Question

If \( \vec{a} = 2\hat{i} - \hat{j} + 3\hat{k} \), \( \vec{b} = \hat{i} + 2\hat{j} - \hat{k} \), then the angle \( \theta \) between \( \vec{a} \) and \( \vec{b} \) is:

Hint:

To find the angle between two vectors, use the dot product formula \( \cos \theta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}| |\vec{b}|} \). Ensure all components are correctly multiplied.

Answer 1

- To find the angle \( \theta \) between vectors \( \vec{a} = 2\hat{i} - \hat{j} + 3\hat{k} \) and \( \vec{b} = \hat{i} + 2\hat{j} - \hat{k} \), use the dot product formula: \[ \cos \theta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}| |\vec{b}|} \]
- Compute the dot product \( \vec{a} \cdot \vec{b} \): \[ \vec{a} \cdot \vec{b} = (2)(1) + (-1)(2) + (3)(-1) = 2 - 2 - 3 = -3 \]
- Compute the magnitudes \( |\vec{a}| \) and \( |\vec{b}| \): \[ |\vec{a}| = \sqrt{2^2 + (-1)^2 + 3^2} = \sqrt{4 + 1 + 9} = \sqrt{14} \] \[ |\vec{b}| = \sqrt{1^2 + 2^2 + (-1)^2} = \sqrt{1 + 4 + 1} = \sqrt{6} \]
- So: \[ \cos \theta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}| |\vec{b}|} = \frac{-3}{\sqrt{14} \cdot \sqrt{6}} = \frac{-3}{\sqrt{14 \times 6}} = \frac{-3}{\sqrt{84}} \]
- Simplify \( \sqrt{84} \): \[ \sqrt{84} = \sqrt{4 \times 21} = 2\sqrt{21} \]
- Thus: \[ \cos \theta = \frac{-3}{2\sqrt{21}} \]