A parallelogram is fromed with`veca` and `vecb` as the sides let `vecd_(1)` and `vecd_(2)` be the diagonals of the parallelogram , them `a^(2) +b^(2)=`

Options

A. `(d_(1)^(2) + d_(2)^(2))/(2)`
B. `(d_(1)^(2) + d_(2)^(2))/(1)`
C. `d_(1)^(2) + d_(2)^(2)`
D. `d_(1)^(2) - d_(2)^(2)`

Explanation

To solve the problem, we need to establish the relationship between the sides of the parallelogram formed by vectors \(\vec{a}\) and \(\vec{b}\) and its diagonals \(\vec{d_1}\) and \(\vec{d_2}\).

### Step-by-Step Solution:

1. **Identify the Diagonals**:
In a parallelogram formed by vectors \(\vec{a}\) and \(\vec{b}\), the diagonals can be expressed as:
\[ \vec{d_1} = \vec{a} + \vec{b} \]
\[ \vec{d_2} = \vec{a} - \vec{b} \]

2. **Calculate the Magnitudes of the Diagonals**:
To find the squares of the magnitudes of the diagonals, we use the following formulas:
\[ |\vec{d_1}|^2 = |\vec{a} + \vec{b}|^2 = |\vec{a}|^2 + |\vec{b}|^2 + 2 |\vec{a}| |\vec{b}| \cos(\theta) \]
\[ |\vec{d_2}|^2 = |\vec{a} - \vec{b}|^2 = |\vec{a}|^2 + |\vec{b}|^2 - 2 |\vec{a}| |\vec{b}| \cos(\theta) \]

3. **Add the Squares of the Diagonals**:
Now, we add the squares of the magnitudes of the diagonals:
\[ |\vec{d_1}|^2 + |\vec{d_2}|^2 = (|\vec{a}|^2 + |\vec{b}|^2 + 2 |\vec{a}| |\vec{b}| \cos(\theta)) + (|\vec{a}|^2 + |\vec{b}|^2 - 2 |\vec{a}| |\vec{b}| \cos(\theta)) \]
Simplifying this, we find:
\[ |\vec{d_1}|^2 + |\vec{d_2}|^2 = 2|\vec{a}|^2 + 2|\vec{b}|^2 \]

4. **Factor Out the Common Terms**:
We can factor out the common factor of 2:
\[ |\vec{d_1}|^2 + |\vec{d_2}|^2 = 2(|\vec{a}|^2 + |\vec{b}|^2) \]

5. **Solve for \( |\vec{a}|^2 + |\vec{b}|^2 \)**:
Dividing both sides by 2 gives us:
\[ |\vec{a}|^2 + |\vec{b}|^2 = \frac{|\vec{d_1}|^2 + |\vec{d_2}|^2}{2} \]

### Final Result:
Thus, we conclude that:
\[ |\vec{a}|^2 + |\vec{b}|^2 = \frac{|\vec{d_1}|^2 + |\vec{d_2}|^2}{2} \]