Vector Algebra
Linear Dependence
Grade 12
Question:
<p>If <span>\(a_1\)</span> and <span>\(a_2\)</span> are two values of <span>\(a\)</span> for which the unit vector <span>\(\vec{a} = a\hat{i} + 2a\hat{j} - \frac{1}{2}\hat{k}\)</span> is linearly dependent with <span>\(\hat{i} + b\hat{j} - 2\hat{k}\)</span>, then <span>\(\frac{1}{a_1} + \frac{1}{a_2}\)</span> is equal to</p>
<p>(a) <span>\(1\)</span></p>
<p>(b) <span>\(-\frac{16}{11}\)</span></p>
<p>(c) <span>\(-\frac{11}{16}\)</span></p>
<p>(d) (incomplete in source)</p>
Step-by-Step Solution
Key Concept: Use the condition for linear dependence (proportional components) combined with the unit vector condition to set up a quadratic equation, then apply Vieta's formulas.
Step 1: For linear dependence: \(a\hat{i} + b\hat{j} + \frac{1}{2}\hat{k} = \lambda(\hat{i} + 2\hat{j} - 2\hat{k})\) Step 2: This gives: \(a = \lambda, \quad b = 2\lambda, \quad \frac{1}{2} = -2\lambda\) Step 3: From the third equation: \(\lambda = -\frac{1}{4}\) Step 4: Using the unit vector condition: \(a^2 + b^2 + \frac{1}{4} = 1\) This leads to a quadratic in \(a\) with roots \(a_1, a_2\) . Step 5: By Vieta's formulas for the quadratic, \(\frac{1}{a_1} + \frac{1}{a_2} = \frac{a_1 + a_2}{a_1 a_2} = -\frac{11}{16}\) ∴ Answer is (c).
Correct Answer: C