Two vectors `vecA` and `vecB` are defined as `vecA=ahati` and `vecB=a( cos omegahati+sin omega hatj)`, were a is a constant and `omega=pi//6 rads^(-1)`. If `|vecA+vecB|=sqrt(3)|vecA-vecB|` at time `t=tau` for the first time, the value of `tau`, in seconds , is _________

Explanation

To solve the problem, we need to find the value of \( \tau \) such that the condition \( |\vec{A} + \vec{B}| = \sqrt{3} |\vec{A} - \vec{B}| \) holds true for the first time.

### Step-by-Step Solution:

1. **Define the vectors**:
\[ \vec{A} = a \hat{i} \]
\[ \vec{B} = a (\cos(\omega t) \hat{i} + \sin(\omega t) \hat{j}) \]
where \( \omega = \frac{\pi}{6} \, \text{rad/s} \).

2. **Calculate \( \vec{A} + \vec{B} \)**:
\[ \vec{A} + \vec{B} = a \hat{i} + a (\cos(\omega t) \hat{i} + \sin(\omega t) \hat{j}) = a(1 + \cos(\omega t)) \hat{i} + a \sin(\omega t) \hat{j} \]

3. **Calculate the magnitude of \( \vec{A} + \vec{B} \)**:
\[ |\vec{A} + \vec{B}| = \sqrt{(a(1 + \cos(\omega t)))^2 + (a \sin(\omega t))^2} \]
\[ = a \sqrt{(1 + \cos(\omega t))^2 + \sin^2(\omega t)} \]
\[ = a \sqrt{1 + 2\cos(\omega t) + \cos^2(\omega t) + \sin^2(\omega t)} \]
\[ = a \sqrt{2 + 2\cos(\omega t)} = a \sqrt{2(1 + \cos(\omega t))} = a \sqrt{4 \cos^2\left(\frac{\omega t}{2}\right)} = 2a \cos\left(\frac{\omega t}{2}\right) \]

4. **Calculate \( \vec{A} - \vec{B} \)**:
\[ \vec{A} - \vec{B} = a \hat{i} - a (\cos(\omega t) \hat{i} + \sin(\omega t) \hat{j}) = a(1 - \cos(\omega t)) \hat{i} - a \sin(\omega t) \hat{j} \]

5. **Calculate the magnitude of \( \vec{A} - \vec{B} \)**:
\[ |\vec{A} - \vec{B}| = \sqrt{(a(1 - \cos(\omega t)))^2 + (-a \sin(\omega t))^2} \]
\[ = a \sqrt{(1 - \cos(\omega t))^2 + \sin^2(\omega t)} \]
\[ = a \sqrt{1 - 2\cos(\omega t) + \cos^2(\omega t) + \sin^2(\omega t)} \]
\[ = a \sqrt{2 - 2\cos(\omega t)} = a \sqrt{4 \sin^2\left(\frac{\omega t}{2}\right)} = 2a \sin\left(\frac{\omega t}{2}\right) \]

6. **Set up the equation**:
According to the problem statement:
\[ |\vec{A} + \vec{B}| = \sqrt{3} |\vec{A} - \vec{B}| \]
Substituting the magnitudes we found:
\[ 2a \cos\left(\frac{\omega t}{2}\right) = \sqrt{3} \cdot 2a \sin\left(\frac{\omega t}{2}\right) \]
Dividing both sides by \( 2a \) (assuming \( a \neq 0 \)):
\[ \cos\left(\frac{\omega t}{2}\right) = \sqrt{3} \sin\left(\frac{\omega t}{2}\right) \]

7. **Use the tangent function**:
\[ \frac{\cos\left(\frac{\omega t}{2}\right)}{\sin\left(\frac{\omega t}{2}\right)} = \sqrt{3} \implies \tan\left(\frac{\omega t}{2}\right) = \frac{1}{\sqrt{3}} \]
The angle for which \( \tan(\theta) = \frac{1}{\sqrt{3}} \) is \( \frac{\pi}{6} \):
\[ \frac{\omega t}{2} = \frac{\pi}{6} \implies \omega t = \frac{\pi}{3} \]

8. **Solve for \( t \)**:
Since \( \omega = \frac{\pi}{6} \):
\[ \frac{\pi}{6} t = \frac{\pi}{3} \implies t = 2 \, \text{seconds} \]

Thus, the value of \( \tau \) is \( \boxed{2} \) seconds.