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, we will follow the instructions given in the video transcript.

### Step 1: Define the vector
Let \(\vec{a}\) be a vector in the xy-plane making an angle of \(60^\circ\) with the positive x-axis. We can express this vector in terms of its components:
\[ \vec{a} = |a| \cos(60^\circ) \hat{i} + |a| \sin(60^\circ) \hat{j} \]
Since \(\cos(60^\circ) = \frac{1}{2}\) and \(\sin(60^\circ) = \frac{\sqrt{3}}{2}\), we have:
\[ \vec{a} = |a| \cdot \frac{1}{2} \hat{i} + |a| \cdot \frac{\sqrt{3}}{2} \hat{j} \]
Let \( |a| = x \), then:
\[ \vec{a} = \frac{x}{2} \hat{i} + \frac{\sqrt{3}x}{2} \hat{j} \]

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

### Step 3: 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:
\[ |\vec{a} - 2\hat{i}| = \sqrt{\left(\frac{x}{2} - 2\right)^2 + \left(\frac{\sqrt{3}x}{2}\right)^2} \]

### Step 4: Set up the geometric mean equation
According to the problem, \(|\vec{a} - \hat{i}|\) is the geometric mean of \(|\vec{a}|\) and \(|\vec{a} - 2\hat{i}|\):
\[ |\vec{a} - \hat{i}|^2 = |\vec{a}| \cdot |\vec{a} - 2\hat{i}| \]

### Step 5: Substitute and simplify
Substituting the magnitudes we calculated:
\[ \left(\sqrt{\left(\frac{x}{2} - 1\right)^2 + \left(\frac{\sqrt{3}x}{2}\right)^2}\right)^2 = x \cdot \sqrt{\left(\frac{x}{2} - 2\right)^2 + \left(\frac{\sqrt{3}x}{2}\right)^2} \]
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 the equation
Now we will solve the equation derived in Step 5. First, we expand both sides and simplify:
1. Expand \(\left(\frac{x}{2} - 1\right)^2 + \frac{3x^2}{4}\).
2. Expand \(\left(\frac{x}{2} - 2\right)^2 + \frac{3x^2}{4}\).
3. Set the two sides equal and solve for \(x\).

After solving, we find:
\[ x = 2(1 - \sqrt{2}) \]

### Step 7: Find \(|\vec{a}|\)
Finally, we calculate \(|\vec{a}|\):
\[ |\vec{a}| = x = 2(1 - \sqrt{2}) \]

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