Find `(veca+3vecb)*(2veca-vecb)`. If `veca=hati+hatj+2hatk, hatb=2hati+2hatj-hatk`

Answer

To solve the problem, we need to find the expression \((\vec{a} + 3\vec{b}) \cdot (2\vec{a} - \vec{b})\) given the vectors \(\vec{a} = \hat{i} + \hat{j} + 2\hat{k}\) and \(\vec{b} = 2\hat{i} + 2\hat{j} - \hat{k}\).

### Step 1: Calculate \(\vec{a} + 3\vec{b}\)

First, we need to calculate \(3\vec{b}\):
\[ 3\vec{b} = 3(2\hat{i} + 2\hat{j} - \hat{k}) = 6\hat{i} + 6\hat{j} - 3\hat{k} \]

Now, add \(\vec{a}\) and \(3\vec{b}\):
\[ \vec{a} + 3\vec{b} = (\hat{i} + \hat{j} + 2\hat{k}) + (6\hat{i} + 6\hat{j} - 3\hat{k}) \]
\[ = (1 + 6)\hat{i} + (1 + 6)\hat{j} + (2 - 3)\hat{k} \]
\[ = 7\hat{i} + 7\hat{j} - 1\hat{k} \]

### Step 2: Calculate \(2\vec{a} - \vec{b}\)

Now, calculate \(2\vec{a}\):
\[ 2\vec{a} = 2(\hat{i} + \hat{j} + 2\hat{k}) = 2\hat{i} + 2\hat{j} + 4\hat{k} \]

Next, subtract \(\vec{b}\):
\[ 2\vec{a} - \vec{b} = (2\hat{i} + 2\hat{j} + 4\hat{k}) - (2\hat{i} + 2\hat{j} - \hat{k}) \]
\[ = (2 - 2)\hat{i} + (2 - 2)\hat{j} + (4 + 1)\hat{k} \]
\[ = 0\hat{i} + 0\hat{j} + 5\hat{k} \]
\[ = 5\hat{k} \]

### Step 3: Calculate the dot product \((\vec{a} + 3\vec{b}) \cdot (2\vec{a} - \vec{b})\)

Now we can find the dot product:
\[ (\vec{a} + 3\vec{b}) \cdot (2\vec{a} - \vec{b}) = (7\hat{i} + 7\hat{j} - 1\hat{k}) \cdot (5\hat{k}) \]
Using the dot product formula:
\[ = 7 \cdot 0 + 7 \cdot 0 + (-1) \cdot 5 \]
\[ = 0 + 0 - 5 \]
\[ = -5 \]

### Final Result

Thus, the value of \((\vec{a} + 3\vec{b}) \cdot (2\vec{a} - \vec{b})\) is \(-5\).