Let \(|\vec{a}|= 5,|\vec{b}|=2\) and \(\vec{a}.\vec{b}= 6\), then the value of \(|\vec{a}\times\vec{b}|\) is equal to
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 8
Since \(\vec{a}\cdot\vec{b} = |\vec{a}||\vec{b}|\cos\theta\) and \(|\vec{a}\times\vec{b}| = |\vec{a}||\vec{b}|\sin\theta\), adding their squares gives Lagrange’s identity:
\(|\vec{a}\times\vec{b}|^2 + (\vec{a}\cdot\vec{b})^2 = |\vec{a}|^2|\vec{b}|^2\)
\(|\vec{a}\times\vec{b}|^2 = 25 \times 4 - 36 = 64\)
\(|\vec{a}\times\vec{b}| = 8\)