Question

If \( \bar{a} = \frac{1}{\sqrt{10}}(3\hat{i} + \hat{k}) \) and \( \bar{b} = \frac{1}{7}(2\hat{i} + 3\hat{j} - 6\hat{k}) \), then the value of \( (2\bar{a} - \bar{b}) \cdot ((\bar{a} \times \bar{b}) \times (\bar{a} + 2\bar{b})) = \)}

• A. 3
• B. -3
• C. 5
• D. -5

Hint:

Vector Triple Product: $(\vec{u} \times \vec{v}) \times \vec{w} = (\vec{u} \cdot \vec{w})\vec{v} - (\vec{v} \cdot \vec{w})\vec{u}$.

Answer 1

The Correct Option is D

Step 1: Simplify Cross Product
$(\bar{a} \times \bar{b}) \times (\bar{a} + 2\bar{b}) = (\bar{a} \times \bar{b}) \times \bar{a} + 2((\bar{a} \times \bar{b}) \times \bar{b})$.
$= [|\bar{a}|^2 \bar{b} - (\bar{a} \cdot \bar{b})\bar{a}] + 2[(\bar{a} \cdot \bar{b})\bar{b} - |\bar{b}|^2 \bar{a}]$.
Step 2: Magnitudes
$|\bar{a}|^2 = 1$, $|\bar{b}|^2 = 1$.
$\bar{a} \cdot \bar{b} = \frac{1}{7\sqrt{10}}(6 + 0 - 6) = 0$. (Orthogonal!)
Step 3: Substitute
The vector becomes $[1\bar{b} - 0] + 2[0 - 1\bar{a}] = \bar{b} - 2\bar{a}$.
Step 4: Final Dot Product
$(2\bar{a} - \bar{b}) \cdot (\bar{b} - 2\bar{a}) = -|2\bar{a} - \bar{b}|^2$.
Since $\bar{a} \cdot \bar{b} = 0$: $-[4|\bar{a}|^2 + |\bar{b}|^2] = -[4(1) + 1] = -5$.
Final Answer:(D)