For some non zero vector `vecv`, if the sum of `vecv` and the vector obtained from `vecv` by rotating it by an angle `2alpha` equals to the vector obtained from `vecv` by rotating it by `alpha` then the value of `alpha`, is

Options

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

Answer

To solve the problem, we need to analyze the given condition involving the vector \( \vec{v} \) and its rotations. Let's break it down step by step.

### Step 1: Understanding the Problem
We are given a non-zero vector \( \vec{v} \). The problem states that the sum of \( \vec{v} \) and the vector obtained by rotating \( \vec{v} \) by \( 2\alpha \) equals the vector obtained by rotating \( \vec{v} \) by \( \alpha \).

### Step 2: Expressing the Vectors
Using the rotation formula for vectors, we can express the rotated vectors:
- The vector \( \vec{v} \) can be represented as \( \vec{v} = v(\cos(0) \hat{i} + \sin(0) \hat{j}) = v \hat{i} \).
- The vector obtained by rotating \( \vec{v} \) by \( \alpha \) is:
\[ \vec{v}_\alpha = v(\cos(\alpha) \hat{i} + \sin(\alpha) \hat{j}) \]
- The vector obtained by rotating \( \vec{v} \) by \( 2\alpha \) is:
\[ \vec{v}_{2\alpha} = v(\cos(2\alpha) \hat{i} + \sin(2\alpha) \hat{j}) \]

### Step 3: Setting Up the Equation
According to the problem, we have:
\[ \vec{v} + \vec{v}_{2\alpha} = \vec{v}_\alpha \]
Substituting the expressions for the vectors:
\[ v \hat{i} + v(\cos(2\alpha) \hat{i} + \sin(2\alpha) \hat{j}) = v(\cos(\alpha) \hat{i} + \sin(\alpha) \hat{j}) \]

### Step 4: Simplifying the Equation
Factoring out \( v \) (since \( v \neq 0 \)):
\[ \hat{i} + (\cos(2\alpha) \hat{i} + \sin(2\alpha) \hat{j}) = (\cos(\alpha) \hat{i} + \sin(\alpha) \hat{j}) \]
This leads to:
\[ (1 + \cos(2\alpha)) \hat{i} + \sin(2\alpha) \hat{j} = \cos(\alpha) \hat{i} + \sin(\alpha) \hat{j} \]

### Step 5: Equating Components
From the equation above, we can equate the components:
1. For the \( \hat{i} \) component:
\[ 1 + \cos(2\alpha) = \cos(\alpha) \]
2. For the \( \hat{j} \) component:
\[ \sin(2\alpha) = \sin(\alpha) \]

### Step 6: Solving the Equations
**From the first equation:**
Using the identity \( \cos(2\alpha) = 2\cos^2(\alpha) - 1 \):
\[ 1 + (2\cos^2(\alpha) - 1) = \cos(\alpha) \]
This simplifies to:
\[ 2\cos^2(\alpha) = \cos(\alpha) \]
Rearranging gives:
\[ 2\cos^2(\alpha) - \cos(\alpha) = 0 \]
Factoring out \( \cos(\alpha) \):
\[ \cos(\alpha)(2\cos(\alpha) - 1) = 0 \]
Thus, \( \cos(\alpha) = 0 \) or \( 2\cos(\alpha) - 1 = 0 \).

**From the second equation:**
Using the double angle identity \( \sin(2\alpha) = 2\sin(\alpha)\cos(\alpha) \):
\[ 2\sin(\alpha)\cos(\alpha) = \sin(\alpha) \]
If \( \sin(\alpha) \neq 0 \), we can divide by \( \sin(\alpha) \):
\[ 2\cos(\alpha) = 1 \quad \Rightarrow \quad \cos(\alpha) = \frac{1}{2} \]
Thus, \( \alpha = \frac{\pi}{3} \) (since \( \alpha \) is acute).

### Conclusion
The value of \( \alpha \) is:
\[ \alpha = \frac{\pi}{3} \]