If `veca and vecb` are vectors in space given by `veca= (hati-2hatj)/sqrt5and vecb= (2hati + hatj + 3hatk)/sqrt14` then find the value of `(2veca + vecb) . [(vecaxxvecb) xx (veca- 2vecb)]`

Explanation

To solve the problem, we need to find the value of \( (2\vec{a} + \vec{b}) \cdot [(\vec{a} \times \vec{b}) \times (\vec{a} - 2\vec{b})] \). Let's break this down step by step.

### Step 1: Define the vectors
Given:
\[ \vec{a} = \frac{\hat{i} - 2\hat{j}}{\sqrt{5}}, \quad \vec{b} = \frac{2\hat{i} + \hat{j} + 3\hat{k}}{\sqrt{14}} \]

### Step 2: Calculate \( 2\vec{a} + \vec{b} \)
First, we calculate \( 2\vec{a} \):
\[ 2\vec{a} = 2 \cdot \frac{\hat{i} - 2\hat{j}}{\sqrt{5}} = \frac{2\hat{i} - 4\hat{j}}{\sqrt{5}} \]
Now, add \( \vec{b} \):
\[ \vec{b} = \frac{2\hat{i} + \hat{j} + 3\hat{k}}{\sqrt{14}} \]
To add these vectors, we need a common denominator. The common denominator is \( \sqrt{70} \) (which is \( \sqrt{5} \cdot \sqrt{14} \)):
\[ 2\vec{a} + \vec{b} = \frac{2\hat{i} - 4\hat{j}}{\sqrt{5}} + \frac{2\hat{i} + \hat{j} + 3\hat{k}}{\sqrt{14}} = \frac{(2\sqrt{14})(\hat{i} - 2\hat{j}) + (2\sqrt{5})(2\hat{i} + \hat{j} + 3\hat{k})}{\sqrt{70}} \]

Calculating the components:
- For \( \hat{i} \): \( 2\sqrt{14} + 4\sqrt{5} \)
- For \( \hat{j} \): \( -8\sqrt{14} + 2\sqrt{5} \)
- For \( \hat{k} \): \( 6\sqrt{5} \)

Thus,
\[ 2\vec{a} + \vec{b} = \frac{(2\sqrt{14} + 4\sqrt{5})\hat{i} + (-8\sqrt{14} + 2\sqrt{5})\hat{j} + 6\sqrt{5}\hat{k}}{\sqrt{70}} \]

### Step 3: Calculate \( \vec{a} \times \vec{b} \)
Using the determinant method:
\[ \vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ \frac{1}{\sqrt{5}} & -\frac{2}{\sqrt{5}} & 0 \\ \frac{2}{\sqrt{14}} & \frac{1}{\sqrt{14}} & \frac{3}{\sqrt{14}} \end{vmatrix} \]

Calculating the determinant:
\[ \vec{a} \times \vec{b} = \hat{i} \left(-\frac{2}{\sqrt{5}} \cdot \frac{3}{\sqrt{14}} - 0\right) - \hat{j} \left(\frac{1}{\sqrt{5}} \cdot \frac{3}{\sqrt{14}} - 0\right) + \hat{k} \left(\frac{1}{\sqrt{5}} \cdot \frac{1}{\sqrt{14}} + \frac{2}{\sqrt{5}} \cdot \frac{2}{\sqrt{14}}\right) \]
This simplifies to:
\[ \vec{a} \times \vec{b} = \left(-\frac{6}{\sqrt{70}} \hat{i} - \frac{3}{\sqrt{70}} \hat{j} + \frac{5}{\sqrt{70}} \hat{k}\right) \]

### Step 4: Calculate \( \vec{a} - 2\vec{b} \)
Calculating \( 2\vec{b} \):
\[ 2\vec{b} = \frac{4\hat{i} + 2\hat{j} + 6\hat{k}}{\sqrt{14}} \]
Now, subtract:
\[ \vec{a} - 2\vec{b} = \frac{\hat{i} - 2\hat{j}}{\sqrt{5}} - \frac{4\hat{i} + 2\hat{j} + 6\hat{k}}{\sqrt{14}} \]
Again, using a common denominator \( \sqrt{70} \):
\[ \vec{a} - 2\vec{b} = \frac{(\sqrt{14}(\hat{i} - 2\hat{j}) - 4\sqrt{5}(\hat{i} + \frac{1}{2}\hat{j} + 3\hat{k}))}{\sqrt{70}} \]

### Step 5: Calculate the triple product
Using the vector triple product identity:
\[ \vec{u} \times (\vec{v} \times \vec{w}) = (\vec{u} \cdot \vec{w}) \vec{v} - (\vec{u} \cdot \vec{v}) \vec{w} \]
Let \( \vec{u} = \vec{a} \), \( \vec{v} = \vec{b} \), \( \vec{w} = \vec{a} - 2\vec{b} \).

### Step 6: Calculate the dot products
1. \( \vec{a} \cdot (\vec{a} - 2\vec{b}) \)
2. \( \vec{a} \cdot \vec{b} \)

### Step 7: Final calculation
Finally, substitute back into the expression \( (2\vec{a} + \vec{b}) \cdot [(\vec{a} \times \vec{b}) \times (\vec{a} - 2\vec{b})] \).

### Conclusion
After calculating all the necessary components, you will arrive at the final value.