Question

If \( \vec{a}, \vec{b}, \vec{c} \) are three vectors such that \[ \vec{a} \times \vec{b} = 2(\vec{a} \times \vec{c}), \] \( |\vec{a}| = 1,\; |\vec{b}| = 4,\; |\vec{c}| = 2 \) and the angle between \( \vec{b} \) and \( \vec{c} \) is \( 60^\circ \), then find \( |\vec{a} \cdot \vec{c}| \):

• A. 4
• B. 1
• C. 2
• D. \( \dfrac{1}{2} \)

Hint:

When cross products of a common vector are proportional, the angles made with that vector are equal.

Answer 1

The Correct Option is B

Step 1: Use the given vector identity.
From \[ \vec{a} \times \vec{b} = 2(\vec{a} \times \vec{c}), \] taking magnitudes on both sides, \[ |\vec{a} \times \vec{b}| = 2|\vec{a} \times \vec{c}| \]
Step 2: Expand the magnitudes.
Using \( |\vec{a} \times \vec{b}| = |\vec{a}||\vec{b}|\sin\theta \), \[ 1 \cdot 4 \sin\theta = 2(1 \cdot 2 \sin\phi) \Rightarrow \sin\theta = \sin\phi \]
Step 3: Relate angles.
This implies that vectors \( \vec{b} \) and \( \vec{c} \) make equal angles with \( \vec{a} \).
Step 4: Find \( \vec{a} \cdot \vec{c} \).
Using \[ |\vec{a} \cdot \vec{c}| = |\vec{a}||\vec{c}|\cos\phi \] Since \( \angle(\vec{b}, \vec{c}) = 60^\circ \) and the magnitudes are given, solving gives \[ |\vec{a} \cdot \vec{c}| = 1 \]