Question

If $\vec{a} + 2\vec{b} - \vec{c} = \vec{0}$ and $\vec{a} \times \vec{b} + \vec{b} \times \vec{c} + \vec{c} \times \vec{a} = \lambda \, \vec{a} \times \vec{b}$, then the value of $\lambda$ is equal to:

• A. $5$
• B. $4$
• C. $2$
• D. $-2$
• E. $-4$

Hint:

Always express one vector in terms of others to simplify cyclic cross product expressions quickly.

Answer 1

The Correct Option is D

Concept: Use properties of cross product: \[ \vec{a} \times \vec{b} = -(\vec{b} \times \vec{a}), \quad \vec{x} \times \vec{x} = 0 \] and distributive law.

Step 1: Express \( \vec{c} \) in terms of \( \vec{a}, \vec{b} \).
\[ \vec{a} + 2\vec{b} - \vec{c} = 0 \Rightarrow \vec{c} = \vec{a} + 2\vec{b} \]

Step 2: Substitute into given expression.
\[ \vec{a} \times \vec{b} + \vec{b} \times \vec{c} + \vec{c} \times \vec{a} \] \[ = \vec{a} \times \vec{b} + \vec{b} \times (\vec{a} + 2\vec{b}) + (\vec{a} + 2\vec{b}) \times \vec{a} \]

Step 3: Expand each term.
\[ \vec{b} \times (\vec{a} + 2\vec{b}) = \vec{b} \times \vec{a} + 2(\vec{b} \times \vec{b}) = -\vec{a} \times \vec{b} \] \[ (\vec{a} + 2\vec{b}) \times \vec{a} = \vec{a} \times \vec{a} + 2(\vec{b} \times \vec{a}) = -2\vec{a} \times \vec{b} \]

Step 4: Combine all terms.
\[ = (\vec{a} \times \vec{b}) - (\vec{a} \times \vec{b}) - 2(\vec{a} \times \vec{b}) \] \[ = -2 \vec{a} \times \vec{b} \]

Step 5: Compare with given form.
\[ \vec{a} \times \vec{b} + \vec{b} \times \vec{c} + \vec{c} \times \vec{a} = \lambda (\vec{a} \times \vec{b}) \] \[ \Rightarrow \lambda = -2 \]