<p>If \([\vec{a} \times \vec{b} \quad \vec{b} \times \vec{c} \quad \vec{c} \times \vec{a}] = \lambda[\vec{a} \; \vec{b} \; \vec{c}]^2\), then \(\lambda\) is equal to</p>
Step-by-Step Solution
Key Concept: Use the scalar triple product identity: [u v w] = u·(v×w), and recognize that [a×b b×c c×a] can be expressed using the BAC-CAB rule and properties of determinants to relate it to [a b c]².
Step 1: Recall that the scalar triple product [u v w] = u·(v×w) = det[u v w]. Step 2: For the left side, we need (a×b)·(b×c × c×a). Using the cyclic property of scalar triple product and the vector identity for cross products. Step 3: Apply the BAC-CAB rule: (b×c)×(c×a) = [b·a·c - b·c·a]c. After careful expansion using the identity (u×v)·(w×x) = (u·w)(v·x) - (u·x)(v·w). Step 4: The scalar triple product satisfies: [a×b b×c c×a] = [a b c]^2 (this is a known identity in vector algebra). Step 5: Comparing with the given equation [a×b b×c c×a] = λ[a b c]^2, we get λ = 1. ∴ Answer: B (λ = 1)
Correct Answer: B