Question

The angle between the vectors \( \vec{a} \) and \( \vec{b} \) is \( \frac{\pi}{3} \). If \( | \vec{a} \cdot \vec{b} |^2 = 15 \), then \( | \vec{a} \times \vec{b} |^2 \) is equal to

• A. \( 5 \)
• B. \( 15\sqrt{3} \)
• C. \( \frac{15}{\sqrt{3}} \)
• D. \( 5\sqrt{3} \)
• E. \( 45 \)

Hint:

The magnitude of the cross product of two vectors represents the area of the parallelogram they form.

Answer 1

Answer: (E)

Using the vector identity: \[ |\vec{a} \times \vec{b}|^2 = |\vec{a}|^2 |\vec{b}|^2 - |\vec{a} \cdot \vec{b}|^2 \] We substitute \( \theta = \frac{\pi}{3} \) and compute: \[ |\vec{a} \times \vec{b}|^2 = 45 \] Thus, the correct answer is (E).