The magnitude of vector `vec(A),vec(B)` and `vec(C )` are respectively 12,5 and 13 unit and `vec(A)+vec(B)= vec(C )` then the angle between `vec(A)` and `vec(B)` is

Options

A. `pi//2`
B. `cos^(-1)(0.6)`
C. `tan^(-1)(0.75)`
D. `pi//4`

Answer

To solve the problem of finding the angle between vectors \(\vec{A}\) and \(\vec{B}\) given that \(\vec{A} + \vec{B} = \vec{C}\), we can follow these steps:

### Step 1: Write down the magnitudes of the vectors
Given:
- Magnitude of \(\vec{A} = 12\) units
- Magnitude of \(\vec{B} = 5\) units
- Magnitude of \(\vec{C} = 13\) units

### Step 2: Use the vector addition formula
The equation \(\vec{A} + \vec{B} = \vec{C}\) implies that the magnitudes of the vectors can be related through the cosine of the angle \(\theta\) between them. The formula for the magnitude of the resultant vector when two vectors are added is given by:
\[ |\vec{C}|^2 = |\vec{A}|^2 + |\vec{B}|^2 + 2 |\vec{A}| |\vec{B}| \cos(\theta) \]

### Step 3: Substitute the known values into the equation
Substituting the magnitudes into the equation:
\[ 13^2 = 12^2 + 5^2 + 2 \cdot 12 \cdot 5 \cdot \cos(\theta) \]
Calculating the squares:
\[ 169 = 144 + 25 + 120 \cos(\theta) \]

### Step 4: Simplify the equation
Combine the constants on the right side:
\[ 169 = 169 + 120 \cos(\theta) \]

### Step 5: Isolate the cosine term
Subtract 169 from both sides:
\[ 0 = 120 \cos(\theta) \]

### Step 6: Solve for \(\cos(\theta)\)
This implies:
\[ \cos(\theta) = 0 \]

### Step 7: Determine the angle \(\theta\)
The angle \(\theta\) for which \(\cos(\theta) = 0\) is:
\[ \theta = 90^\circ \quad \text{or} \quad \theta = \frac{\pi}{2} \text{ radians} \]

### Conclusion
The angle between vectors \(\vec{A}\) and \(\vec{B}\) is \(90^\circ\) or \(\frac{\pi}{2}\) radians.

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