Find the angle between the pair of vectors to the nearest tenth of a degree.
--8, 9 , -9, -6
A) 41.1°
B) 82.1°
C) 31.1°
D) 92.1°

Question

Quizplus · Accepted Answer

The angle θ between two vectors $ \mathbf{a} = (a_1, a_2) $ and $ \mathbf{b} = (b_1, b_2) $ can be found using the dot product formula: $ \mathbf{a} \cdot \mathbf{b} = \|a\| \|b\| \cos(\theta) $, where $ \|a\| $ and $ \|b\| $ are the magnitudes of vectors $ \mathbf{a} $ and $ \mathbf{b} $, respectively. For vectors $ \mathbf{a} = (-8, 9) $ and $ \mathbf{b} = (-9, -6) $, the dot product $ \mathbf{a} \cdot \mathbf{b} = (-8)(-9) + (9)(-6) = 72 - 54 = 18 $. The magnitudes are $ \|a\| = \sqrt{(-8)^2 + 9^2} = \sqrt{64 + 81} = \sqrt{145} $ and $ \|b\| = \sqrt{(-9)^2 + (-6)^2} = \sqrt{81 + 36} = \sqrt{117} $. Thus, $ \cos(\theta) = \frac{18}{\sqrt{145} \sqrt{117}} $. Solving for $ \theta $ gives $ \theta \approx 82.1^\circ $.

Find the Angle Between the Pair of Vectors to the Nearest