vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos(\theta)$, where $|\vec{a}|$ and $|\vec{b}|$ are the magnitudes of the vectors. For vectors $\langle -5, -7 \rangle$ and $\langle 6, -4 \rangle$, the dot product is $(-5)(6) + (-7)(-4) = -30 + 28 = -2$. The magnitudes are $|\vec{a}| = \sqrt{(-5)^2 + (-7)^2} = \sqrt{25 + 49} = \sqrt{74}$ and $|\vec{b}| = \sqrt{6^2 + (-4)^2} = \sqrt{36 + 16} = \sqrt{52}$. Substituting these into the formula gives $-2 = \sqrt{74} \sqrt{52} \cos(\theta)$, solving for $\theta$ gives approximately $91.8^\circ$.

Find the Angle Between the Pair of Vectors to the Nearest