The magnitudes of vectors `vec(A),vec(B) and vec(C)` are 3,4 and 5 unit respectively. If `vec(A)+vec(B)=vec(C),` the angle between `vec(A) and vec(B)` is

Options

A. 0
B. 180
C. 90
D. 45

Answer

To solve the problem, we need to find the angle between vectors A and B given that the magnitudes of vectors A, B, and C are 3, 4, and 5 units respectively, and that \( \vec{A} + \vec{B} = \vec{C} \).

### Step-by-Step Solution:

1. **Identify the Magnitudes**:
- Let the magnitude of \( \vec{A} = 3 \) units.
- Let the magnitude of \( \vec{B} = 4 \) units.
- Let the magnitude of \( \vec{C} = 5 \) units.

2. **Use the Vector Addition Formula**:
According to the problem, we have:
\[ \vec{A} + \vec{B} = \vec{C} \]
Taking the magnitude on both sides, we have:
\[ |\vec{A} + \vec{B}| = |\vec{C}| \]

3. **Apply the Law of Cosines**:
The magnitude of the resultant vector from two vectors can be expressed as:
\[ |\vec{A} + \vec{B}| = \sqrt{|\vec{A}|^2 + |\vec{B}|^2 + 2 |\vec{A}| |\vec{B}| \cos \theta} \]
Substituting the magnitudes:
\[ 5 = \sqrt{3^2 + 4^2 + 2 \cdot 3 \cdot 4 \cdot \cos \theta} \]

4. **Square Both Sides**:
Squaring both sides to eliminate the square root gives:
\[ 25 = 9 + 16 + 24 \cos \theta \]

5. **Simplify the Equation**:
Combine the constants:
\[ 25 = 25 + 24 \cos \theta \]
This simplifies to:
\[ 0 = 24 \cos \theta \]

6. **Solve for Cosine**:
Dividing both sides by 24 gives:
\[ \cos \theta = 0 \]

7. **Determine the Angle**:
The angle \( \theta \) for which \( \cos \theta = 0 \) is:
\[ \theta = 90^\circ \]

### Conclusion:
The angle between vectors \( \vec{A} \) and \( \vec{B} \) is \( 90^\circ \).