If \(|\vec{a}| = 2\), \(|\vec{b}| = 5\) and \(\vec{a}\cdot\vec{b} = 6\), then \(|\vec{a}\times\vec{b}|\) is equal to
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 8
Let \(\theta\) be the angle between \(\vec{a}\) and \(\vec{b}\). The scalar product gives \(\cos\theta\):
\(\vec{a}\cdot\vec{b} = |\vec{a}|\,|\vec{b}|\cos\theta\)
\(6 = 2\times5\times\cos\theta\)
\( \displaystyle \cos\theta = \frac{3}{5} \)
Since \(0 \le \theta \le \pi\), \(\sin\theta\) is not negative:
\( \displaystyle \sin\theta = \sqrt{1-\frac{9}{25}} = \frac{4}{5} \)
The magnitude of the vector product is
\( \displaystyle |\vec{a}\times\vec{b}| = |\vec{a}|\,|\vec{b}|\sin\theta = 2\times5\times\frac{4}{5} = 8 \)
Shortcut: \(|\vec{a}\times\vec{b}|^2+(\vec{a}\cdot\vec{b})^2 = |\vec{a}|^2\,|\vec{b}|^2\), so \(|\vec{a}\times\vec{b}|^2 = 100-36 = 64\).
Why the other options are incorrect
Option (1) 6 — Incorrect. This is the scalar product \(\vec{a}\cdot\vec{b}\), not \(|\vec{a}\times\vec{b}|\).
Option (2) 10 — Incorrect. This is \(|\vec{a}|\,|\vec{b}|\); it would need \(\sin\theta = 1\), but \(\sin\theta = \displaystyle\frac{4}{5}\).
Option (4) \(2\sqrt{34}\) — Incorrect. This is \(\sqrt{100+36}\); the square of the scalar product must be subtracted, not added.