Question

Find the angle between the vectors \[ \mathbf{A} = 5\vec{i} + 3\vec{j} + 4\vec{k} \quad \text{and} \quad \mathbf{B} = 6\vec{i} - 8\vec{j} - \vec{k}. \]

Hint:

Use the formula \( \cos \theta = \frac{\mathbf{A} \cdot \mathbf{B}}{\|\mathbf{A}\| \|\mathbf{B}\|} \) to find the angle between two vectors.

Answer 1

The angle \( \theta \) between two vectors \(\mathbf{A}\) and \(\mathbf{B}\) is given by: \[ \cos \theta = \frac{\mathbf{A} \cdot \mathbf{B}}{\|\mathbf{A}\| \|\mathbf{B}\|}. \] Calculate the dot product: \[ \mathbf{A} \cdot \mathbf{B} = 5 \times 6 + 3 \times (-8) + 4 \times (-1) = 30 - 24 - 4 = 2. \] Calculate magnitudes: \[ \|\mathbf{A}\| = \sqrt{5^2 + 3^2 + 4^2} = \sqrt{25 + 9 + 16} = \sqrt{50} = 5 \sqrt{2}. \] \[ \|\mathbf{B}\| = \sqrt{6^2 + (-8)^2 + (-1)^2} = \sqrt{36 + 64 + 1} = \sqrt{101}. \] Therefore, \[ \cos \theta = \frac{2}{5 \sqrt{2} \times \sqrt{101}} = \frac{2}{5 \sqrt{202}}. \] Hence, \[ \boxed{ \theta = \cos^{-1} \left( \frac{2}{5 \sqrt{202}} \right). } \]