Find the magnitude and direction of the vector, $0 \leq \theta< 2 \pi$. Answer exactly, or round to at least 2 decimal places.
\[
\begin{array}{l}
v=< 0,3> \\
|v|=3 \\
\theta=3
\end{array}
\]

Step 1 ：Given the vector v = <0,3>.

Step 2 ：The magnitude of a vector is calculated by taking the square root of the sum of the squares of its components. In this case, the vector v has components 0 and 3, so its magnitude is \(\sqrt{0^2 + 3^2} = 3\).

Step 3 ：The direction of a vector is the angle it makes with the positive x-axis. Since the x-component of the vector is 0 and the y-component is positive, the vector points straight up along the y-axis. Therefore, the direction of the vector is 90 degrees or \(\frac{\pi}{2}\) radians.

Step 4 ：Final Answer: The magnitude of the vector is \(\boxed{3}\) and its direction is \(\boxed{90}\) degrees or \(\boxed{\frac{\pi}{2}}\) radians.