The vector `(vec(a)+3vec(b))` is perpendicular to `(7 vec(a)-5vec(b))` and `(vec(a)-4vec(b))` is perpendicular to `(7vec(a)-2vec(b))`. The angle between `vec(a)` and `vec(b)` is :

Options

A. `30^(@)`
B. `45^(@)`
C. `60^(@)`
D. None of these

Explanation

To solve the problem, we need to find the angle between the vectors \(\vec{a}\) and \(\vec{b}\) given the conditions of perpendicularity between certain combinations of these vectors.

### Step-by-Step Solution:

1. **Understanding Perpendicularity**:
The vectors \((\vec{a} + 3\vec{b})\) and \((7\vec{a} - 5\vec{b})\) are perpendicular. This means their dot product is zero:
\[ (\vec{a} + 3\vec{b}) \cdot (7\vec{a} - 5\vec{b}) = 0 \]

2. **Calculating the Dot Product**:
Expanding the dot product:
\[ \vec{a} \cdot (7\vec{a}) + \vec{a} \cdot (-5\vec{b}) + 3\vec{b} \cdot (7\vec{a}) + 3\vec{b} \cdot (-5\vec{b}) = 0 \]
This simplifies to:
\[ 7\vec{a} \cdot \vec{a} - 5\vec{a} \cdot \vec{b} + 21\vec{b} \cdot \vec{a} - 15\vec{b} \cdot \vec{b} = 0 \]
Combining like terms:
\[ 7|\vec{a}|^2 + 16\vec{a} \cdot \vec{b} - 15|\vec{b}|^2 = 0 \quad \text{(Equation 1)} \]

3. **Second Condition of Perpendicularity**:
The vectors \((\vec{a} - 4\vec{b})\) and \((7\vec{a} - 2\vec{b})\) are also perpendicular:
\[ (\vec{a} - 4\vec{b}) \cdot (7\vec{a} - 2\vec{b}) = 0 \]

4. **Calculating the Second Dot Product**:
Expanding this dot product:
\[ \vec{a} \cdot (7\vec{a}) + \vec{a} \cdot (-2\vec{b}) - 4\vec{b} \cdot (7\vec{a}) + 8\vec{b} \cdot \vec{b} = 0 \]
This simplifies to:
\[ 7|\vec{a}|^2 - 2\vec{a} \cdot \vec{b} - 28\vec{b} \cdot \vec{a} + 8|\vec{b}|^2 = 0 \]
Combining like terms:
\[ 7|\vec{a}|^2 - 30\vec{a} \cdot \vec{b} + 8|\vec{b}|^2 = 0 \quad \text{(Equation 2)} \]

5. **Solving the Equations**:
Now we have two equations:
- \(7|\vec{a}|^2 + 16\vec{a} \cdot \vec{b} - 15|\vec{b}|^2 = 0\) (Equation 1)
- \(7|\vec{a}|^2 - 30\vec{a} \cdot \vec{b} + 8|\vec{b}|^2 = 0\) (Equation 2)

We can solve these equations simultaneously. Subtract Equation 1 from Equation 2:
\[ (7|\vec{a}|^2 - 30\vec{a} \cdot \vec{b} + 8|\vec{b}|^2) - (7|\vec{a}|^2 + 16\vec{a} \cdot \vec{b} - 15|\vec{b}|^2) = 0 \]
This simplifies to:
\[ 23|\vec{b}|^2 - 46\vec{a} \cdot \vec{b} = 0 \]
Thus, we have:
\[ 23|\vec{b}|^2 = 46\vec{a} \cdot \vec{b} \quad \Rightarrow \quad |\vec{b}|^2 = 2\vec{a} \cdot \vec{b} \quad \text{(Equation 3)} \]

6. **Substituting Back**:
From Equation 3, we can express \(\vec{a} \cdot \vec{b}\) in terms of magnitudes:
\[ \vec{a} \cdot \vec{b} = |\vec{a}||\vec{b}|\cos\theta \]
Substituting this into Equation 3 gives:
\[ |\vec{b}|^2 = 2|\vec{a}||\vec{b}|\cos\theta \]

7. **Finding the Angle**:
If we assume \(|\vec{a}| = |\vec{b}|\) (from the earlier derived equations), let \(|\vec{a}| = |\vec{b}| = k\):
\[ k^2 = 2k^2\cos\theta \quad \Rightarrow \quad 1 = 2\cos\theta \quad \Rightarrow \quad \cos\theta = \frac{1}{2} \]
Therefore, the angle \(\theta\) is:
\[ \theta = \cos^{-1}\left(\frac{1}{2}\right) = 60^\circ \]

### Final Answer:
The angle between \(\vec{a}\) and \(\vec{b}\) is \(60^\circ\).