`A_(1)` and `A_(2)` are two vectors such that `|A_(1)| = 3 , |A_(2)| = 5` and `|A_(1)+A_(2)| = 5` the value of `(2A_(1)+3A_(2)).(2A_(1)-2A_(2))` is

Options

A. `(237)/(2)`
B. `-123`
C. `(-337)/(2)`
D. `(337)/(2)`

Answer

To solve the problem, we need to find the value of \((2A_1 + 3A_2) \cdot (2A_1 - 2A_2)\) given the magnitudes of the vectors \(A_1\) and \(A_2\) and the magnitude of their sum.

### Step-by-Step Solution:

1. **Given Information:**
- \(|A_1| = 3\)
- \(|A_2| = 5\)
- \(|A_1 + A_2| = 5\)

2. **Using the formula for the magnitude of the sum of two vectors:**
\[ |A_1 + A_2|^2 = |A_1|^2 + |A_2|^2 + 2 A_1 \cdot A_2 \]
Substituting the known values:
\[ 5^2 = 3^2 + 5^2 + 2 A_1 \cdot A_2 \]
\[ 25 = 9 + 25 + 2 A_1 \cdot A_2 \]
\[ 25 = 34 + 2 A_1 \cdot A_2 \]
Rearranging gives:
\[ 2 A_1 \cdot A_2 = 25 - 34 = -9 \]
Therefore:
\[ A_1 \cdot A_2 = -\frac{9}{2} \]

3. **Now, calculate \((2A_1 + 3A_2) \cdot (2A_1 - 2A_2)\):**
Using the distributive property of the dot product:
\[ (2A_1 + 3A_2) \cdot (2A_1 - 2A_2) = 2A_1 \cdot 2A_1 + 2A_1 \cdot (-2A_2) + 3A_2 \cdot 2A_1 + 3A_2 \cdot (-2A_2) \]
Simplifying each term:
\[ = 4A_1 \cdot A_1 - 4A_1 \cdot A_2 + 6A_2 \cdot A_1 - 6A_2 \cdot A_2 \]
\[ = 4|A_1|^2 - 4A_1 \cdot A_2 + 6A_1 \cdot A_2 - 6|A_2|^2 \]
\[ = 4|A_1|^2 + 2A_1 \cdot A_2 - 6|A_2|^2 \]

4. **Substituting the values:**
\[ = 4(3^2) + 2\left(-\frac{9}{2}\right) - 6(5^2) \]
\[ = 4(9) - 9 - 6(25) \]
\[ = 36 - 9 - 150 \]
\[ = 36 - 9 - 150 = 36 - 159 = -123 \]

### Final Answer:
The value of \((2A_1 + 3A_2) \cdot (2A_1 - 2A_2)\) is \(-123\).