Question

If \( \vec{a}, \vec{b}, \vec{c} \) are vectors such that \( \vec{a} \ne 0 \), \( \vec{a} \times \vec{b} = 2(\vec{a} \times \vec{c}) \), \( |\vec{a}| = 1 \), \( |\vec{c}| = 1 \), \( |\vec{b}| = 4 \), and the angle between \( \vec{b} \) and \( \vec{c} \) is \( \cos^{-1} \left( \frac{1}{4} \right) \). If \( \vec{b} - 2\vec{c} = \lambda \vec{a} \), then \( \lambda = \):

• A. \( 4 \)
• B. \( 3 \)
• C. \( 2 \)
• D. \( 1 \)

Hint:

Cross Product and Parallel Vectors}
If cross product is zero, vectors are parallel
Use dot product to compute magnitude of projection
Always verify using geometric identity: \( \vec{a} \cdot \vec{b} = |\vec{a}||\vec{b}|\cos\theta \)

Answer 1

The Correct Option is A

Given: \[ \vec{a} \times \vec{b} = 2(\vec{a} \times \vec{c}) \Rightarrow \vec{a} \times (\vec{b} - 2\vec{c}) = \vec{0} \Rightarrow \vec{b} - 2\vec{c} \parallel \vec{a} \Rightarrow \vec{b} - 2\vec{c} = \lambda \vec{a} \] Take dot product of both sides with themselves: \[ |\vec{b} - 2\vec{c}|^2 = \lambda^2 \Rightarrow |\vec{b}|^2 + 4|\vec{c}|^2 - 4\vec{b} \cdot \vec{c} = \lambda^2 \] Use: \[ |\vec{b}| = 4, \quad |\vec{c}| = 1, \quad \vec{b} \cdot \vec{c} = 4 \cdot 1 \cdot \frac{1}{4} = 1 \Rightarrow 16 + 4 - 4 = \lambda^2 \Rightarrow \lambda^2 = 16 \Rightarrow \lambda = 4 \]