Vector Algebra
Magnitude and Scalar Product
Grade None
Question:
<p>Given <span>|</span><strong>a</strong><span>|</span> = <span>|</span><strong>b</strong><span>|</span> = <span>|</span><strong>c</strong><span>|</span> = 1 and |<strong>a</strong> − <strong>b</strong>|² + |<strong>b</strong> − <strong>c</strong>|² + |<strong>c</strong> − <strong>a</strong>|² = 9, find <span>|</span>2<strong>a</strong> + 5<strong>b</strong> + 5<strong>c</strong><span>|</span>.</p>
Step-by-Step Solution
Key Concept: Use the constraint |a| = |b| = |c| = 1 and the given condition to find that a + b + c = 0, then substitute to find the magnitude.
Step 1: Expand | a − b |^2 + | b − c |^2 + | c − a |^2:
= | a |^2 + | b |^2 − 2 a · b + | b |^2 + | c |^2 − 2 b · c + | c |^2 + | a |^2 − 2 c · a
= 6 − 2( a · b + b · c + c · a ) = 9 Step 2: Solve for the sum of dot products:
a · b + b · c + c · a = −3/2 Step 3: Use the constraint that a · b + b · c + c · a ≥ −(1/2)(| a |^2 + | b |^2 + | c |^2) = −3/2, with equality when a + b + c = 0 . Step 4: Since a + b + c = 0 , we have b + c = − a . Step 5: Calculate |2 a + 5 b + 5 c | = |2 a + 5( b + c )| = |2 a + 5(− a )| = |−3 a | = 3| a | = 3 ∴ Answer is 3.
Correct Answer: 3