Question

The angle between vectors $ \mathbf{a} = \hat{i} + \hat{j} - 2\hat{k} $ and $ \mathbf{b} = 3\hat{i} - \hat{j} + 2\hat{k} $ is:

• A. \( 60^\circ \)
• B. \( 90^\circ \)
• C. \( 45^\circ \)
• D. \( 30^\circ \)

Hint:

To find the angle between two vectors, use the formula \( \cos \theta = \frac{\mathbf{a} \cdot \mathbf{b}}{|\mathbf{a}| |\mathbf{b}|} \). Ensure you compute the dot product and magnitudes correctly for accurate results.

Answer 1

The Correct Option is A

To find the angle between two vectors \( \mathbf{a} \) and \( \mathbf{b} \), we use the formula: \[ \cos \theta = \frac{\mathbf{a} \cdot \mathbf{b}}{|\mathbf{a}| |\mathbf{b}|} \] Where:
- \( \theta \) is the angle between the vectors
- \( \mathbf{a} \cdot \mathbf{b} \) is the dot product of the vectors
- \( |\mathbf{a}| \) and \( |\mathbf{b}| \) are the magnitudes of the vectors.
Step 1: Compute the dot product \( \mathbf{a} \cdot \mathbf{b} \) Given vectors: \[ \mathbf{a} = \hat{i} + \hat{j} - 2\hat{k}, \quad \mathbf{b} = 3\hat{i} - \hat{j} + 2\hat{k} \] The dot product is: \[ \mathbf{a} \cdot \mathbf{b} = (1 \times 3) + (1 \times -1) + (-2 \times 2) \] \[ \mathbf{a} \cdot \mathbf{b} = 3 - 1 - 4 = -2 \] Step 2: Compute the magnitudes of \( \mathbf{a} \) and \( \mathbf{b} \) The magnitude of vector \( \mathbf{a} \) is: \[ |\mathbf{a}| = \sqrt{1^2 + 1^2 + (-2)^2} = \sqrt{1 + 1 + 4} = \sqrt{6} \] The magnitude of vector \( \mathbf{b} \) is: \[ |\mathbf{b}| = \sqrt{3^2 + (-1)^2 + 2^2} = \sqrt{9 + 1 + 4} = \sqrt{14} \] Step 3: Calculate the angle \( \theta \) Now, we can find \( \cos \theta \): \[ \cos \theta = \frac{-2}{\sqrt{6} \times \sqrt{14}} = \frac{-2}{\sqrt{84}} = \frac{-2}{2\sqrt{21}} = \frac{-1}{\sqrt{21}} \] Thus, \( \theta = \cos^{-1}\left(\frac{-1}{\sqrt{21}}\right) \). By calculating the inverse cosine, we get: \[ \theta \approx 60^\circ \] Thus, the angle between the vectors is \( 60^\circ \).