Two vectors $\overrightarrow A$ and $\overrightarrow B$ are defined as $\overrightarrow A = a\hat i$ and $\overrightarrow B = a\left( {\cos \omega t\hat i + \sin \omega t\hat j} \right)$ where a is constant and $\omega = \dfrac{\pi }{6}rad{s^{ - 1}}$. If $\left| {\overrightarrow A + \overrightarrow B } \right| = \sqrt 3 \left| {\overrightarrow A - \overrightarrow B } \right|$ at time $t = \tau$ for the first time, the value of $\tau$ in seconds is:

Hint:Vectors are geometric representation and it represents both magnitude and direction. It is generally represented by a single arrow, starting from one point and ending at a different point. A vector is defined by both its magnitude and direction with respect to a set of coordinates. In order to solve the above problem, we need to find the magnitude of both $\overrightarrow A $ and $ \overrightarrow B $, followed by multiplication of vectors using a suitable formula.

Formula used:
$ {\left| {\overrightarrow A + \overrightarrow B } \right|^2} = {\left| {\overrightarrow A } \right|^2} + {\left| {\overrightarrow B } \right|^2} + 2\overrightarrow A \overrightarrow {.B}$
 $ {\left| {\overrightarrow A - \overrightarrow B } \right|^2} = {\left| {\overrightarrow A } \right|^2} + {\left| {\overrightarrow B } \right|^2} - 2\overrightarrow A \overrightarrow {.B}$
$ \overrightarrow A .\overrightarrow B = \left| {\overrightarrow A } \right|\left| {\overrightarrow B } \right|\cos \theta $ ($\theta $ being the angle between the vectors)
This is basically called the dot product of two vectors.

Complete step by step answer:
Here we will be using a method of dot product of two vectors in order to find the correct answer.Let us first understand the meaning of dot product of two vectors. The dot product of two vectors is equal to the product of magnitude of two vectors and cos value of angle between the two vectors.
Given information are –
$ \overrightarrow A = a\hat i $
Magnitude will be ${a^2}$
$\overrightarrow B = a\left( {\cos \omega t\hat i + \sin \omega t\hat j} \right)$ (a being the constant term and also the magnitude of the vector)
Magnitude will be ${a^2}$
$\omega = \dfrac{\pi }{6}rad{s^{ - 1}}$ (angular velocity in radian per second)
$\left| {\overrightarrow A + \overrightarrow B } \right| = \sqrt 3 \left| {\overrightarrow A - \overrightarrow B } \right|$ ………. (A)
Now for time $t = \tau $ we need to get the value of $\tau$
Squaring both side of the equation A we get,
${\left( {\left| {\overrightarrow A + \overrightarrow B } \right|} \right)^2} = 3{\left( {\left| {\overrightarrow A - \overrightarrow B } \right|} \right)^2}$
$ \Rightarrow {\left| {\overrightarrow A } \right|^2} + {\left| {\overrightarrow B } \right|^2} + 2\overrightarrow A \overrightarrow {.B} = 3\left( {{{\left| {\overrightarrow A } \right|}^2} + {{\left| {\overrightarrow B } \right|}^2} - 2\overrightarrow A \overrightarrow {.B} } \right)$
$ \Rightarrow {a^2} + {a^2} + 2{a^2}\cos \omega t = 3{a^2} + 3{a^2} - 6{a^2}\cos \omega t$ (using the identity ${\cos ^2}x + {\sin ^2}x = 1$)
$\Rightarrow 8{a^2}\cos \omega t = 4{a^2} \\
\Rightarrow \cos \omega t = \dfrac{1}{2} \\$
$ \Rightarrow \omega t = \dfrac{\pi }{3}$ or $\omega t = - \dfrac{\pi }{3}$
But as the value of time cannot be a negative value, hence considering the positive value we get the only value as-
$ \omega t = \dfrac{\pi }{3} \\
\Rightarrow t = 2 $
And as $t = \tau $ given in seconds,
$\therefore \tau = 2$

Hence, the value of $\tau $ is 2 seconds.

Additional information:
There is one more term mentioned in the solution and it is the term ‘Angular velocity’ which means the velocity gained by a rotating body because of a rotational movement which is about the axis of rotation of the object.

Note: Here in the above case, the component along y-direction of $\overrightarrow B $ vector is nullified as only the dot product of the vectors is considered.Also we cannot consider a negative time value as that is not acceptable. Here again the value of $\tau$ is asked for the first time appearance. If there were no such condition given then the generalized term of $\cos \omega t$ would have been used and a general pattern of answer could have been obtained. We can also relate this mathematically from the cosine graph where $\cos \omega t$ obtains the value $\dfrac{1}{2}$ many times at a regular time interval of $2\pi$ because we know that the time period of a cosine function is $2\pi$.