Vector Algebra
Vector Decomposition and Cross Product
Grade 12
Question:
<p><strong>64.</strong> Let \(\vec{\alpha} = 3\hat{i} + \hat{j}\) and \(\vec{\beta} = 2\hat{i} - \hat{j} + 3\hat{k}\). If \(\vec{\beta} = \vec{\beta}_1 - \vec{\beta}_2\), where \(\vec{\beta}_1\) 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>\(-3\hat{i} + 9\hat{j} + 5\hat{k}\)</p>
<p>\(3\hat{i} - 9\hat{j} - 5\hat{k}\)</p>
<p>\(\dfrac{1}{2}(-3\hat{i} + 9\hat{j} + 5\hat{k})\)</p>
<p>\(\dfrac{1}{2}(3\hat{i} - 9\hat{j} + 5\hat{k})\)</p>
Step-by-Step Solution
Key Concept: Decompose β into parallel and perpendicular components using projection formulas: β₁ = (β·α/|α|²)α and β₂ = β - β₁, then compute their cross product directly.
Step 1: Find β_1 (component of β parallel to α) β_1 = (β·α/|α|^2)α β·α = (2)(3) + (-1)(1) + (3)(0) = 6 - 1 = 5 |α|^2 = 3^2 + 1^2 = 10 β_1 = (5/10)(3î + ĵ) = (1/2)(3î + ĵ) = (3/2)î + (1/2)ĵ Step 2: Find β_2 (component of β perpendicular to α) β_2 = β - β_1 = (2î - ĵ + 3k̂) - (3/2î + 1/2ĵ) β_2 = (1/2)î - (3/2)ĵ + 3k̂ Step 3: Calculate 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 ∴ Answer: A
Correct Answer: A