The magnitude of vectors `vec A, vec B and vec C` are 3, 4 and 5 units respectively. If `vec A + vec B= vec C`, the angle between `vec A and vec B` is

Options

A. `(pi)/(2)`
B. `(pi)/(4)`
C. `tan^(-1)((7)/(5))`
D. `cos^(-1)(0.6)`

Explanation

To find the angle between vectors \(\vec{A}\) and \(\vec{B}\), we can use the information given in the problem. The magnitudes of the vectors are as follows:

- Magnitude of \(\vec{A} = 3\) units
- Magnitude of \(\vec{B} = 4\) units
- Magnitude of \(\vec{C} = 5\) units

We know from the problem that:

\[ \vec{A} + \vec{B} = \vec{C} \]

According to the triangle law of vector addition, if the vectors \(\vec{A}\) and \(\vec{B}\) add up to \(\vec{C}\), we can apply the cosine rule to find the angle between them. The cosine rule states:

\[ |\vec{C}|^2 = |\vec{A}|^2 + |\vec{B}|^2 + 2 |\vec{A}| |\vec{B}| \cos(\theta) \]

Where:
- \(|\vec{C}|\) is the magnitude of vector \(\vec{C}\),
- \(|\vec{A}|\) is the magnitude of vector \(\vec{A}\),
- \(|\vec{B}|\) is the magnitude of vector \(\vec{B}\),
- \(\theta\) is the angle between vectors \(\vec{A}\) and \(\vec{B}\).

Substituting the magnitudes into the equation:

\[ 5^2 = 3^2 + 4^2 + 2 \cdot 3 \cdot 4 \cdot \cos(\theta) \]

Calculating the squares:

\[ 25 = 9 + 16 + 24 \cos(\theta) \]

Simplifying:

\[ 25 = 25 + 24 \cos(\theta) \]

Subtracting 25 from both sides:

\[ 0 = 24 \cos(\theta) \]

This implies:

\[ \cos(\theta) = 0 \]

The angle \(\theta\) for which the cosine is zero is:

\[ \theta = 90^\circ \quad \text{or} \quad \theta = \frac{\pi}{2} \text{ radians} \]

Thus, the angle between vectors \(\vec{A}\) and \(\vec{B}\) is \(90^\circ\) or \(\frac{\pi}{2}\) radians.

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

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