Question

Let \( \vec{a} \) and \( \vec{b} \) be two non-zero vectors. Prove that \( |\vec{a} \times \vec{b}| \leq |\vec{a}| |\vec{b}| \). State the condition under which equality holds, i.e., \( |\vec{a} \times \vec{b}| = |\vec{a}| |\vec{b}| \).

Hint:

The cross product reaches its maximum magnitude when the vectors are perpendicular.

Answer 1

1. Expression for the magnitude of the cross product: \[ |\vec{a} \times \vec{b}| = |\vec{a}| |\vec{b}| \sin \theta, \] where \( \theta \) is the angle between \( \vec{a} \) and \( \vec{b} \). 2. Inequality: The sine function satisfies: \[ 0 \leq \sin \theta \leq 1. \] Thus: \[ |\vec{a} \times \vec{b}| = |\vec{a}| |\vec{b}| \sin \theta \leq |\vec{a}| |\vec{b}|. \] 3. Equality condition: Equality holds when: \[ \sin \theta = 1 \quad \Rightarrow \quad \theta = \frac{\pi}{2}. \] This means \( \vec{a} \) and \( \vec{b} \) are perpendicular. Final Answer: \[ |\vec{a} \times \vec{b}| \leq |\vec{a}| |\vec{b}|, \quad {equality holds when \( \vec{a} \perp \vec{b} \)}. \]