Question

If two vectors \( \vec{a} = \cos \alpha \hat{i} + \sin \alpha \hat{j} + \sin \frac{\alpha}{2} \hat{k} \) and \( \vec{b} = \sin \alpha \hat{i} - \cos \alpha \hat{j} + \cos \frac{\alpha}{2} \hat{k} \) are perpendicular, then the values of \( \alpha \) are:

• A. \( 0 \) and \( \frac{\pi}{2} \)
• B. \( \frac{\pi}{4} \) and \( \frac{\pi}{2} \)
• C. \( 0 \) and \( \pi \)
• D. \( \frac{\pi}{2} \) and \( \frac{3\pi}{2} \)
• E. \( 0 \) and \( \frac{\pi}{4} \)

Hint:

To check perpendicularity, compute the dot product and solve for values that make it zero.

Answer 1

The Correct Option is C

For the vectors to be perpendicular, their dot product must be zero: \[ \vec{a} \cdot \vec{b} = (\cos \alpha)(\sin \alpha) + (\sin \alpha)(-\cos \alpha) + (\sin \frac{\alpha}{2})(\cos \frac{\alpha}{2}) \] Simplifying, \[ \vec{a} \cdot \vec{b} = \cos \alpha \sin \alpha - \cos \alpha \sin \alpha + \sin \frac{\alpha}{2} \cos \frac{\alpha}{2} \] Using the identity \( \sin 2x = 2\sin x \cos x \), we get: \[ \sin \frac{\alpha}{2} \cos \frac{\alpha}{2} = \frac{1}{2} \sin \alpha \] For the vectors to be perpendicular, \[ \frac{1}{2} \sin \alpha = 0 \Rightarrow \sin \alpha = 0 \] This gives \( \alpha = 0 \) or \( \pi \).