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

Step 1: Understanding the Concept:
We are given \(\vec{c}\) as a linear combination of two known vectors \(\vec{a}\) and \(\vec{b}\).
By applying the distributive property of the dot product, we can generate a system of two linear equations in terms of the unknown scalars \(\lambda\) and \(\mu\).
Once \(\lambda\) and \(\mu\) are found, we can construct \(\vec{c}\) and compute its squared magnitude.
Step 2: Key Formula or Approach:
The dot product distributes over vector addition: \((\lambda \vec{a} + \mu \vec{b}) \cdot \vec{v} = \lambda(\vec{a} \cdot \vec{v}) + \mu(\vec{b} \cdot \vec{v})\).
Let \(\vec{v}_1 = 3\hat{i} - 6\hat{j} + 2\hat{k}\) and \(\vec{v}_2 = \hat{i} + \hat{j} + \hat{k}\).
The given conditions are \(\vec{c} \cdot \vec{v}_1 = 10\) and \(\vec{c} \cdot \vec{v}_2 = -2\).
Step 3: Detailed Explanation:
First, calculate the dot products of \(\vec{a}\) and \(\vec{b}\) with \(\vec{v}_1\).
\[ \vec{a} \cdot \vec{v}_1 = (-1)(3) + (1)(-6) + (3)(2) = -3 - 6 + 6 = -3 \] \[ \vec{b} \cdot \vec{v}_1 = (1)(3) + (3)(-6) + (1)(2) = 3 - 18 + 2 = -13 \] Substitute these into the first condition.
\[ \lambda(-3) + \mu(-13) = 10 \implies -3\lambda - 13\mu = 10 \quad \text{--- (Eq 1)} \] Next, calculate the dot products of \(\vec{a}\) and \(\vec{b}\) with \(\vec{v}_2\).
\[ \vec{a} \cdot \vec{v}_2 = (-1)(1) + (1)(1) + (3)(1) = -1 + 1 + 3 = 3 \] \[ \vec{b} \cdot \vec{v}_2 = (1)(1) + (3)(1) + (1)(1) = 1 + 3 + 1 = 5 \] Substitute these into the second condition.
\[ \lambda(3) + \mu(5) = -2 \implies 3\lambda + 5\mu = -2 \quad \text{--- (Eq 2)} \] Now, solve the linear system by adding (Eq 1) and (Eq 2).
\[ (-3\lambda - 13\mu) + (3\lambda + 5\mu) = 10 - 2 \] \[ -8\mu = 8 \implies \mu = -1 \] Substitute \(\mu = -1\) into (Eq 2) to find \(\lambda\).
\[ 3\lambda + 5(-1) = -2 \implies 3\lambda = 3 \implies \lambda = 1 \] Now that we have \(\lambda\) and \(\mu\), we construct the vector \(\vec{c}\).
\[ \vec{c} = (1)\vec{a} + (-1)\vec{b} = \vec{a} - \vec{b} \] \[ \vec{c} = (-\hat{i} + \hat{j} + 3\hat{k}) - (\hat{i} + 3\hat{j} + \hat{k}) = -2\hat{i} - 2\hat{j} + 2\hat{k} \] Finally, calculate the squared magnitude \(|\vec{c}|^2\).
\[ |\vec{c}|^2 = (-2)^2 + (-2)^2 + 2^2 = 4 + 4 + 4 = 12 \] Step 4: Final Answer:
The value of \(|\vec{c}|^2\) is \(12\).