Let `vecA=(hati+hatj)and,vecB=(2hati-hatj)`. The magnitude of a coplanar vector `vecC` such that `vecA*vecC=vecB*vecC=vecA*vecB`, is given by :

Options

A. `sqrt((9)/(12))`
B. `sqrt((5)/(9))`
C. `sqrt((20)/(9))`
D. `sqrt((10)/(9))`

Answer

To solve the problem, we need to find the magnitude of a coplanar vector \(\vec{C}\) such that \(\vec{A} \cdot \vec{C} = \vec{B} \cdot \vec{C} = \vec{A} \cdot \vec{B}\).

### Step-by-Step Solution:

1. **Define the vectors**:
\[ \vec{A} = \hat{i} + \hat{j} \]
\[ \vec{B} = 2\hat{i} - \hat{j} \]

2. **Calculate \(\vec{A} \cdot \vec{B}\)**:
\[ \vec{A} \cdot \vec{B} = (1)(2) + (1)(-1) = 2 - 1 = 1 \]
Thus, \(\vec{A} \cdot \vec{B} = 1\).

3. **Assume the form of vector \(\vec{C}\)**:
Let \(\vec{C} = x\hat{i} + y\hat{j}\).

4. **Calculate \(\vec{A} \cdot \vec{C}\)**:
\[ \vec{A} \cdot \vec{C} = (\hat{i} + \hat{j}) \cdot (x\hat{i} + y\hat{j}) = x + y \]

5. **Calculate \(\vec{B} \cdot \vec{C}\)**:
\[ \vec{B} \cdot \vec{C} = (2\hat{i} - \hat{j}) \cdot (x\hat{i} + y\hat{j}) = 2x - y \]

6. **Set up the equations**:
Since \(\vec{A} \cdot \vec{C} = \vec{B} \cdot \vec{C} = \vec{A} \cdot \vec{B}\), we have:
\[ x + y = 1 \quad \text{(1)} \]
\[ 2x - y = 1 \quad \text{(2)} \]

7. **Solve the equations**:
- From equation (1): \(y = 1 - x\).
- Substitute \(y\) in equation (2):
\[ 2x - (1 - x) = 1 \]
\[ 2x - 1 + x = 1 \]
\[ 3x - 1 = 1 \]
\[ 3x = 2 \implies x = \frac{2}{3} \]

8. **Find \(y\)**:
Substitute \(x\) back into equation (1):
\[ y = 1 - \frac{2}{3} = \frac{1}{3} \]

9. **Magnitude of vector \(\vec{C}\)**:
Now, we find the magnitude of \(\vec{C}\):
\[ |\vec{C}| = \sqrt{x^2 + y^2} = \sqrt{\left(\frac{2}{3}\right)^2 + \left(\frac{1}{3}\right)^2} \]
\[ = \sqrt{\frac{4}{9} + \frac{1}{9}} = \sqrt{\frac{5}{9}} = \frac{\sqrt{5}}{3} \]

### Final Answer:
The magnitude of vector \(\vec{C}\) is \(\frac{\sqrt{5}}{3}\).