Question

If \[ \mathbf{a} = \frac{1}{\sqrt{10}} (3i + k), \quad \mathbf{b} = \frac{1}{7} (2i + 3j - 6k), \quad \text{then the value of} \] \[ (2\mathbf{a} - \mathbf{b}) \cdot \left[ (\mathbf{a} \times \mathbf{b}) \times (\mathbf{a} + 2\mathbf{b}) \right] \]

• A. 7
• B. -5
• C. 5
• D. -7

Hint:

For vector problems involving cross and dot products, always carefully compute the cross product first, then use the dot product to find the final result.

Answer 1

The Correct Option is B

Step 1: Compute \( 2\mathbf{a} - \mathbf{b} \).
\[ 2\mathbf{a} = \frac{2}{\sqrt{10}} (3i + k) = \frac{6}{\sqrt{10}} i + \frac{2}{\sqrt{10}} k \] \[ \mathbf{b} = \frac{1}{7} (2i + 3j - 6k) = \frac{2}{7} i + \frac{3}{7} j - \frac{6}{7} k \] Thus, \[ 2\mathbf{a} - \mathbf{b} = \left( \frac{6}{\sqrt{10}} - \frac{2}{7} \right) i + \left( -\frac{3}{7} \right) j + \left( \frac{2}{\sqrt{10}} + \frac{6}{7} \right) k \]
Step 2: Compute the cross product \( \mathbf{a} \times \mathbf{b} \).
We use the determinant formula for the cross product: \[ \mathbf{a} \times \mathbf{b} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ \frac{3}{\sqrt{10}} & 0 & \frac{1}{\sqrt{10}} \\ \frac{2}{7} & \frac{3}{7} & -\frac{6}{7} \end{vmatrix} \] After calculating the determinant, we get: \[ \mathbf{a} \times \mathbf{b} = \left( \frac{-3}{7\sqrt{10}} \right) \mathbf{i} + \left( \frac{-6}{7\sqrt{10}} \right) \mathbf{j} + \left( \frac{9}{7\sqrt{10}} \right) \mathbf{k} \]
Step 3: Compute \( (\mathbf{a} \times \mathbf{b}) \times (\mathbf{a} + 2\mathbf{b}) \).
Next, calculate the cross product of \( \mathbf{a} \times \mathbf{b} \) with \( \mathbf{a} + 2\mathbf{b} \). The result simplifies, and after performing the operations, we get: \[ (\mathbf{a} \times \mathbf{b}) \times (\mathbf{a} + 2\mathbf{b}) = \mathbf{C} \quad (\text{a vector result}) \]
Step 4: Take the dot product.
Finally, take the dot product of \( 2\mathbf{a} - \mathbf{b} \) with the resulting vector \( \mathbf{C} \). After simplifying, the final value is \( -5 \). Therefore, the correct answer is option (B).