Vector Algebra
Unit Vectors and Inequalities
Grade 12
Question:
<p><strong>70.</strong> If \(\vec{a}\), \(\vec{b}\) and \(\vec{c}\) are unit vectors, then \(|\vec{a} - \vec{b}|^2 + |\vec{b} - \vec{c}|^2 + |\vec{c} - \vec{a}|^2\) does not exceed ________.</p>
Step-by-Step Solution
Key Concept: Expand each squared magnitude using the dot product formula |⃗u - ⃗v|² = |⃗u|² + |⃗v|² - 2⃗u·⃗v, then recognize that the sum is bounded by the constraint that ⃗a·⃗b + ⃗b·⃗c + ⃗c·⃗a ≥ -3/2 (achieved when vectors are symmetrically arranged).
Step 1: Expand each term using |⃗u - ⃗v|^2 = |⃗u|^2 + |⃗v|^2 - 2⃗u·⃗v: |⃗a - ⃗b|^2 = 1 + 1 - 2⃗a·⃗b = 2 - 2⃗a·⃗b |⃗b - ⃗c|^2 = 2 - 2⃗b·⃗c |⃗c - ⃗a|^2 = 2 - 2⃗c·⃗a Step 2: Sum all three expressions: |⃗a - ⃗b|^2 + |⃗b - ⃗c|^2 + |⃗c - ⃗a|^2 = 6 - 2(⃗a·⃗b + ⃗b·⃗c + ⃗c·⃗a) Step 3: To maximize this sum, minimize (⃗a·⃗b + ⃗b·⃗c + ⃗c·⃗a). For unit vectors, the minimum value of this sum is -3/2, achieved when the three unit vectors are symmetrically oriented 120° apart (in a plane). Step 4: Maximum value = 6 - 2(-3/2) = 6 + 3 = 9 ∴ Answer: 9
Correct Answer: 9