The resultant of two vectors \(\vec{A}\) and \(\vec{B}\) is perpendicular to \(\overrightarrow{\mathrm{A}}\) and its magnitude is half that of \(\vec{B}\). The angle between vectors \(\vec{A}\) and \(\vec{B}\) is _______.

Explanation:
Let the angle between vector \(\vec{A}\) and vector \(\vec{B}\) be \(\alpha\), and let the resultant vector be \(\vec{R}\). We can analyze the vector components geometrically or analytically:

1. Analytical Method:
The magnitude of the component of \(\vec{B}\) perpendicular to \(\vec{A}\) must balance out to form the resultant vector \(\vec{R}\), since \(\vec{R}\) has no component along the direction of \(\vec{A}\). This means that the components can be split as:
- Along the direction of \(\vec{A}\): \(A + B \cos\alpha = 0 \implies \cos\alpha = -\frac{A}{B}\)
- Perpendicular to \(\vec{A}\): \(R = B \sin\alpha\)

2. Using the Given Conditions:
We are given that the magnitude of the resultant is half of \(\vec{B}\), so \(R = \frac{B}{2}\). Substituting this into the perpendicular component equation yields:
\( \frac{B}{2} = B \sin\alpha \implies \sin\alpha = \frac{1}{2} \)

3. Determining the Angle (\(\alpha\)):
Since \(\cos\alpha = -\frac{A}{B}\) is negative (as magnitudes \(A\) and \(B\) are positive), the angle \(\alpha\) must lie in the second quadrant (\(90^\circ < \alpha < 180^\circ\)).
For \(\sin\alpha = \frac{1}{2}\) in the second quadrant:
\( \alpha = 180^\circ - 30^\circ = 150^\circ \)

4. Geometric Verification:
Consider a right-angled triangle formed by the vectors where \(\vec{R}\) is perpendicular to \(\vec{A}\). Let \(\theta\) be the angle that \(\vec{B}\) makes with the vector \(\vec{R}\).
From the geometry of the components:
\( \cos\theta = \frac{R}{B} = \frac{\left(\frac{B}{2}\right)}{B} = \frac{1}{2} \implies \theta = 60^\circ \)
Since \(\vec{R}\) is at \(90^\circ\) relative to \(\vec{A}\), the total angle \(\alpha\) between \(\vec{A}\) and \(\vec{B}\) is:
\( \alpha = 90^\circ + \theta = 90^\circ + 60^\circ = 150^\circ \)

Hence, Option C is the correct choice.