A

To find the resultant of the three vectors \\( \vec{OA}, \vec{OB}, \vec{OC} \\), we start by defining the vectors and their magnitudes. Given the angles and the radius of the circle (let's denote it as \( r \\)), we can use trigonometric functions to resolve the components of each vector.
Step 1: Decompose each vector into its x and y components:
- For \( \vec{OA} \):
\( OA_x = r \cdot \cos(90^\circ) = 0 \)
\( OA_y = r \cdot \sin(90^\circ) = r \)
- For \( \vec{OB} \):
\( OB_x = r \cdot \cos(45^\circ) = \frac{r}{\sqrt{2}} \)
\( OB_y = r \cdot \sin(45^\circ) = \frac{r}{\sqrt{2}} \)
- For \( \vec{OC} \):
\( OC_x = r \cdot \cos(0^\circ) = r \)
\( OC_y = r \cdot \sin(0^\circ) = 0 \)
Step 2: Now, sum the components:
\( R_x = OA_x + OB_x + OC_x = 0 + \frac{r}{\sqrt{2}} + r = r + \frac{r}{\sqrt{2}} \)
\( R_y = OA_y + OB_y + OC_y = r + \frac{r}{\sqrt{2}} + 0 = r + \frac{r}{\sqrt{2}} \)
Step 3: Calculate the magnitude of the resultant vector \( R \):
\( R = \sqrt{R_x^2 + R_y^2}
R = \sqrt{ \left( r + \frac{r}{\sqrt{2}} \right)^2 + \left( r + \frac{r}{\sqrt{2}} \right)^2 } = \sqrt{2 \left( r + \frac{r}{\sqrt{2}} \right)^2} = \sqrt{2} \left( r + \frac{r}{\sqrt{2}} \right)
= \sqrt{2}r \left( 1 + \frac{1}{\sqrt{2}} \right)
Therefore, the resultant vector \( \vec{R} \) can be calculated and simplified further if needed.