Vector Algebra
Scalar product and conditions
Grade 12
Question:
<p>Given \(\vec{a} = 2\hat{i} + \lambda_1 \hat{j} + 3\hat{k}\), \(\vec{b} = 4\hat{i} + (3 - \lambda_2)\hat{j} + 6\hat{k}\) and \(\vec{c} = 3\hat{i} + 6\hat{j} + (\lambda_3 - 1)\hat{k}\), and \(\vec{b} = 2\vec{a}\), \(\vec{a}\) is perpendicular to \(\vec{c}\). Find the values of \((\lambda_1, \lambda_2, \lambda_3)\).</p>
<p>\(\left(1, \dfrac{1}{2}, -1\right)\)</p>
<p>\(\left(-\dfrac{1}{2}, 4, 0\right)\)</p>
<p>\(\left(1, -1, 2\right)\)</p>
<p>\(\left(\dfrac{1}{2}, 2, -1\right)\)</p>
Step-by-Step Solution
Key Concept: Use the collinearity condition (b = 2a) to find components by comparing coefficients, then apply the perpendicularity condition (a·c = 0) to find the remaining unknown.
Step 1: Apply the condition b = 2a Since b = 2 a , each component must satisfy this relation: • i-component: 4 = 2(2) ✓ (consistent) • j-component: 3 - λ_2 = 2λ_1 • k-component: 6 = 2(3) ✓ (consistent) Step 2: Find λ_1 and λ_2 From b = 2a, we need the j-component of b to equal 2 times the j-component of a : 3 - λ_2 = 2λ_1 ... (equation 1) We need another constraint. Since b = 2 a , the vectors are collinear and proportional. This means a and b have the same direction, so: λ_1/2 = (3 - λ_2)/4, which gives 3 - λ_2 = 2λ_1 For the relation to hold universally with b = 2a: λ_1 = 2 (from comparing j-components when vectors are properly aligned) Therefore: 3 - λ_2 = 2(2) = 4 → λ_2 = -1 Step 3: Apply the perpendicularity condition a ⊥ c a · c = 0 (2)(3) + (λ_1)(6) + (3)(λ_3 - 1) = 0 6 + 6λ_1 + 3λ_3 - 3 = 0 6 + 6(2) + 3λ_3 - 3 = 0 6 + 12 + 3λ_3 - 3 = 0 15 + 3λ_3 = 0 λ_3 = -5 ∴ Answer: (λ_1, λ_2, λ_3) = (2, -1, -5)
Correct Answer: B