Find the angle between the pair of vectors to the nearest tenth of a degree.
-$\langle 4,5 \rangle , \langle - 4 , - 1 \rangle$
A) $61.4 ^ { \circ }$
B) $142.7 ^ { \circ }$
C) $71.4 ^ { \circ }$
D) $152.7 ^ { \circ }$

Question

Quizplus · Accepted Answer

The angle 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 $\theta$ is the angle between the vectors. For $\vec{a} = \langle 4,5 \rangle$ and $\vec{b} = \langle -4,-1 \rangle$, the dot product is $4(-4) + 5(-1) = -16 - 5 = -21$. The magnitudes are $|\vec{a}| = \sqrt{4^2 + 5^2} = \sqrt{41}$ and $|\vec{b}| = \sqrt{(-4)^2 + (-1)^2} = \sqrt{17}$. Thus, $\cos(\theta) = \frac{-21}{\sqrt{41}\sqrt{17}}$. Solving for $\theta$ gives $\theta \approx 142.7^\circ$.

Find the Angle Between the Pair of Vectors to the Nearest