Let `vec(a)` and `vec(b)` he two unit vectors. If the vectors : `vec(c )=vec(a)+2vec(b)` and `vec(d)=5vec(a)-4vec(b)` are perpendicular to eqach other, then the angle between `vec(a)` and `vec(b)` is :

Options

A. `(pi)/(6)`
B. `(pi)/(2)`
C. `(pi)/(3)`
D. `(pi)/(4)`

Answer

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

### Step-by-step Solution:

1. **Understanding the Condition of Perpendicularity**:
Since \( \vec{c} \) and \( \vec{d} \) are perpendicular, their dot product must be zero:
\[ \vec{c} \cdot \vec{d} = 0 \]

2. **Substituting the Expressions for \( \vec{c} \) and \( \vec{d} \)**:
Substitute \( \vec{c} \) and \( \vec{d} \) into the dot product:
\[ (\vec{a} + 2\vec{b}) \cdot (5\vec{a} - 4\vec{b}) = 0 \]

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

4. **Using Properties of Unit Vectors**:
Since \( \vec{a} \) and \( \vec{b} \) are unit vectors:
\[ \vec{a} \cdot \vec{a} = 1 \quad \text{and} \quad \vec{b} \cdot \vec{b} = 1 \]
Therefore, we can substitute:
\[ 5(1) - 4(\vec{a} \cdot \vec{b}) + 10(\vec{a} \cdot \vec{b}) - 8(1) = 0 \]

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

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

7. **Relating Dot Product to Angle**:
The dot product of two vectors is 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:
\[ \cos \theta = \frac{1}{2} \]

8. **Finding the Angle \( \theta \)**:
The angle \( \theta \) for which \( \cos \theta = \frac{1}{2} \) is:
\[ \theta = 60^\circ \quad \text{or} \quad \theta = \frac{\pi}{3} \text{ radians} \]

### Final Answer:
The angle between \( \vec{a} \) and \( \vec{b} \) is \( 60^\circ \) or \( \frac{\pi}{3} \) radians.

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