The resultant of two vectors vec A and vec B is perpendicular to overrightarrow A and its magnitude is half that of vec B . The angle between vectors vec A and vec B is

Physics · Vectors and Scalars

JEE Main 2024 — 9 April, Shift 2 — Question 52

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

Answer: 150

Numerical answer — enter this value.

Step-by-step solution

\mathrm{We\; are\; given:}\quad \vec{R}=\vec{A}+\vec{B},\quad \vec{R}\perp \vec{A},\quad |\vec{R}|=\tfrac{1}{2}|\vec{B}|.

\mathrm{Condition\ 1:}\quad \vec{R}\cdot\vec{A}=0\quad\Rightarrow\quad (\vec{A}+\vec{B})\cdot\vec{A}=0.

|\vec{A}|^2+\vec{A}\cdot\vec{B}=0\quad\Rightarrow\quad \vec{A}\cdot\vec{B}=-|\vec{A}|^2.

\mathrm{Let\ }|\vec{A}|=a,\;|\vec{B}|=b,\;\theta=\angle(\vec{A},\vec{B}).

\vec{A}\cdot\vec{B}=ab\cos\theta=-a^2\quad\Rightarrow\quad \cos\theta=-\frac{a}{b}.

\mathrm{Condition\ 2:}\quad |\vec{R}|=\tfrac{1}{2}b.

|\vec{R}|^2=|\vec{A}+\vec{B}|^2=a^2+b^2+2ab\cos\theta.

=\;a^2+b^2+2(-a^2)=b^2-a^2.

\therefore\ |\vec{R}|^2=b^2-a^2=\left(\tfrac{b}{2}\right)^2=\tfrac{b^2}{4}.

\Rightarrow\ b^2-a^2=\tfrac{b^2}{4}\quad\Rightarrow\quad \tfrac{3b^2}{4}=a^2.

\therefore\ \frac{a}{b}=\frac{\sqrt{3}}{2}.

\cos\theta=-\frac{a}{b}=-\frac{\sqrt{3}}{2}.

\therefore\ \theta=150^\circ.

\boxed{150^\circ}

Answer key and solution verified before publishing.

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Exam: JEE Main 2024
Paper: 9 April, Shift 2
Subject: Physics
Chapter: Vectors and Scalars
Topic: Product of Vectors and Applications