To solve this problem, we'll analyze the conditions given about the vectors $\vec{A}$ and $\vec{B}$, and their resultant $\vec{R}$.

Given:

1. $\vec{R} = \vec{A} + \vec{B}$ is perpendicular to $\vec{A}$.


2. The magnitude of $\vec{R}$ is half that of $\vec{B}$: $|\vec{R}| = \frac{1}{2}|\vec{B}|$.

Approach:

We know that if $\vec{R}$ is perpendicular to $\vec{A}$, then their dot product is zero:

$ \vec{R} \cdot \vec{A} = 0 $

Substitute $\vec{R} = \vec{A} + \vec{B}$:

$ (\vec{A} + \vec{B}) \cdot \vec{A} = 0 $

$$ \Rightarrow $$$ \vec{A} \cdot \vec{A} + \vec{B} \cdot \vec{A} = 0 $

$$ \Rightarrow $$$ |\vec{A}|^2 + |\vec{A}||\vec{B}|\cos \theta = 0 $

$$ \Rightarrow $$$ \cos \theta = -\frac{|\vec{A}|^2}{|\vec{A}||\vec{B}|} $

$$ \Rightarrow $$$ \cos \theta = -\frac{|\vec{A}|}{|\vec{B}|} $

Next, we use the second condition involving magnitudes:

$ |\vec{R}| = |\vec{A} + \vec{B}| = \frac{1}{2}|\vec{B}| $

$$ \Rightarrow $$$ \sqrt{|\vec{A}|^2 + |\vec{B}|^2 + 2|\vec{A}||\vec{B}|\cos \theta} = \frac{1}{2}|\vec{B}| $

$$ \Rightarrow $$$ |\vec{A}|^2 + |\vec{B}|^2 - 2|\vec{A}||\vec{B}|\frac{|\vec{A}|}{|\vec{B}|} = \frac{1}{4}|\vec{B}|^2 $

$$ \Rightarrow $$$ |\vec{A}|^2 + |\vec{B}|^2 - 2|\vec{A}|^2 = \frac{1}{4}|\vec{B}|^2 $

$$ \Rightarrow $$$ |\vec{B}|^2 - |\vec{A}|^2 = \frac{1}{4}|\vec{B}|^2 $

$$ \Rightarrow $$$ \frac{3}{4}|\vec{B}|^2 = |\vec{A}|^2 $

$$ \Rightarrow $$$ |\vec{A}| = \frac{\sqrt{3}}{2}|\vec{B}| $

Finally, substitute this back into the cosine formula:

$ \cos \theta = -\frac{\frac{\sqrt{3}}{2}|\vec{B}|}{|\vec{B}|} $

$ \cos \theta = -\frac{\sqrt{3}}{2} $

This value of $\cos \theta$ corresponds to an angle of $150^\circ$ because $\cos 150^\circ = -\frac{\sqrt{3}}{2}$.

Answer:

The angle between $\vec{A}$ and $\vec{B}$ is $150^\circ$.