Two vectors vec A and vec B are defined as vec A =a hat i and vec B =a(cos ω t hat i +sin ω t hat j ) , where a is a constant and ω=π / 6 rad s s -1 . If vec A +vec B =√(3) vec A -vec B at time t=τ for the first time, the value of τ , in seconds, is \ \ \ \ -.

Physics · Vectors and Scalars

JEE Advanced 2018 — Paper 1 — Question 7

Two vectors $\vec{A}$ and $\vec{B}$ are defined as $\vec{A}=a \hat{i}$ and $\vec{B}=a(\cos \omega t \hat{i}+\sin \omega t \hat{j})$ , where $a$ is a constant and

$\omega=\pi / 6 \mathrm{rad} \mathrm{s} s^{-1}$ . If $|\vec{A}+\vec{B}|=\sqrt{3}|\vec{A}-\vec{B}|$ at time $t=\tau$ for the first time, the value of $\tau$ , in seconds,

is $\_\_\_\_$ -.

Answer: 2

Numerical answer — enter this value.

Step-by-step solution

$\therefore 2 a \cos \frac{\omega t}{2}=\sqrt{3}(2 a) \sin \frac{\omega t}{2}$

$\therefore \tan \frac{\omega t}{2}=\frac{1}{\sqrt{3}}$

for the first time $\Rightarrow \frac{\pi t}{12}=\frac{\pi}{6} \Rightarrow t=2 \mathrm{sec}$

Answer key and solution verified before publishing.

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Exam: JEE Advanced 2018
Paper: Paper 1
Subject: Physics
Chapter: Vectors and Scalars
Topic: Product of Vectors and Applications