A magnitude of vector `vecA,vecB` and `vecC` are respectively `12, 5` and `13` units and `vecA+vecB=vecC` then the angle between `vecA` and `vecB` is

Options

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

Answer

To find the angle between vectors \(\vec{A}\) and \(\vec{B}\) given that \(\vec{A} + \vec{B} = \vec{C}\) with magnitudes \(|\vec{A}| = 12\), \(|\vec{B}| = 5\), and \(|\vec{C}| = 13\), we can use the cosine rule for vectors.

### Step-by-Step Solution:

1. **Write the equation for vector addition**:
\[ |\vec{C}|^2 = |\vec{A}|^2 + |\vec{B}|^2 + 2|\vec{A}||\vec{B}|\cos\theta \]
where \(\theta\) is the angle between \(\vec{A}\) and \(\vec{B}\).

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

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

4. **Combine the constants on the right side**:
\[ 169 = 169 + 120\cos\theta \]

5. **Rearrange the equation**:
\[ 169 - 169 = 120\cos\theta \]
\[ 0 = 120\cos\theta \]

6. **Solve for \(\cos\theta\)**:
\[ \cos\theta = 0 \]

7. **Find the angle \(\theta\)**:
\[ \theta = \cos^{-1}(0) = 90^\circ \]

Thus, the angle between vectors \(\vec{A}\) and \(\vec{B}\) is \(90^\circ\).