Time for Vector Magnitude Ratio Condition under Uniform Rotation

To find the value of $\tau$, we analyze the vectors $\vec{A}$ and $\vec{B}$:

Given: $\vec{A} = a\,\hat{i}$ $\vec{B} = a\cos(\omega t)\,\hat{i} + a\sin(\omega t)\,\hat{j}$

The magnitudes of the vectors are: $|\vec{A}| = a$ $|\vec{B}| = \sqrt{(a\cos\omega t)^2 + (a\sin\omega t)^2} = a$

Since $\vec{A}$ lies along the positive $x$-axis and $\vec{B}$ makes an angle $\omega t$ with the positive $x$-axis, the angle between $\vec{A}$ and $\vec{B}$ is: $\theta = \omega t$

The magnitudes of the sum and difference of the two vectors are: $|\vec{A} + \vec{B}|^2 = |\vec{A}|^2 + |\vec{B}|^2 + 2|\vec{A}||\vec{B}|\cos\theta = a^2 + a^2 + 2a^2\cos(\omega t) = 2a^2(1 + \cos\omega t)$ $|\vec{A} - \vec{B}|^2 = |\vec{A}|^2 + |\vec{B}|^2 - 2|\vec{A}||\vec{B}|\cos\theta = a^2 + a^2 - 2a^2\cos(\omega t) = 2a^2(1 - \cos\omega t)$

We are given that at time $t = \tau$: $|\vec{A} + \vec{B}| = \sqrt{3}|\vec{A} - \vec{B}|$

Squaring both sides: $|\vec{A} + \vec{B}|^2 = 3|\vec{A} - \vec{B}|^2$

Substitute the expressions in terms of $\omega \tau$: $2a^2(1 + \cos\omega\tau) = 3 \times 2a^2(1 - \cos\omega\tau)$

Dividing both sides by $2a^2$: $1 + \cos\omega\tau = 3(1 - \cos\omega\tau)$ $1 + \cos\omega\tau = 3 - 3\cos\omega\tau$ $4\cos\omega\tau = 2$ $\cos\omega\tau = \frac{1}{2}$

For the condition to be satisfied for the first time ($t > 0$), the smallest positive angle is: $\omega\tau = \frac{\pi}{3}$

Given that $\omega = \frac{\pi}{6}\text{ rad s}^{-1}$: $\left(\frac{\pi}{6}\right)\tau = \frac{\pi}{3}$ $\tau = \frac{\pi/3}{\pi/6} = 2\text{ s}$

Thus, the value of $\tau$ is $2$ (or $2.00$).