Vector Algebra
Inequalities involving unit vectors
Grade 12
Question:
<p>Given that \(\vec{a}\), \(\vec{b}\) and \(\vec{c}\) are unit vectors. We know that \(|\vec{a}+\vec{b}+\vec{c}|^2 = |\vec{a}|^2 + |\vec{b}|^2 + |\vec{c}|^2 + 2(\vec{a}\cdot\vec{b}+\vec{b}\cdot\vec{c}+\vec{c}\cdot\vec{a}) \geq 0\). Find the minimum value of \(|\vec{a}-\vec{b}|^2+|\vec{b}-\vec{c}|^2+|\vec{c}-\vec{a}|^2\).</p>
Step-by-Step Solution
Key Concept: Expand each squared magnitude using the dot product formula, then use the constraint that unit vectors satisfy |a|² = |b|² = |c|² = 1 to simplify the sum into a form involving dot products, recognizing that the dot products are bounded by the condition given.
Step 1: Expand each squared difference using dot product. | a - b |^2 = | a |^2 + | b |^2 - 2 a · b = 1 + 1 - 2 a · b = 2 - 2 a · b | b - c |^2 = 2 - 2 b · c | c - a |^2 = 2 - 2 c · 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: Apply the given constraint. From | a + b + c |^2 ≥ 0: 3 + 2( a · b + b · c + c · a ) ≥ 0 Therefore: a · b + b · c + c · a ≥ -3/2 Step 4: Find minimum value. To minimize the sum in Step 2, we maximize the dot product sum at its constraint boundary: Minimum = 6 - 2(-3/2) = 6 + 3 = 9 Note: This minimum is achieved when a + b + c = 0 (equality case), which is geometrically possible with three unit vectors at 120° angles in a plane.
Correct Answer: 9