The maqunitudes of vecotr `overset(rarrA), overset(rarr)B` and `overset(rarr)C` are respectively 12,5 and 13 units and `overset(rarr)A+overset(rarr)B=overset(rarr)C` then the angle between `overset(rarr)A` and `overset(rarr)B` is :

Options

A. 0
B. `pi`
C. `pi//2`
D. `pi//4`

Explanation

To find the angle between the vectors \( \vec{A} \) and \( \vec{B} \) given that \( \vec{A} + \vec{B} = \vec{C} \), we can use the law of cosines. The magnitudes of the vectors are given as follows:

- \( |\vec{A}| = 12 \) units
- \( |\vec{B}| = 5 \) units
- \( |\vec{C}| = 13 \) units

### Step-by-Step Solution:

1. **Write the equation based on vector addition**:
\[ \vec{A} + \vec{B} = \vec{C} \]

2. **Use the law of cosines**:
According to the law of cosines, we can express the relationship between the magnitudes of the vectors as:
\[ |\vec{C}|^2 = |\vec{A}|^2 + |\vec{B}|^2 + 2 |\vec{A}| |\vec{B}| \cos \theta \]
where \( \theta \) is the angle between \( \vec{A} \) and \( \vec{B} \).

3. **Substitute the magnitudes into the equation**:
\[ 13^2 = 12^2 + 5^2 + 2 \cdot 12 \cdot 5 \cdot \cos \theta \]

4. **Calculate the squares**:
\[ 169 = 144 + 25 + 120 \cos \theta \]

5. **Combine the constants**:
\[ 169 = 169 + 120 \cos \theta \]

6. **Rearranging the equation**:
\[ 169 - 169 = 120 \cos \theta \]
\[ 0 = 120 \cos \theta \]

7. **Solve for \( \cos \theta \)**:
Since \( 120 \) is not zero, we can conclude:
\[ \cos \theta = 0 \]

8. **Determine the angle \( \theta \)**:
The cosine of an angle is zero at \( \theta = \frac{\pi}{2} \) radians (or 90 degrees).

### Final Answer:
The angle between \( \vec{A} \) and \( \vec{B} \) is \( \frac{\pi}{2} \) radians.

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