Q123 / 160 · AP EAMCET 2023 (17 May Shift 1) · Physics
The resultant of two vectors $\vec{A}$ and $\vec{B}$ is perpendicular to vector $\vec{A}$ and the resultant magnitude is equal to half of the magnitude of $\vec{B}$, then the angle between $\vec{A}$ and $\vec{B}$ is
- A$30^{\circ}$
- B$60^{\circ}$
- Correct answerC$150^{\circ}$
- D$120^{\circ}$
Solution
$\begin{aligned}
& \vec{R}=\vec{A}+\vec{B} \\
& R=\frac{|\vec{B}|}{2} \\
& \frac{|\vec{B}|}{2}=\sqrt{A^2+B^2+2 A B \cos \theta} \\
& A^2+B^2+2 A B \cos \theta=\frac{B^2}{4} \\
& A^2+2 A B \cos \theta+\frac{3 B^2}{4}=0...(i) \\
& \vec{R} \perp \text { to } \vec{A} \\
& \vec{A} \cdot(\vec{A}+\vec{B})=0 \\
& A^2+A B \cos \theta=0\\
& A B \cos \theta=-A^2...(ii)
\end{aligned}$
$\begin{aligned}
& \text { put this value in equation (1) } \\
& \mathrm{A}^2-2 \mathrm{~A}^2+\frac{3 \mathrm{~B}^2}{4}=0 \\
& \mathrm{~A}^2-\frac{3 \mathrm{~B}^2}{4}=0 \Rightarrow \mathrm{A}=\frac{\sqrt{3} \mathrm{~B}}{2}
\end{aligned}$
from equation (2),
$\begin{aligned}
& \cos \theta=-\frac{A}{B}=\frac{\sqrt{\frac{3}{2}} B}{B} \\
& \cos \theta=\cos 150 \Rightarrow \theta=150^{\circ}
\end{aligned}$
Hence, the correct answer is option: 'C'.