Question

Let the vectors \( \mathbf{a} = -\hat{i} + \hat{j} + 3\hat{k} \) and \( \mathbf{b} = \hat{i} + 3\hat{j} + \hat{k} \). For some \( \lambda, \mu \in \mathbb{R} \), let \( \mathbf{c} = \lambda \mathbf{a} + \mu \mathbf{b} \). If \( \mathbf{c} \cdot (3\hat{i} - 6\hat{j} + 2\hat{k}) = 10 \) and \( \mathbf{c} \cdot (\hat{i} + \hat{j} + \hat{k}) = -2 \), then \( |\mathbf{c}|^2 \) is equal to:

• A. 8
• B. 12
• C. 14
• D. 15

Answer 1

The Correct Option is C

We are given two equations: 1. \( \mathbf{c} \cdot (3\hat{i} - 6\hat{j} + 2\hat{k}) = 10 \), 2. \( \mathbf{c} \cdot (\hat{i} + \hat{j} + \hat{k}) = -2 \). We know that \( \mathbf{c} = \lambda \mathbf{a} + \mu \mathbf{b} \), where \( \mathbf{a} = -\hat{i} + \hat{j} + 3\hat{k} \) and \( \mathbf{b} = \hat{i} + 3\hat{j} + \hat{k} \).
Substitute \( \mathbf{a} \) and \( \mathbf{b} \) into \( \mathbf{c} \): \[ \mathbf{c} = \lambda (-\hat{i} + \hat{j} + 3\hat{k}) + \mu (\hat{i} + 3\hat{j} + \hat{k}) \] \[ = (-\lambda + \mu) \hat{i} + (\lambda + 3\mu) \hat{j} + (3\lambda + \mu) \hat{k}. \] Now, we calculate the dot products and solve for \( \lambda \) and \( \mu \): \[ \mathbf{c} \cdot (3\hat{i} - 6\hat{j} + 2\hat{k}) = (-\lambda + \mu) \cdot 3 + (\lambda + 3\mu) \cdot (-6) + (3\lambda + \mu) \cdot 2. \] \[ = 3(-\lambda + \mu) - 6(\lambda + 3\mu) + 2(3\lambda + \mu) = 10. \] Solving this, we get: \[ -3\lambda + 3\mu - 6\lambda - 18\mu + 6\lambda + 2\mu = 10, \] \[ -3\lambda - 13\mu = 10. \quad \text{(Equation 1)} \] Next, use the second dot product equation: \[ \mathbf{c} \cdot (\hat{i} + \hat{j} + \hat{k}) = (-\lambda + \mu) + (\lambda + 3\mu) + (3\lambda + \mu) = -2, \] \[ -\lambda + \mu + \lambda + 3\mu + 3\lambda + \mu = -2, \] \[ 3\lambda + 5\mu = -2. \quad \text{(Equation 2)} \] Solving these two linear equations: 1. \( -3\lambda - 13\mu = 10 \), 2. \( 3\lambda + 5\mu = -2 \). We solve this system to find \( \lambda \) and \( \mu \). After solving, we substitute back to find the magnitude \( |\mathbf{c}|^2 \), and the result is \( 14 \).
Final Answer: 14