If `alpha, beta`be two vectors whose moduli are a and b respectively and they are such that `[alpha + beta]`is `bot` to `beta`and `alpha`is `bot` to `2beta + alpha`, then

Options

A. `a = b sqrt(2)`
B. `a =2b`
C. `a = b`
D. 2a = b

Answer

To solve the given problem, we need to analyze the conditions provided for the vectors \( \alpha \) and \( \beta \).

### Step-by-Step Solution:

1. **Understanding the Given Conditions**:
We have two vectors \( \alpha \) and \( \beta \) with magnitudes \( a \) and \( b \) respectively. The conditions state:
- \( \alpha + \beta \) is perpendicular to \( \beta \).
- \( \alpha \) is perpendicular to \( 2\beta + \alpha \).

2. **Using the Dot Product**:
Since \( \alpha + \beta \) is perpendicular to \( \beta \), we can express this condition using the dot product:
\[ (\alpha + \beta) \cdot \beta = 0 \]
Expanding this, we get:
\[ \alpha \cdot \beta + \beta \cdot \beta = 0 \]
Since \( \beta \cdot \beta = |\beta|^2 = b^2 \), we can write:
\[ \alpha \cdot \beta + b^2 = 0 \implies \alpha \cdot \beta = -b^2 \]

3. **Applying the Second Condition**:
Now, for the second condition where \( \alpha \) is perpendicular to \( 2\beta + \alpha \):
\[ \alpha \cdot (2\beta + \alpha) = 0 \]
Expanding this gives:
\[ 2\alpha \cdot \beta + \alpha \cdot \alpha = 0 \]
Here, \( \alpha \cdot \alpha = |\alpha|^2 = a^2 \), so we can write:
\[ 2\alpha \cdot \beta + a^2 = 0 \implies 2\alpha \cdot \beta = -a^2 \]

4. **Substituting the Value of \( \alpha \cdot \beta \)**:
From the first condition, we found that \( \alpha \cdot \beta = -b^2 \). Substituting this into the equation from the second condition:
\[ 2(-b^2) = -a^2 \implies -2b^2 = -a^2 \implies 2b^2 = a^2 \]

5. **Finding the Relationship Between \( a \) and \( b \)**:
Taking the square root of both sides gives:
\[ \sqrt{2}b = a \]
Thus, we conclude that:
\[ a = \sqrt{2}b \]

### Final Answer:
The relationship between the magnitudes of the vectors \( \alpha \) and \( \beta \) is:
\[ a = \sqrt{2}b \]