Question

Let \( \vec{a}, \vec{b}, \vec{c} \) be three vectors such that \[ \vec{a} \times \vec{b} = 2(\vec{a} \times \vec{c}). \] If \( |\vec{a}| = 1 \), \( |\vec{b}| = 4 \), \( |\vec{c}| = 2 \), and the angle between \( \vec{b} \) and \( \vec{c} \) is \(60^\circ\), then \( |\vec{a} \cdot \vec{c}| \) is equal to.

• A. 4
• B. 2
• C. 0
• D. 1

Hint:

$|\vec{u}-\vec{v}|^2 = u^2 + v^2 - 2\vec{u}\cdot\vec{v}$ is a fundamental vector identity.

Answer 1

The Correct Option is A

$\vec{a} \times \vec{b} - 2\vec{a} \times \vec{c} = 0 \implies \vec{a} \times (\vec{b} - 2\vec{c}) = 0$.
This implies $\vec{b} - 2\vec{c}$ is parallel to $\vec{a}$, or $\vec{b} - 2\vec{c} = \lambda \vec{a}$.
Squaring both sides: $|\vec{b} - 2\vec{c}|^2 = \lambda^2 |\vec{a}|^2 = \lambda^2$.
Expand LHS: $|\vec{b}|^2 + 4|\vec{c}|^2 - 4\vec{b}\cdot\vec{c}$.
$\vec{b}\cdot\vec{c} = |\vec{b}||\vec{c}|\cos 60^\circ = 4 \cdot 2 \cdot \frac{1}{2} = 4$.
LHS $= 16 + 4(4) - 4(4) = 16$.
So $|\vec{b} - 2\vec{c}|^2 = 16 \implies |\vec{b} - 2\vec{c}| = 4$.