Vector Algebra
Dot Product and Collinear Vectors
Grade 12
Question:
<p>Let <strong>61.</strong> Let \(\vec{a} = 2\hat{i} + \lambda_1\hat{j} + 3\hat{k}\), \(\vec{b} = 4\hat{i} + (3-\lambda_2)\hat{j} + 6\hat{k}\) and \(\vec{c} = 3\hat{i} + 6\hat{j} + (\lambda_3 - 1)\hat{k}\) be three vectors such that \(\vec{b} = 2\vec{a}\) and \(\vec{a}\) is perpendicular to \(\vec{c}\). Then a possible value of \((\lambda_1, \lambda_2, \lambda_3)\) is:</p>
<p>(1, 3, 1)</p>
<p>\(\left(-\dfrac{1}{2}, 4, 0\right)\)</p>
<p>\(\left(\dfrac{1}{2}, 4, -2\right)\)</p>
<p>(1, 5, 1)</p>
Step-by-Step Solution
Key Concept: Use the collinearity condition $\vec{b} = 2\vec{a}$ to find $\lambda_1$ and $\lambda_2$, then apply the perpendicularity condition $\vec{a} \cdot \vec{c} = 0$ to find $\lambda_3$.
Step 1: Apply collinearity condition $\vec{b} = 2\vec{a}$ Comparing components: i-component: $4 = 2(2)$ ✓ j-component: $3 - \lambda_2 = 2\lambda_1$ k-component: $6 = 2(3)$ ✓ From j-component: $\lambda_2 = 3 - 2\lambda_1$ ... (1) Step 2: Apply perpendicularity condition $\vec{a} \perp \vec{c}$ $\vec{a} \cdot \vec{c} = 0$ $(2)(3) + (\lambda_1)(6) + (3)(\lambda_3 - 1) = 0$ $6 + 6\lambda_1 + 3\lambda_3 - 3 = 0$ $6\lambda_1 + 3\lambda_3 = -3$ $2\lambda_1 + \lambda_3 = -1$ $\lambda_3 = -1 - 2\lambda_1$ ... (2) Step 3: Find possible values Choose $\lambda_1 = 0$: From (1): $\lambda_2 = 3 - 0 = 3$ From (2): $\lambda_3 = -1 - 0 = -1$ ∴ A possible value is $(\lambda_1, \lambda_2, \lambda_3) = (0, 3, -1)$ Alternative: $\lambda_1 = 1$ gives $(1, 1, -3)$; $\lambda_1 = -1$ gives $(-1, 5, 1)$, etc.
Correct Answer: C