Question

Given that \( \mathbf{a} \times (2\hat{i} + 3\hat{j} + 4\hat{k}) = (2\hat{i} + 3\hat{j} + 4\hat{k}) \times \mathbf{b} \), \( |\mathbf{a} + \mathbf{b}| = \sqrt{29} \), \( \mathbf{a} \cdot \mathbf{b} = ? \)

• A. 0
• B. 5
• C. 10
• D. 15

Hint:

For problems involving cross products and dot products of parallel vectors, the cross product is zero, and the dot product simplifies based on the magnitudes and directions of the vectors.

Answer 1

The Correct Option is A

The equation \( \mathbf{a} \times (2\hat{i} + 3\hat{j} + 4\hat{k}) = (2\hat{i} + 3\hat{j} + 4\hat{k}) \times \mathbf{b} \) implies that \( \mathbf{a} \) and \( \mathbf{b} \) are parallel to the vector \( (2\hat{i} + 3\hat{j} + 4\hat{k}) \). Hence, \( \mathbf{a} = \lambda (2\hat{i} + 3\hat{j} + 4\hat{k}) \) and \( \mathbf{b} = \mu (2\hat{i} + 3\hat{j} + 4\hat{k}) \) for some scalar constants \( \lambda \) and \( \mu \). The condition \( |\mathbf{a} + \mathbf{b}| = \sqrt{29} \) implies: \[ |\lambda (2\hat{i} + 3\hat{j} + 4\hat{k}) + \mu (2\hat{i} + 3\hat{j} + 4\hat{k})| = \sqrt{29} \] This results in \( |\lambda + \mu| \times \sqrt{29} = \sqrt{29} \), which gives \( \lambda + \mu = 1 \). Finally, the dot product \( \mathbf{a} \cdot \mathbf{b} = \lambda \mu |\mathbf{a}|^2 \). Since \( \mathbf{a} \) and \( \mathbf{b} \) are parallel, \( \mathbf{a} \cdot \mathbf{b} = 0 \).