Question

Let \(\hat{a}, \hat{b}\) be unit vectors. If \(\vec{c}\) be a vector such that the angle between \(\hat{a}\) and \(\vec{c}\) is \(\frac{\pi}{12}\), and
\(\hat{b} = \vec{c} + 2(\vec{c} \times \hat{a})\)
, then \(|6\vec{c}|^2\) is equal to :

• A. \(6(3 - \sqrt{3})\)
• B. \(3 + \sqrt{3}\)
• C. \(6(3 + \sqrt{3})\)
• D. \(6(\sqrt{3} + 1)\)

Hint:

Take the squared magnitude of \(\hat b\); the cross product is perpendicular to \(\vec c\), so the cross term vanishes.

Answer 1

The Correct Option is C

\(|\hat{b}|^2 = 1 = |\vec{c}|^2 + 4|\vec{c}\times\hat{a}|^2 + 4\,\vec{c}\cdot(\vec{c}\times\hat{a})\).
The last term is 0 (scalar triple product with a repeated vector), and \(|\vec{c}\times\hat{a}|^2 = |\vec{c}|^2\sin^2\frac{\pi}{12}\).
\(\sin^2\frac{\pi}{12} = \frac{1 - \cos\frac{\pi}{6}}{2} = \frac{2-\sqrt{3}}{4}\), so \(4\sin^2\frac{\pi}{12} = 2 - \sqrt{3}\).
\(1 = |\vec{c}|^2\left(1 + 2 - \sqrt{3}\right) = |\vec{c}|^2(3 - \sqrt{3}) \Rightarrow |\vec{c}|^2 = \frac{1}{3-\sqrt{3}} = \frac{3+\sqrt{3}}{6}\).
\(|6\vec{c}|^2 = 36|\vec{c}|^2 = 6(3+\sqrt{3})\).
Answer: (3) \(6(3 + \sqrt{3})\)