Find the angle between u and v in radians.
-u = -2i + 8j - 8k, v = 8i + 4j - 5k
A) 1.57
B) 0.50
C) 1.33
D) 1.08

Question

Quizplus · Accepted Answer

The angle between two vectors $ \mathbf{u} $ and $ \mathbf{v} $ can be found using the dot product formula: $$ \cos(\theta) = \frac{\mathbf{u} \cdot \mathbf{v}}{\|\mathbf{u}\| \|\mathbf{v}\|} $$ where $ \mathbf{u} \cdot \mathbf{v} = u_i v_i + u_j v_j + u_k v_k $ and $ \|\mathbf{u}\| $ and $ \|\mathbf{v}\| $ are the magnitudes of $ \mathbf{u} $ and $ \mathbf{v} $, respectively.Given $ \mathbf{u} = -2\mathbf{i} + 8\mathbf{j} - 8\mathbf{k} $ and $ \mathbf{v} = 8\mathbf{i} + 4\mathbf{j} - 5\mathbf{k} $, the dot product $ \mathbf{u} \cdot \mathbf{v} = (-2)(8) + (8)(4) + (-8)(-5) = -16 + 32 + 40 = 56 $.The magnitudes are $ \|\mathbf{u}\| = \sqrt{(-2)^2 + 8^2 + (-8)^2} = \sqrt{4 + 64 + 64} = \sqrt{132} $ and $ \|\mathbf{v}\| = \sqrt{8^2 + 4^2 + (-5)^2} = \sqrt{64 + 16 + 25} = \sqrt{105} $.Thus, $ \cos(\theta) = \frac{56}{\sqrt{132} \sqrt{105}} $.Solving for $ \theta $, we find $ \theta = \cos^{-1}\left(\frac{56}{\sqrt{132} \sqrt{105}}\right) \approx 1.08 $ radians.

Find the Angle Between U and V in Radians