Vector Algebra
Cross product and magnitude
Grade 12
Question:
<p>As \(|\vec{a}| = 1\), \(|\vec{b}| = 1\), \(|\vec{a}+\vec{b}| = \sqrt{3}\), \(\vec{c} = \vec{a} + 2\vec{b} + 3(\vec{a} \times \vec{b})\). Find \(2|\vec{c}|\).</p>
<p>\(\sqrt{55}\)</p>
<p>\(\sqrt{50}\)</p>
<p>\(\sqrt{60}\)</p>
<p>\(\sqrt{45}\)</p>
Step-by-Step Solution
Key Concept: Use the constraint |a+b|=√3 to find a·b, then compute |c|² by expanding the dot product and using the scalar triple product identity (a×b)·(a×b) = |a|²|b|²-(a·b)².
Step 1: Find a·b using the given constraint |a+b|^2 = 3 (a+b)·(a+b) = 3 |a|^2 + 2(a·b) + |b|^2 = 3 1 + 2(a·b) + 1 = 3 a·b = 1/2 Step 2: Find |a×b|^2 |a×b|^2 = |a|^2|b|^2 - (a·b)^2 |a×b|^2 = (1)(1) - (1/2)^2 = 1 - 1/4 = 3/4 Step 3: Compute |c|^2 c = a + 2b + 3(a×b) |c|^2 = (a + 2b + 3(a×b))·(a + 2b + 3(a×b)) |c|^2 = |a|^2 + 4|b|^2 + 9|a×b|^2 + 4(a·b) + 6(a·(a×b)) + 12(b·(a×b)) Since (a·(a×b)) = 0 and (b·(a×b)) = 0: |c|^2 = 1 + 4(1) + 9(3/4) + 4(1/2) |c|^2 = 1 + 4 + 27/4 + 2 = 7 + 27/4 = 55/4 |c| = √(55/4) = √55/2 Step 4: Calculate 2|c| 2|c| = 2 · (√55/2) = √55 ∴ Answer: A
Correct Answer: A