If `veca and vecb` are two unit vectors such that `veca+2vecb and 5veca-4vecb` are perpendicular to each other then the angle between `veca and vecb` is (A) `45^0` (B) `60^0` (C) `cos^-1(1/3)` (D) `cos^-1(2/7)`

Options

A. `45^(@)`
B. `60^(@)`
C. `cos^(-1)(1//3)`
D. `cos^(-1)(2//7)`

Explanation

To solve the problem, we need to find the angle between the two unit vectors \(\vec{a}\) and \(\vec{b}\) given that the vectors \(\vec{a} + 2\vec{b}\) and \(5\vec{a} - 4\vec{b}\) are perpendicular to each other.

### Step-by-Step Solution:

1. **Understanding the Condition of Perpendicularity**:
Two vectors are perpendicular if their dot product is zero. Therefore, we need to set up the equation:
\[ (\vec{a} + 2\vec{b}) \cdot (5\vec{a} - 4\vec{b}) = 0 \]

2. **Expanding the Dot Product**:
Using the distributive property of the dot product, we expand the left-hand side:
\[ \vec{a} \cdot (5\vec{a}) + \vec{a} \cdot (-4\vec{b}) + 2\vec{b} \cdot (5\vec{a}) + 2\vec{b} \cdot (-4\vec{b}) = 0 \]
This simplifies to:
\[ 5(\vec{a} \cdot \vec{a}) - 4(\vec{a} \cdot \vec{b}) + 10(\vec{b} \cdot \vec{a}) - 8(\vec{b} \cdot \vec{b}) = 0 \]

3. **Using the Properties of Unit Vectors**:
Since \(\vec{a}\) and \(\vec{b}\) are unit vectors, we know:
\[ \vec{a} \cdot \vec{a} = 1 \quad \text{and} \quad \vec{b} \cdot \vec{b} = 1 \]
Therefore, substituting these values into the equation gives:
\[ 5(1) - 4(\vec{a} \cdot \vec{b}) + 10(\vec{a} \cdot \vec{b}) - 8(1) = 0 \]

4. **Combining Like Terms**:
This simplifies to:
\[ 5 - 8 + 6(\vec{a} \cdot \vec{b}) = 0 \]
Which further simplifies to:
\[ -3 + 6(\vec{a} \cdot \vec{b}) = 0 \]

5. **Solving for \(\vec{a} \cdot \vec{b}\)**:
Rearranging the equation gives:
\[ 6(\vec{a} \cdot \vec{b}) = 3 \implies \vec{a} \cdot \vec{b} = \frac{1}{2} \]

6. **Finding the Angle**:
The dot product of two vectors is also given by:
\[ \vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos \theta \]
Since both \(\vec{a}\) and \(\vec{b}\) are unit vectors, we have:
\[ \vec{a} \cdot \vec{b} = 1 \cdot 1 \cdot \cos \theta = \cos \theta \]
Therefore:
\[ \cos \theta = \frac{1}{2} \]

7. **Determining the Angle \(\theta\)**:
The angle \(\theta\) that satisfies \(\cos \theta = \frac{1}{2}\) is:
\[ \theta = 60^\circ \]

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