Vector Algebra
Unit vectors and dot product
Grade 12
Question:
<p>If <em>x̂</em>, <em>ŷ</em> and <em>ẑ</em> are three unit vectors in three-dimensional space, then the minimum value of \(|\hat{x}+\hat{y}|^2+|\hat{y}+\hat{z}|^2+|\hat{z}+\hat{x}|^2\) is</p>
<p>\(\dfrac{3}{2}\)</p>
<p>3</p>
<p>\(3\sqrt{3}\)</p>
<p>6</p>
Step-by-Step Solution
Key Concept: Expand each squared magnitude using the dot product formula |a+b|² = |a|² + |b|² + 2a·b, then minimize the sum by recognizing that dot products between unit vectors range from -1 to 1.
Step 1: Expand each squared magnitude using |a+b|^2 = |a|^2 + |b|^2 + 2(a·b). |x̂+ŷ|^2 = 1 + 1 + 2(x̂·ŷ) = 2 + 2(x̂·ŷ) |ŷ+ẑ|^2 = 1 + 1 + 2(ŷ·ẑ) = 2 + 2(ŷ·ẑ) |ẑ+x̂|^2 = 1 + 1 + 2(ẑ·x̂) = 2 + 2(ẑ·x̂) Step 2: Sum all three expressions: Total = 6 + 2[(x̂·ŷ) + (ŷ·ẑ) + (ẑ·x̂)] Step 3: To minimize, we need to minimize (x̂·ŷ) + (ŷ·ẑ) + (ẑ·x̂). Since each dot product ≥ -1, the minimum value occurs when the three unit vectors are symmetrically arranged (120° apart in a plane or optimally in 3D). Step 4: For three unit vectors equally spaced at 120° angles: each dot product = cos(120°) = -1/2, giving sum = -3/2. Minimum value = 6 + 2(-3/2) = 6 - 3 = 3 ∴ Answer: B (or the numerical answer is 3)
Correct Answer: B