Question

The resultant of two vectors $\vec{A}$ and $\vec{B}$ is perpendicular to $\vec{A}$ and its magnitude is half that of $\vec{B}$. The angle between vectors $\vec{A}$ and $\vec{B}$ is ________ .

Answer 1

The resultant vector $\vec{R}$ of $\vec{A}$ and $\vec{B}$ is perpendicular to $\vec{A}$. The magnitude of $\vec{R}$ is given as:
$|\vec{R}| = \frac{|\vec{B}|}{2}.$

Using the vector projection formula, the component of $\vec{B}$ along $\vec{A}$ is:
$B \cos \theta = \frac{B}{2}.$
Simplify: $\cos \theta = \frac{1}{2}.$

From this, $\theta = 60^\circ$. Since $\vec{R}$ is perpendicular to $\vec{A}$, the angle between $\vec{A}$ and $\vec{B}$ is:
$\text{Angle between } \vec{A} \text{ and } \vec{B} = 90^\circ + 60^\circ = 150^\circ.$

Answer 2

The resultant vector $\vec{R}$ of $\vec{A}$ and $\vec{B}$ is perpendicular to $\vec{A}$. The magnitude of $\vec{R}$ is given as:
$|\vec{R}| = \frac{|\vec{B}|}{2}.$

Using the vector projection formula, the component of $\vec{B}$ along $\vec{A}$ is:
$B \cos \theta = \frac{B}{2}.$
Simplify: $\cos \theta = \frac{1}{2}.$

From this, $\theta = 60^\circ$. Since $\vec{R}$ is perpendicular to $\vec{A}$, the angle between $\vec{A}$ and $\vec{B}$ is:
$\text{Angle between } \vec{A} \text{ and } \vec{B} = 90^\circ + 60^\circ = 150^\circ.$