Question 14: Motion in 2D & Vectors

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 _________$^\circ$.

Accepted Answer

To solve this problem, we'll analyze the conditions given about the vectors $\vec{A}$ and $\vec{B}$, and their resultant $\vec{R}$.
Given:
$\vec{R} = \vec{A} + \vec{B}$ is perpendicular to $\vec{A}$.
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 	heta = 0 $
$ \Rightarrow $$ \cos 	heta = -\frac{|\vec{A}|^2}{|\vec{A}||\vec{B}|} $
$ \Rightarrow $$ \cos 	heta = -\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 $$ \sqrt{|\vec{A}|^2 + |\vec{B}|^2 + 2|\vec{A}||\vec{B}|\cos 	heta} = \frac{1}{2}|\vec{B}| $
$ \Rightarrow $$ |\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 	heta = -\frac{\frac{\sqrt{3}}{2}|\vec{B}|}{|\vec{B}|} $
$ \cos 	heta = -\frac{\sqrt{3}}{2} $
This value of $\cos 	heta$ 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 $150^\circ$.

Motion in 2D & Vectors

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 _________^\circ.

Integer / Numeric Answer Type