Question

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

• A. 5
• B. -5
• C. 3
• D. -3

Hint:

Always check if given vectors are unit vectors or orthogonal ($\vec{a} \cdot \vec{b} = 0$) first to simplify complex vector products.

Answer 1

The Correct Option is B

Step 1: Simplify the triple product
Let $\vec{V} = (\vec{a} \times \vec{b}) \times (2\vec{a} + \vec{b})$.
Using Vector Triple Product formula $\vec{X} \times (\vec{Y} \times \vec{Z})$ or expanding directly:
$\vec{V} = [(\vec{a} \times \vec{b}) \cdot \vec{b}] 2\vec{a} - \dots$ simplifies to $(\vec{a} \times \vec{b}) \times 2\vec{a} + (\vec{a} \times \vec{b}) \times \vec{b}$.

Step 2: Geometric Analysis
The entire expression is a Scalar Triple Product: $(\vec{a} - 2\vec{b}) \cdot [(\vec{a} \times \vec{b}) \times (2\vec{a} + \vec{b})]$.
Since $(\vec{a} \times \vec{b})$ is perpendicular to the plane containing $\vec{a}$ and $\vec{b}$, the result is based on $|a|^2|b|^2 - (a \cdot b)^2$.

Step 3: Magnitude calculation
$|\vec{a}|^2 = \frac{1}{10}(9+1) = 1$.
$|\vec{b}|^2 = \frac{1}{49}(4+9+36) = 1$.
$\vec{a} \cdot \vec{b} = \frac{1}{7\sqrt{10}}(6 + 0 - 6) = 0$.

Step 4: Final value
The expression simplifies to $-5 |\vec{a} \times \vec{b}|^2$. Since $\vec{a} \perp \vec{b}$ and both are unit vectors, $|\vec{a} \times \vec{b}| = 1$. Result is $-5$.
Final Answer: (B)