Step 1: Understanding the Question:
We need to evaluate a complex scalar triple product involving vector cross products.
Step 2: Key Formula or Approach:
Use the properties of vector products:
1. \( \vec{u} \cdot (\vec{v} \times \vec{w}) = [\vec{u}, \vec{v}, \vec{w}] \).
2. \( (\vec{u} \times \vec{v}) \times \vec{w} = (\vec{u} \cdot \vec{w})\vec{v} - (\vec{v} \cdot \vec{w})\vec{u} \).
Alternatively, observe the structure: \( \vec{v} \cdot (\vec{u} \times \vec{w}) = \vec{u} \cdot (\vec{w} \times \vec{v}) \).
Step 3: Detailed Explanation:
The expression is \( (2\bar{a} - \bar{b}) \cdot ((\bar{a} \times \bar{b}) \times (\bar{a} + 2\bar{b})) \).
This is a scalar triple product of vectors \( \vec{A} = 2\bar{a} - \bar{b} \), \( \vec{B} = \bar{a} \times \bar{b} \), and \( \vec{C} = \bar{a} + 2\bar{b} \).
We can rewrite it as \( (\bar{a} \times \bar{b}) \cdot ((2\bar{a} - \bar{b}) \times (\bar{a} + 2\bar{b})) \).
Calculate the cross product:
\[ (2\bar{a} - \bar{b}) \times (\bar{a} + 2\bar{b}) = 2(\bar{a} \times \bar{a}) + 4(\bar{a} \times \bar{b}) - (\bar{b} \times \bar{a}) - 2(\bar{b} \times \bar{b}) \]
Since \( \bar{a} \times \bar{a} = 0 \), \( \bar{b} \times \bar{b} = 0 \), and \( \bar{b} \times \bar{a} = -(\bar{a} \times \bar{b}) \):
\[ = 0 + 4(\bar{a} \times \bar{b}) + (\bar{a} \times \bar{b}) - 0 = 5(\bar{a} \times \bar{b}) \]
The final result is \( (\bar{a} \times \bar{b}) \cdot 5(\bar{a} \times \bar{b}) = 5 |\bar{a} \times \bar{b}|^2 \).
Check \( \bar{a} \) and \( \bar{b} \):
\( |\bar{a}|^2 = \frac{1}{10}(3^2 + 1^2) = 1 \).
\( |\bar{b}|^2 = \frac{1}{49}(2^2 + 3^2 + (-6)^2) = 1 \).
\( \bar{a} \cdot \bar{b} = \frac{1}{7\sqrt{10}}(3(2) + 0(3) + 1(-6)) = 0 \).
Since \( \bar{a} \perp \bar{b} \) and both are unit vectors, \( |\bar{a} \times \bar{b}| = |\bar{a}||\bar{b}|\sin 90^\circ = 1 \).
Result \( = 5(1)^2 = 5 \).
Step 4: Final Answer:
The value is 5.