`(vecA+2vecB).(2vecA-3vecB)`:-

Options

A. 0
B. `2AB cos theta-6B^(2)`
C. 8
D. `2A^(2)+Abcostheta-6B^(2)`

Answer

To solve the expression \((\vec{A} + 2\vec{B}) \cdot (2\vec{A} - 3\vec{B})\), we will use the distributive property of the dot product. Here’s the step-by-step solution:

### Step 1: Write the expression
We start with the expression:
\[ (\vec{A} + 2\vec{B}) \cdot (2\vec{A} - 3\vec{B}) \]

### Step 2: Apply the distributive property
Using the distributive property of the dot product, we can expand the expression:
\[ = \vec{A} \cdot (2\vec{A}) + \vec{A} \cdot (-3\vec{B}) + 2\vec{B} \cdot (2\vec{A}) + 2\vec{B} \cdot (-3\vec{B}) \]

### Step 3: Calculate each dot product
Now, we calculate each term:
1. \(\vec{A} \cdot (2\vec{A}) = 2(\vec{A} \cdot \vec{A}) = 2|\vec{A}|^2\)
2. \(\vec{A} \cdot (-3\vec{B}) = -3(\vec{A} \cdot \vec{B})\)
3. \(2\vec{B} \cdot (2\vec{A}) = 4(\vec{B} \cdot \vec{A}) = 4(\vec{A} \cdot \vec{B})\)
4. \(2\vec{B} \cdot (-3\vec{B}) = -6(\vec{B} \cdot \vec{B}) = -6|\vec{B}|^2\)

### Step 4: Combine the results
Now, we combine all the results:
\[ = 2|\vec{A}|^2 - 3(\vec{A} \cdot \vec{B}) + 4(\vec{A} \cdot \vec{B}) - 6|\vec{B}|^2 \]
\[ = 2|\vec{A}|^2 + (4 - 3)(\vec{A} \cdot \vec{B}) - 6|\vec{B}|^2 \]
\[ = 2|\vec{A}|^2 + 1(\vec{A} \cdot \vec{B}) - 6|\vec{B}|^2 \]

### Step 5: Final expression
Thus, the final expression is:
\[ = 2|\vec{A}|^2 + (\vec{A} \cdot \vec{B}) - 6|\vec{B}|^2 \]