Two vectors A and B are defined as A $=$ a i and $B =$ a$(\cos ωt i + \sin ωt j),$ where a is a constant and $ω = π/6$ rad $s^{-1}.$ If |A $+ B| = √3 |$A - B| at time $t = τ$ for the first time, the value of τ, in seconds, is ______.

Step 1: Determine the magnitudes of the vectors.

The magnitudes of $\vec{A}$ and $\vec{B}$ are constant and equal.

$$|\vec{A}| = a$$ $$|\vec{B}| = \sqrt{(a\cos\omega t)^2 + (a\sin\omega t)^2} = a$$

Step 2: Express the magnitudes of the sum and difference.

Let $\theta = \omega t$ be the angle between $\vec{A}$ and $\vec{B}$.

$$|\vec{A} + \vec{B}|^2 = |\vec{A}|^2 + |\vec{B}|^2 + 2|\vec{A}||\vec{B}|\cos\theta$$ $$|\vec{A} + \vec{B}|^2 = 2a^2(1 + \cos\theta)$$ $$|\vec{A} - \vec{B}|^2 = |\vec{A}|^2 + |\vec{B}|^2 - 2|\vec{A}||\vec{B}|\cos\theta$$ $$|\vec{A} - \vec{B}|^2 = 2a^2(1 - \cos\theta)$$

Step 3: Apply the given condition.

Square both sides of the given relation $|\vec{A} + \vec{B}| = \sqrt{3}\,|\vec{A} - \vec{B}|$.

$$|\vec{A} + \vec{B}|^2 = 3|\vec{A} - \vec{B}|^2$$

Step 4: Solve for the angle $\theta$.

Substitute the squared magnitudes into the relation.

$$2a^2(1 + \cos\theta) = 3 \cdot 2a^2(1 - \cos\theta)$$ $$1 + \cos\theta = 3 - 3\cos\theta$$ $$4\cos\theta = 2$$ $$\cos\theta = \frac{1}{2}$$

For the first occurrence, the angle is:

$$\theta = \frac{\pi}{3}$$

Step 5: Calculate the time $\tau$.

Use the relation $\theta = \omega \tau$ to find the time.

$$\tau = \frac{\theta}{\omega}$$ $$\tau = \frac{\pi/3}{\pi/6}$$ $$\tau = 2\ \text{s}$$

Answer: 2