Find the angle between the given vectors to the nearest tenth of a degree.
-<5, 4>, <-5, -5>

Question

Find the angle between the given vectors to the nearest tenth of a degree. -<5, 4>, <-5, -5> A) 183.7° B) 76.9° C) 173.7° D) 86.9°

Quizplus · Accepted Answer

The angle between two vectors $\vec{a}$ and $\vec{b}$ can be found using the dot product formula: $\theta = \cos^{-1}\left(\frac{\vec{a} \cdot \vec{b}}{\|\vec{a}\| \|\vec{b}\|}\right)$. For vectors $\vec{a} = <5, 4>$ and $\vec{b} = <-5, -5>$, the dot product $\vec{a} \cdot \vec{b} = (5)(-5) + (4)(-5) = -25 - 20 = -45$, and the magnitudes are $\|\vec{a}\| = \sqrt{5^2 + 4^2} = \sqrt{25 + 16} = \sqrt{41}$ and $\|\vec{b}\| = \sqrt{(-5)^2 + (-5)^2} = \sqrt{25 + 25} = \sqrt{50}$. Thus, $\theta = \cos^{-1}\left(\frac{-45}{\sqrt{41}\sqrt{50}}\right)$, which calculates to approximately 173.7°.

Find the Angle Between the Given Vectors to the Nearest