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NEET 2015
Exam
NEET
Year
2015
Subject
Physics
Chapter
Mathematics in Physics

NEET 2015 · Physics · Mathematics in Physics

If vectors → A = cos ^ i + sin ^ j and → B = cos( )/(2) ^ i + sin (...

If vectors \overset{\to}{A} = cos \omegat\overset{^}{i} + sin\omegat\overset{^}{j} and \overset{\to}{B} = cos\dfrac{\omegat}{2}\overset{^}{i} + sin \dfrac{\omegat}{2}\overset{^}{j} are functions of time, then the value of tt at which they are orthogonal to each other is:

Options

  1. At=π2ωt = \dfrac{\pi}{2\omega}
  2. Bt=πωt = \dfrac{\pi}{\omega}
  3. Ct=0t = 0
  4. Dt=π4ωt = \dfrac{\pi}{4\omega}

Correct answer

B. t=πωt = \dfrac{\pi}{\omega}

Step-by-step solution

\overset{\to}{A} = coswt\overset{^}{i} + sinwt\overset{^}{j}
\overset{\to}{B} = cos\dfrac{wt}{2}\overset{^}{i} + sin\dfrac{wt}{2}\overset{^}{j}
For these vectors to be orthogonal, A→⋅B→=0\overset{\to}{A} \cdot \overset{\to}{B} = 0
A→⋅B→=0=coswt.coswt2+sinwt.sinwt2\overset{\to}{A} \cdot \overset{\to}{B} = 0 = coswt . cos\dfrac{wt}{2} + sinwt . sin\dfrac{wt}{2}
=cos⁡(wt−wt2)=cos⁡(wt2)=0= \cos \left(wt - \dfrac{wt}{2}\right) = \cos \left(\dfrac{wt}{2}\right) = 0
So, as we know, (\cos (90)=0).
So, wt2=π2⇒t=πw\dfrac{wt}{2} = \dfrac{\pi}{2} ⇒ t = \dfrac{\pi}{w}
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