Question

If vectors $\vec A = \cos \omega t\,\hat i + \sin \omega t\,\hat j$ and $\vec B = \cos \dfrac{{\omega t}}{2}\,\hat i + \sin \dfrac{{\omega t}}{2}\,\hat j$ are functions of time, then the value of $t$ at which they are orthogonal to each other is:
(A) $t = \dfrac{\pi }{\omega }$
(B) $t = 0$
(C) $t = \dfrac{\pi }{{4\omega }}$
(D) $t = \dfrac{\pi }{{2\omega }}$

Answer 1

First of all, we will find the dot of the two vectors which are given in the question and equate the resultant to zero. The dot product of two vectors in orthogonal position is always zero. We will manipulate the expression accordingly and obtain the result.

Complete step by step solution:
In the given problem, we are given two vectors which are $\vec A$ and $\vec B$ .
The representation of the vectors are as follows:
$\vec A = \cos \omega t\,\hat i + \sin \omega t\,\hat j$
And, $\vec B = \cos \dfrac{{\omega t}}{2}\,\hat i + \sin \dfrac{{\omega t}}{2}\,\hat j$
We are also told that both the vectors will be orthogonal which means that the angle between the two vectors is at right angles.We are asked to find the value of $t$ at which they are orthogonal to each other.
To begin with, we will first try to remember that the dot product of two vectors which are right angles to each other is zero. This is due to involvement of the trigonometric ratio of the cosine component. The cosine of the right angle is always zero. We will use this technique to solve the problem.Let us begin to solve the problem.We know, when the two vectors are at orthogonal, their dot product is zero.So, we can write:
$\vec A \cdot \vec B = 0 \\\ \Rightarrow \left( {\cos \omega t\,\hat i + \sin \omega t\,\hat j} \right) \cdot \left( {\cos \dfrac{{\omega t}}{2}\,\hat i + \sin \dfrac{{\omega t}}{2}\,\hat j} \right) = 0 \\\ \Rightarrow \cos \omega t\,\cos \dfrac{{\omega t}}{2} + \sin \omega t\,\sin \dfrac{{\omega t}}{2} = 0 \\\ \Rightarrow \cos \left( {\omega t\, - \dfrac{{\omega t}}{2}} \right) = 0$
Again, we manipulate further and we get:
$\Rightarrow \cos \left( {\omega t\, - \dfrac{{\omega t}}{2}} \right) = \cos \dfrac{\pi }{2} \\\ \Rightarrow \cos \dfrac{{\omega t}}{2} = \cos \dfrac{\pi }{2} \\\ \Rightarrow \dfrac{{\omega t}}{2} = \dfrac{\pi }{2} \\\ \Rightarrow t = \dfrac{\pi }{\omega }$
Hence, the value of $t$ at which they are orthogonal to each other is $\dfrac{\pi }{\omega }$ .

The correct option is (A).

Note: This problem is purely based on the operations of vector quantities. Most of the students seem to have a confusion regarding the dot product and the cross product. The dot product is also known as a scalar product while the cross product is called a vector product. In the dot product, the resultant is maximum when the angle between the vectors is zero and minimum when they are at right angles. But the scenario is opposite in the case of cross product. In cross product, the resultant is maximum.