Given `|vec(A)_(1)|=2,|vec(A)_(2)|=3` and `|vec(A)_(1)+vec(A)_(2)|=3`. Find the value of `(vec(A)_(1)+2vec(A)_(2)).(3vec(A)_(1)-4vec(A)_(2))`.

Options

A. -64
B. 60
C. -60
D. 64

Explanation

To solve the problem step by step, we need to find the value of \((\vec{A}_1 + 2\vec{A}_2) \cdot (3\vec{A}_1 - 4\vec{A}_2)\) given the magnitudes of \(\vec{A}_1\) and \(\vec{A}_2\) and the magnitude of their sum.

### Step 1: Write down the given information
We have:
- \(|\vec{A}_1| = 2\)
- \(|\vec{A}_2| = 3\)
- \(|\vec{A}_1 + \vec{A}_2| = 3\)

### Step 2: Use the formula for the magnitude of the sum of two vectors
The magnitude of the sum of two vectors can be expressed as:
\[ |\vec{A}_1 + \vec{A}_2|^2 = |\vec{A}_1|^2 + |\vec{A}_2|^2 + 2 |\vec{A}_1| |\vec{A}_2| \cos \theta \]
where \(\theta\) is the angle between \(\vec{A}_1\) and \(\vec{A}_2\).

### Step 3: Substitute the known values into the formula
Substituting the known values:
\[ 3^2 = 2^2 + 3^2 + 2 \cdot 2 \cdot 3 \cos \theta \]
This simplifies to:
\[ 9 = 4 + 9 + 12 \cos \theta \]
\[ 9 = 13 + 12 \cos \theta \]

### Step 4: Solve for \(\cos \theta\)
Rearranging the equation:
\[ 12 \cos \theta = 9 - 13 \]
\[ 12 \cos \theta = -4 \]
\[ \cos \theta = -\frac{1}{3} \]

### Step 5: Calculate the dot product
Now we need to find the dot product \((\vec{A}_1 + 2\vec{A}_2) \cdot (3\vec{A}_1 - 4\vec{A}_2)\).

Using the distributive property of the dot product:
\[ (\vec{A}_1 + 2\vec{A}_2) \cdot (3\vec{A}_1 - 4\vec{A}_2) = \vec{A}_1 \cdot (3\vec{A}_1) + \vec{A}_1 \cdot (-4\vec{A}_2) + 2\vec{A}_2 \cdot (3\vec{A}_1) + 2\vec{A}_2 \cdot (-4\vec{A}_2) \]

### Step 6: Simplify the expression
Calculating each term:
1. \(\vec{A}_1 \cdot (3\vec{A}_1) = 3|\vec{A}_1|^2 = 3 \cdot 4 = 12\)
2. \(\vec{A}_1 \cdot (-4\vec{A}_2) = -4 \vec{A}_1 \cdot \vec{A}_2\)
3. \(2\vec{A}_2 \cdot (3\vec{A}_1) = 6 \vec{A}_2 \cdot \vec{A}_1\)
4. \(2\vec{A}_2 \cdot (-4\vec{A}_2) = -8 |\vec{A}_2|^2 = -8 \cdot 9 = -72\)

Combining these:
\[ = 12 - 4 \vec{A}_1 \cdot \vec{A}_2 + 6 \vec{A}_2 \cdot \vec{A}_1 - 72 \]
\[ = 12 + 2 \vec{A}_1 \cdot \vec{A}_2 - 72 \]

### Step 7: Substitute \(\vec{A}_1 \cdot \vec{A}_2\)
Now, we need to find \(\vec{A}_1 \cdot \vec{A}_2\):
\[ \vec{A}_1 \cdot \vec{A}_2 = |\vec{A}_1| |\vec{A}_2| \cos \theta = 2 \cdot 3 \cdot \left(-\frac{1}{3}\right) = -2 \]

### Step 8: Substitute back into the expression
Substituting \(\vec{A}_1 \cdot \vec{A}_2 = -2\):
\[ = 12 + 2(-2) - 72 \]
\[ = 12 - 4 - 72 \]
\[ = 8 - 72 = -64 \]

### Final Answer
Thus, the value of \((\vec{A}_1 + 2\vec{A}_2) \cdot (3\vec{A}_1 - 4\vec{A}_2)\) is \(-64\).