<p>Let \(\vec{a}\) and \(\vec{b}\) be two unit vectors such that \(|\vec{a}+\vec{b}|=\sqrt{3}\). If \(\vec{c}=\vec{a}+2\vec{b}+3(\vec{a}\times\vec{b})\), then \(2|\vec{c}|\) is equal to</p>
Step-by-Step Solution
Key Concept: Use the condition |a+b|=√3 with unit vectors to find a·b, then express |c|² by expanding c using the dot product and cross product properties, noting that (a×b)·a = 0 and (a×b)·b = 0.
Step 1: Since |a| = |b| = 1 (unit vectors), use |a+b|^2 = 3: |a+b|^2 = |a|^2 + |b|^2 + 2(a·b) = 1 + 1 + 2(a·b) = 3 Therefore: a·b = 1/2 Step 2: Find |c|^2 where c = a + 2b + 3(a×b): |c|^2 = (a + 2b + 3(a×b))·(a + 2b + 3(a×b)) Step 3: Expand using distributive property. Key insight: (a×b)⊥a and (a×b)⊥b, so cross terms vanish: |c|^2 = |a|^2 + 4|b|^2 + 9|a×b|^2 + 4(a·b) Step 4: Calculate |a×b|^2: |a×b|^2 = |a|^2|b|^2 - (a·b)^2 = 1·1 - (1/2)^2 = 3/4 Step 5: Substitute values: |c|^2 = 1 + 4(1) + 9(3/4) + 4(1/2) = 1 + 4 + 27/4 + 2 = 7 + 27/4 = 55/4 |c| = √(55/4) = √55/2 Step 6: Therefore: 2|c| = 2·(√55/2) = √55 ∴ Answer: A
Correct Answer: A