If `|hata-hatb|= sqrt2` then calculate the value of `|hata-sqrt3hatb|`.

Explanation

To solve the problem step-by-step, let's break it down clearly:

### Step 1: Understand the given information
We are given that:
\[ | \hat{a} - \hat{b} | = \sqrt{2} \]
We need to find:
\[ | \hat{a} - \sqrt{3} \hat{b} | \]

### Step 2: Use the formula for the magnitude of the difference of two vectors
The magnitude of the difference between two vectors can be expressed using the cosine of the angle between them:
\[ | \hat{a} - \hat{b} | = \sqrt{|\hat{a}|^2 + |\hat{b}|^2 - 2 |\hat{a}| |\hat{b}| \cos \theta} \]
Given that \( |\hat{a}| = 1 \) and \( |\hat{b}| = 1 \) (since they are unit vectors), we can substitute these values into the equation:
\[ | \hat{a} - \hat{b} | = \sqrt{1^2 + 1^2 - 2 \cdot 1 \cdot 1 \cdot \cos \theta} \]
This simplifies to:
\[ | \hat{a} - \hat{b} | = \sqrt{2 - 2 \cos \theta} \]

### Step 3: Set up the equation
Since we know that \( | \hat{a} - \hat{b} | = \sqrt{2} \), we can set the equations equal to each other:
\[ \sqrt{2 - 2 \cos \theta} = \sqrt{2} \]

### Step 4: Square both sides
Squaring both sides gives:
\[ 2 - 2 \cos \theta = 2 \]
This simplifies to:
\[ -2 \cos \theta = 0 \]
Thus:
\[ \cos \theta = 0 \]
This implies that \( \theta = 90^\circ \), meaning the vectors \( \hat{a} \) and \( \hat{b} \) are perpendicular.

### Step 5: Calculate \( | \hat{a} - \sqrt{3} \hat{b} | \)
We can now calculate \( | \hat{a} - \sqrt{3} \hat{b} | \) using the same formula:
\[ | \hat{a} - \sqrt{3} \hat{b} | = \sqrt{|\hat{a}|^2 + |\sqrt{3} \hat{b}|^2 - 2 |\hat{a}| |\sqrt{3} \hat{b}| \cos \theta} \]
Substituting the values:
\[ | \hat{a} - \sqrt{3} \hat{b} | = \sqrt{1^2 + (\sqrt{3})^2 - 2 \cdot 1 \cdot \sqrt{3} \cdot \cos(90^\circ)} \]
Since \( \cos(90^\circ) = 0 \), the equation simplifies to:
\[ | \hat{a} - \sqrt{3} \hat{b} | = \sqrt{1 + 3} \]
\[ | \hat{a} - \sqrt{3} \hat{b} | = \sqrt{4} \]
\[ | \hat{a} - \sqrt{3} \hat{b} | = 2 \]

### Final Answer
Thus, the value of \( | \hat{a} - \sqrt{3} \hat{b} | \) is:
\[ \boxed{2} \]