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Question: If vectors \( \vec A = \cos \omega t\hat i + \sin \omega t\hat j \) and \( \vec B = \cos \dfrac{{\om...

If vectors A⃗=cos⁡ωti^+sin⁡ωtj^\vec A = \cos \omega t\hat i + \sin \omega t\hat j and B⃗=cos⁡ωt2i^+sin⁡ωt2j^\vec B = \cos \dfrac{{\omega t}}{2}\hat i + \sin \dfrac{{\omega t}}{2}\hat j are functions of time, then the value of t at which they are orthogonal to each other.
\left( A \right)t = 0 \\\ \left( B \right)t = \dfrac{\pi }{{4\omega }}\,\,\,\,\,\,\,\,\, \\\ \left( C \right)t = \dfrac{\pi }{{2\omega }}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \\\ \left( D \right)t = \dfrac{\pi }{\omega }\,\, \\\

Explanation

Solution

Hint : In order to solve this question, we are going to first see the condition of the orthogonality of the two vectors and then, simplifying that equation, the value is found for the time tt , at which, the two vectors given are orthogonal and then the correct value is chosen from the given options.
The condition of the orthogonality of two vectors is given by the equation is
The cosine of the function is zero when,
θ=(2n+1)π2\theta = \left( {2n + 1} \right)\dfrac{\pi }{2}

Complete Step By Step Answer:
Given
\vec A = \cos \omega t\hat i + \sin \omega t\hat j \\\ \vec B = \cos \dfrac{{\omega t}}{2}\hat i + \sin \dfrac{{\omega t}}{2}\hat j \\\
Now, as these two vectors are given to be orthogonal, this implies that the dot product of these two vectors is going to be zero,
Now, solving the dot product on the right hand side to form the equation
cos⁡ωtcos⁡ωt2+sin⁡ωtsin⁡ωt2=0\cos \omega t\cos \dfrac{{\omega t}}{2} + \sin \omega t\sin \dfrac{{\omega t}}{2} = 0
This formula can be written as for
cos⁡(ωt−ωt2)=0\cos \left( {\omega t - \dfrac{{\omega t}}{2}} \right) = 0
Now, cosine function is zero, when
\omega t - \dfrac{{\omega t}}{2} = \dfrac{\pi }{2} \\\ \Rightarrow \dfrac{{\omega t}}{2} = \dfrac{\pi }{2} \\\ \Rightarrow t = \dfrac{\pi }{\omega } \\\
Hence, the option (D)t=πω  \left( D \right)t = \dfrac{\pi }{\omega }\,\, is the correct answer.

Note :
The two vectors are orthogonal if they are perpendicular to each other. This means that the dot product of the two vectors is equal to zero. Because if two vectors are perpendicular to each other, then,
θ=π2\theta = \dfrac{\pi }{2}
Thus, the cosine of this angle is
cos⁡θ=0\cos \theta = 0
Hence, making the dot product also equal to zero.