Question

If \( \vec{a} \) and \( \vec{b} \) are two vectors such that \( |\vec{a}|=|\vec{b}|=\sqrt{6} \) and \( \vec{a} \cdot \vec{b} = -1 \), then \( |\vec{a} \times \vec{b}| \sin(\vec{a}, \vec{b}) = \)

• A. \( (|\vec{a}|^2-1)(|\vec{b}|^2+1) \)
• B. \( \frac{1}{6} \)
• C. \( (|\vec{a}|^2-1)(1+\frac{1}{|\vec{b}|^2}) \)
• D. \( \frac{\sqrt{35}}{6} \)

Hint:

Remember the fundamental relationships between dot product, cross product, and the angle \( \theta \) between vectors: \( \vec{a} \cdot \vec{b} = |\vec{a}||\vec{b}|\cos\theta \) and \( |\vec{a} \times \vec{b}| = |\vec{a}||\vec{b}|\sin\theta \). These allow you to switch between geometric and algebraic representations.

Answer 1

The Correct Option is C