Let `vecu` be a vector on rectangular coodinate system with sloping angle `60^(@)` suppose that `|vecu-hati|` is geomtric mean of `|vecu| and |vecu-2hati|`, where `hati` is the unit vector along the x-axis . Then find the value of `(sqrt2- 1)/ |vecu|`

Explanation

To solve the problem step-by-step, we will follow the reasoning outlined in the video transcript while providing a clear mathematical derivation.

### Step 1: Understand the Given Information
We have a vector \( \vec{u} \) that makes an angle of \( 60^\circ \) with the positive x-axis (represented by the unit vector \( \hat{i} \)). We also know that \( |\vec{u} - \hat{i}| \) is the geometric mean of \( |\vec{u}| \) and \( |\vec{u} - 2\hat{i}| \).

### Step 2: Use the Dot Product
The dot product of \( \vec{u} \) and \( \hat{i} \) can be expressed as:
\[ \vec{u} \cdot \hat{i} = |\vec{u}| \cdot |\hat{i}| \cdot \cos(60^\circ) \]
Since \( |\hat{i}| = 1 \) and \( \cos(60^\circ) = \frac{1}{2} \), we have:
\[ \vec{u} \cdot \hat{i} = \frac{|\vec{u}|}{2} \]

### Step 3: Set Up the Geometric Mean Condition
According to the problem, we have:
\[ |\vec{u} - \hat{i}|^2 = |\vec{u}| \cdot |\vec{u} - 2\hat{i}| \]
Let’s denote \( |\vec{u}| = u \). We can express the left-hand side:
\[ |\vec{u} - \hat{i}|^2 = |\vec{u}|^2 + |\hat{i}|^2 - 2 \vec{u} \cdot \hat{i} = u^2 + 1 - 2 \cdot \frac{u}{2} = u^2 + 1 - u = u^2 - u + 1 \]

### Step 4: Calculate the Right-Hand Side
Now, we calculate \( |\vec{u} - 2\hat{i}| \):
\[ |\vec{u} - 2\hat{i}|^2 = |\vec{u}|^2 + |2\hat{i}|^2 - 2 \vec{u} \cdot (2\hat{i}) = u^2 + 4 - 2 \cdot 2 \cdot \frac{u}{2} = u^2 + 4 - 2u = u^2 - 2u + 4 \]
Thus, the right-hand side becomes:
\[ |\vec{u}| \cdot |\vec{u} - 2\hat{i}| = u \cdot \sqrt{u^2 - 2u + 4} \]

### Step 5: Set the Two Sides Equal
Now we equate both sides:
\[ u^2 - u + 1 = u \cdot \sqrt{u^2 - 2u + 4} \]

### Step 6: Square Both Sides
Squaring both sides gives:
\[ (u^2 - u + 1)^2 = u^2(u^2 - 2u + 4) \]

### Step 7: Expand Both Sides
Expanding the left-hand side:
\[ (u^2 - u + 1)^2 = u^4 - 2u^3 + 3u^2 - 2u + 1 \]
Expanding the right-hand side:
\[ u^2(u^2 - 2u + 4) = u^4 - 2u^3 + 4u^2 \]

### Step 8: Set the Equation to Zero
Setting both expansions equal:
\[ u^4 - 2u^3 + 3u^2 - 2u + 1 = u^4 - 2u^3 + 4u^2 \]
This simplifies to:
\[ 3u^2 - 2u + 1 - 4u^2 = 0 \implies -u^2 - 2u + 1 = 0 \implies u^2 + 2u - 1 = 0 \]

### Step 9: Solve the Quadratic Equation
Using the quadratic formula:
\[ u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-2 \pm \sqrt{2^2 - 4 \cdot 1 \cdot (-1)}}{2 \cdot 1} = \frac{-2 \pm \sqrt{4 + 4}}{2} = \frac{-2 \pm 2\sqrt{2}}{2} = -1 \pm \sqrt{2} \]
Since \( |\vec{u}| \) must be positive, we take:
\[ |\vec{u}| = -1 + \sqrt{2} \]

### Step 10: Find the Final Value
We need to find:
\[ \frac{\sqrt{2} - 1}{|\vec{u}|} = \frac{\sqrt{2} - 1}{-1 + \sqrt{2}} = 1 \]

### Final Answer
Thus, the value of \( \frac{\sqrt{2} - 1}{|\vec{u}|} \) is:
\[ \boxed{1} \]