1. Write the vectors explicitly

Given $\vec A = a\hat i$ and $\vec B = a(\cos \omega t\,\hat i + \sin \omega t\,\hat j), \qquad \omega = \frac{\pi}{6}\;\text{rad s}^{-1}.$

So, $\vec A + \vec B = a(1+\cos \omega t)\hat i + a\sin \omega t\,\hat j$

and $\vec A - \vec B = a(1-\cos \omega t)\hat i - a\sin \omega t\,\hat j.$

2. Find their magnitudes

For $\vec A+\vec B$: $|\vec A+\vec B|^2 = a^2\big[(1+\cos \omega t)^2 + \sin^2 \omega t\big].$

Using $\sin^2\theta+\cos^2\theta=1$, $|\vec A+\vec B|^2 = a^2\big[1+2\cos \omega t + \cos^2\omega t + \sin^2\omega t\big]$ $= a^2(2+2\cos \omega t).$

Thus, $|\vec A+\vec B| = a\sqrt{2(1+\cos \omega t)}.$

Similarly, for $\vec A-\vec B$: $|\vec A-\vec B|^2 = a^2\big[(1-\cos \omega t)^2 + \sin^2 \omega t\big]$ $= a^2(2-2\cos \omega t).$

Hence, $|\vec A-\vec B| = a\sqrt{2(1-\cos \omega t)}.$

3. Use the given condition

Given $|\vec A+\vec B| = \sqrt3\,|\vec A-\vec B|.$

Substitute the expressions: $a\sqrt{2(1+\cos \omega t)} = \sqrt3\,a\sqrt{2(1-\cos \omega t)}.$

Cancel $a$ and $\sqrt2$: $\sqrt{1+\cos \omega t} = \sqrt3\sqrt{1-\cos \omega t}.$

Square both sides: $1+\cos \omega t = 3(1-\cos \omega t).$

So, $1+\cos \omega t = 3-3\cos \omega t$ $4\cos \omega t = 2$ $\cos \omega t = \frac12.$

4. Find the first time

The first positive time when $\cos \omega t = \frac12$ is when $\omega t = \frac{\pi}{3}.$

Given $\omega = \frac{\pi}{6},$ so $t = \frac{\pi/3}{\pi/6} = 2\;\text{s}.$

5. Final answer

$\boxed{\tau = 2}$

6. Comparison with stored answer

Stored correct answer: $2$

My derived answer is also $2$, so they agree.