Vector Algebra
Vector Decomposition and Cross Product
Grade 12
Question:
<p>Given \(\vec{\alpha} = 3\hat{i} + \hat{j}\), \(\vec{\beta} = 2\hat{i} - \hat{j} + 3\hat{k}\). If \(\vec{\beta} = \vec{\beta}_1 - \vec{\beta}_2\) and \(\vec{\beta}_1\) is parallel to \(\vec{\alpha}\) and \(\vec{\beta}_2\) is perpendicular to \(\vec{\alpha}\), then \(\vec{\beta}_1 \times \vec{\beta}_2\) is equal to</p>
<p>\(\dfrac{1}{2}(-3\hat{i} + 9\hat{j} + 5\hat{k})\)</p>
<p>\(\dfrac{1}{2}(3\hat{i} - 9\hat{j} + 5\hat{k})\)</p>
<p>\(3\hat{i} - 9\hat{j} - 5\hat{k}\)</p>
<p>\(-3\hat{i} + 9\hat{j} + 5\hat{k}\)</p>
Step-by-Step Solution
Key Concept: Decompose β into components parallel and perpendicular to α using projection formula: β₁ = (β·α/|α|²)α, then β₂ = β - β₁. The cross product of orthogonal components yields maximum magnitude proportional to both vectors' magnitudes.
Step 1: Find β_1 (component of β parallel to α) Since β_1 is parallel to α: β_1 = λα for some scalar λ Using the condition that β = β_1 + β_2 and β_2 ⊥ α: β · α = β_1 · α = λ|α|^2 β · α = (2)(3) + (-1)(1) + (3)(0) = 6 - 1 = 5 |α|^2 = 9 + 1 = 10 λ = 5/10 = 1/2 β_1 = (1/2)(3î + ĵ) = (3/2)î + (1/2)ĵ Step 2: Find β_2 (component perpendicular to α) β_2 = β - β_1 = (2î - ĵ + 3k̂) - ((3/2)î + (1/2)ĵ) β_2 = (1/2)î - (3/2)ĵ + 3k̂ Step 3: Compute β_1 × β_2 β_1 × β_2 = |î ĵ k̂| |(3/2) (1/2) 0| |(1/2) (-3/2) 3| = î[(1/2)(3) - 0(-3/2)] - ĵ[(3/2)(3) - 0(1/2)] + k̂[(3/2)(-3/2) - (1/2)(1/2)] = î(3/2) - ĵ(9/2) + k̂(-9/4 - 1/4) = (3/2)î - (9/2)ĵ - (5/2)k̂ ∴ β_1 × β_2 = (3/2)î - (9/2)ĵ - (5/2)k̂ (or equivalent form)
Correct Answer: A