The magnitudes of vectors `vecA,vecB and vecC` are 3,4 and 5 units respectively. If `vecA+vecB= vecC`, the angle between `vecA` and vecB` is

Options

A. `(pi)/(2)`
B. `cos^(-1)(0.6)`
C. `tan^(-1)((7)/(5))`
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 properties of dot products and magnitudes of vectors. Here’s a step-by-step solution:

### Step 1: Write down the given data
We know the magnitudes of the vectors:
- \(|\vec{A}| = 3\)
- \(|\vec{B}| = 4\)
- \(|\vec{C}| = 5\)

### Step 2: Use the vector addition formula
From the equation \(\vec{A} + \vec{B} = \vec{C}\), we can square both sides to utilize the dot product:
\[ |\vec{A} + \vec{B}|^2 = |\vec{C}|^2 \]

### Step 3: Expand the left side using the dot product
The left side can be expanded as follows:
\[ |\vec{A}|^2 + |\vec{B}|^2 + 2 \vec{A} \cdot \vec{B} = |\vec{C}|^2 \]

### Step 4: Substitute the magnitudes
Substituting the known magnitudes into the equation:
\[ 3^2 + 4^2 + 2 \vec{A} \cdot \vec{B} = 5^2 \]
This simplifies to:
\[ 9 + 16 + 2 \vec{A} \cdot \vec{B} = 25 \]

### Step 5: Simplify the equation
Combining the constants gives:
\[ 25 + 2 \vec{A} \cdot \vec{B} = 25 \]
Subtracting 25 from both sides results in:
\[ 2 \vec{A} \cdot \vec{B} = 0 \]

### Step 6: Solve for the dot product
Dividing both sides by 2 gives:
\[ \vec{A} \cdot \vec{B} = 0 \]

### Step 7: Relate the dot product to the angle
The dot product of two vectors is also given by:
\[ \vec{A} \cdot \vec{B} = |\vec{A}| |\vec{B}| \cos \theta \]
Substituting the magnitudes:
\[ 0 = 3 \cdot 4 \cdot \cos \theta \]
This implies:
\[ \cos \theta = 0 \]

### Step 8: Find the angle
The cosine of an angle is zero at:
\[ \theta = \frac{\pi}{2} \text{ radians} \quad \text{(or 90 degrees)} \]

### Conclusion
Thus, the angle between vectors \(\vec{A}\) and \(\vec{B}\) is:
\[ \theta = \frac{\pi}{2} \text{ radians} \]

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