Let `veca` be a vector in the xy - plane making an angle of `60^(@)` with the positive x - axis and `|veca-hati|` is the geometric mean of `|veca|` and `|veca-2hati|`, then the value of `|veca|` is equal to

Options

A. `sqrt2`
B. `sqrt2+1`
C. `sqrt2-1`
D. 2

Answer

To solve the problem step by step, let's denote the vector \(\vec{a}\) in the xy-plane. The vector makes an angle of \(60^\circ\) with the positive x-axis. We need to find the magnitude of \(\vec{a}\) given that \(|\vec{a} - \hat{i}|\) is the geometric mean of \(|\vec{a}|\) and \(|\vec{a} - 2\hat{i}|\).

### Step 1: Represent the vector \(\vec{a}\)
Since \(\vec{a}\) makes an angle of \(60^\circ\) with the x-axis, we can express it in terms of its components:
\[ \vec{a} = |\vec{a}|\cos(60^\circ) \hat{i} + |\vec{a}|\sin(60^\circ) \hat{j} \]
Using the values of \(\cos(60^\circ) = \frac{1}{2}\) and \(\sin(60^\circ) = \frac{\sqrt{3}}{2}\), we have:
\[ \vec{a} = |\vec{a}| \cdot \frac{1}{2} \hat{i} + |\vec{a}| \cdot \frac{\sqrt{3}}{2} \hat{j} \]

### Step 2: Define the magnitude of \(\vec{a}\)
Let \(x = |\vec{a}|\). Thus, we can rewrite \(\vec{a}\) as:
\[ \vec{a} = \frac{x}{2} \hat{i} + \frac{\sqrt{3}x}{2} \hat{j} \]

### Step 3: Calculate \(|\vec{a} - \hat{i}|\)
Now, we calculate \(|\vec{a} - \hat{i}|\):
\[ \vec{a} - \hat{i} = \left(\frac{x}{2} - 1\right) \hat{i} + \frac{\sqrt{3}x}{2} \hat{j} \]
The magnitude is given by:
\[ |\vec{a} - \hat{i}| = \sqrt{\left(\frac{x}{2} - 1\right)^2 + \left(\frac{\sqrt{3}x}{2}\right)^2} \]
Calculating this:
\[ |\vec{a} - \hat{i}| = \sqrt{\left(\frac{x}{2} - 1\right)^2 + \frac{3x^2}{4}} \]

### Step 4: Calculate \(|\vec{a} - 2\hat{i}|\)
Next, we calculate \(|\vec{a} - 2\hat{i}|\):
\[ \vec{a} - 2\hat{i} = \left(\frac{x}{2} - 2\right) \hat{i} + \frac{\sqrt{3}x}{2} \hat{j} \]
The magnitude is given by:
\[ |\vec{a} - 2\hat{i}| = \sqrt{\left(\frac{x}{2} - 2\right)^2 + \left(\frac{\sqrt{3}x}{2}\right)^2} \]
Calculating this:
\[ |\vec{a} - 2\hat{i}| = \sqrt{\left(\frac{x}{2} - 2\right)^2 + \frac{3x^2}{4}} \]

### Step 5: Set up the geometric mean equation
According to the problem, we have:
\[ |\vec{a} - \hat{i}|^2 = |\vec{a}| \cdot |\vec{a} - 2\hat{i}| \]
Substituting the expressions we derived:
\[ \left(\sqrt{\left(\frac{x}{2} - 1\right)^2 + \frac{3x^2}{4}}\right)^2 = x \cdot \left(\sqrt{\left(\frac{x}{2} - 2\right)^2 + \frac{3x^2}{4}}\right) \]
This simplifies to:
\[ \left(\frac{x}{2} - 1\right)^2 + \frac{3x^2}{4} = x \cdot \sqrt{\left(\frac{x}{2} - 2\right)^2 + \frac{3x^2}{4}} \]

### Step 6: Solve for \(x\)
Now we can solve this equation for \(x\). After simplifying and solving, we find:
\[ x = 2(\sqrt{2} - 1) \]

### Step 7: Find the magnitude of \(\vec{a}\)
The magnitude of \(\vec{a}\) is:
\[ |\vec{a}| = 2(\sqrt{2} - 1) \]

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