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Question: If vectors \(\vec A = \cos \left( {\omega t} \right)\hat i + \sin \left( {\omega t} \right)\hat j\) ...

If vectors A⃗=cos⁡(ωt)i^+sin⁡(ωt)j^\vec A = \cos \left( {\omega t} \right)\hat i + \sin \left( {\omega t} \right)\hat j and B⃗=cos⁡(ωt2)i^+sin⁡(ωt2)j^\vec B = \cos \left( {\dfrac{{\omega t}}{2}} \right)\hat i + \sin \left( {\dfrac{{\omega t}}{2}} \right)\hat j are functions of time, then the value of tt at which they are orthogonal to each other is?

Explanation

Solution

To solve this question, we will use the basic concept of vectors and one who knows what orthogonal means can easily solve this question. Here we know that two vectors are orthogonal if and only if the dot product is zero. So, proceed accordingly to get the required solution.

Complete step by step answer:
As we know that, vectors to be orthogonal to each other their dot or scalar product should be zero and according to the question it is given that,
A⃗=cos⁡(ωt)i^+sin⁡(ωt)j^\vec A = \cos \left( {\omega t} \right)\hat i + \sin \left( {\omega t} \right)\hat j and
B⃗=cos⁡(ωt2)i^+sin⁡(ωt2)j^\vec B = \cos \left( {\dfrac{{\omega t}}{2}} \right)\hat i + \sin \left( {\dfrac{{\omega t}}{2}} \right)\hat j
Now,
[cos⁡(ωt)i^+sin⁡(ωt)j^]cos⁡(ωt2)i^+sin⁡(ωt2)j^=0 ⇒cos⁡ωtcos⁡ωt2+sin⁡ωtsin⁡ωt2=0 \left[ {\cos \left( {\omega t} \right)\hat i + \sin \left( {\omega t} \right)\hat j} \right]\cos \left( {\dfrac{{\omega t}}{2}} \right)\hat i + \sin \left( {\dfrac{{\omega t}}{2}} \right)\hat j = 0 \\\ \Rightarrow \cos \omega t\cos \dfrac{{\omega t}}{2} + \sin \omega t\sin \dfrac{{\omega t}}{2} = 0 \\\
And we know that,
cos⁡(A−B)=cos⁡Acos⁡B+sin⁡Asin⁡B\cos \left( {A - B} \right) = \cos A\cos B + \sin A\sin B
Now solving accordingly,
cos⁡ωtcos⁡ωt2+sin⁡ωtsin⁡ωt2=0 ⇒cos⁡(ωt−ωt2)=0 ⇒cos⁡ωt2=0 \cos \omega t\cos \dfrac{{\omega t}}{2} + \sin \omega t\sin \dfrac{{\omega t}}{2} = 0 \\\ \Rightarrow \cos \left( {\omega t - \dfrac{{\omega t}}{2}} \right) = 0 \\\ \Rightarrow \cos \dfrac{{\omega t}}{2} = 0 \\\
And we know that, cos⁡π2=0\cos \dfrac{\pi }{2} = 0 ,
Therefore,

\Rightarrow \cos \dfrac{{\omega t}}{2} = \cos \dfrac{\pi }{2} \\\ \Rightarrow \dfrac{{\omega t}}{2} = \dfrac{\pi }{2} \\\ $$ Now, solving for $t$ , $$\dfrac{{\omega t}}{2} = \dfrac{\pi }{2} \\\ \Rightarrow t = \dfrac{\pi }{\omega } \\\ $$ **Hence, the value of $t$ at which both the vectors are orthogonal is $$\dfrac{\pi }{\omega }$$.** **Note:** orthogonal means lines intersecting to each other perpendicularly or we can say that they will have perpendicular slopes to the point where they will intersect. Or orthogonal refers to or entails lines that are perpendicular to each other or create right angles, as in Many orthogonal elements are used in this design. This is referred to as orthographic. And remember the trigonometric formula which is used above.