If \(\vec{a}\) and \(\vec{b}\) are two non zero vectors such that \(|\vec{a}| = 10\), \(|\vec{b}| = 2\) and \(\vec{a}\cdot\vec{b} = 12\), then value of \(|\vec{a}\times\vec{b}|\) is:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 16
Let θ be the angle between the vectors. From the dot product:
\(\vec{a}\cdot\vec{b} = |\vec{a}||\vec{b}|\cos\theta\)
\(12 = 10 \times 2 \times \cos\theta\)
\( \displaystyle \cos\theta = \frac{12}{20} = \frac{3}{5} \)
Since \(0 \le \theta \le \pi\), \(\sin\theta \ge 0\):
\( \displaystyle \sin\theta = \sqrt{1 - \frac{9}{25}} = \frac{4}{5} \)
The magnitude of the cross product is
\( \displaystyle |\vec{a}\times\vec{b}| = |\vec{a}||\vec{b}|\sin\theta = 10 \times 2 \times \frac{4}{5} = 16 \)
Shortcut: By Lagrange’s identity \(|\vec{a}\times\vec{b}|^2 + (\vec{a}\cdot\vec{b})^2 = |\vec{a}|^2|\vec{b}|^2\):
\(|\vec{a}\times\vec{b}|^2 = 100 \times 4 - 144 = 256\)
\(|\vec{a}\times\vec{b}| = 16\)
NCERT Exemplar:
Vector Algebra
