Two vectors `vecA` and `vecB` are defined as `vecA=ahati` and `vecB=a( cos omegathati+sin omegat 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 _________

Answer

To solve the problem, we need to analyze the two vectors given and apply the condition provided in the question. Let's break down the solution step by step.

### Step 1: Define the Vectors
We have two vectors defined as:
- \(\vec{A} = a \hat{i}\)
- \(\vec{B} = a (\cos(\omega t) \hat{i} + \sin(\omega t) \hat{j})\)

Here, \(a\) is a constant and \(\omega = \frac{\pi}{6} \, \text{radians/second}\).

### Step 2: Calculate the Magnitudes
The magnitude of \(\vec{A} + \vec{B}\) is given by:
\[ |\vec{A} + \vec{B}| = |a \hat{i} + a (\cos(\omega t) \hat{i} + \sin(\omega t) \hat{j})| \]
This simplifies to:
\[ |\vec{A} + \vec{B}| = |a(1 + \cos(\omega t)) \hat{i} + a \sin(\omega t) \hat{j}| \]
Using the Pythagorean theorem, we can find the magnitude:
\[ |\vec{A} + \vec{B}| = a \sqrt{(1 + \cos(\omega t))^2 + \sin^2(\omega t)} \]

### Step 3: Calculate the Magnitude of \(\vec{A} - \vec{B}\)
Next, we calculate the magnitude of \(\vec{A} - \vec{B}\):
\[ |\vec{A} - \vec{B}| = |a \hat{i} - a (\cos(\omega t) \hat{i} + \sin(\omega t) \hat{j})| \]
This simplifies to:
\[ |\vec{A} - \vec{B}| = |a(1 - \cos(\omega t)) \hat{i} - a \sin(\omega t) \hat{j}| \]
Using the Pythagorean theorem again:
\[ |\vec{A} - \vec{B}| = a \sqrt{(1 - \cos(\omega t))^2 + \sin^2(\omega t)} \]

### Step 4: Set Up the Given Condition
According to the problem, we have:
\[ |\vec{A} + \vec{B}| = \sqrt{3} |\vec{A} - \vec{B}| \]
Substituting the magnitudes we calculated:
\[ a \sqrt{(1 + \cos(\omega t))^2 + \sin^2(\omega t)} = \sqrt{3} a \sqrt{(1 - \cos(\omega t))^2 + \sin^2(\omega t)} \]
Dividing both sides by \(a\) (assuming \(a \neq 0\)):
\[ \sqrt{(1 + \cos(\omega t))^2 + \sin^2(\omega t)} = \sqrt{3} \sqrt{(1 - \cos(\omega t))^2 + \sin^2(\omega t)} \]

### Step 5: Square Both Sides
Squaring both sides gives:
\[ (1 + \cos(\omega t))^2 + \sin^2(\omega t) = 3 \left((1 - \cos(\omega t))^2 + \sin^2(\omega t)\right) \]

### Step 6: Expand and Simplify
Expanding both sides:
\[ 1 + 2\cos(\omega t) + \cos^2(\omega t) + \sin^2(\omega t) = 3(1 - 2\cos(\omega t) + \cos^2(\omega t) + \sin^2(\omega t)) \]
Using \(\sin^2(\omega t) + \cos^2(\omega t) = 1\):
\[ 1 + 2\cos(\omega t) + 1 = 3(1 - 2\cos(\omega t) + 1) \]
This simplifies to:
\[ 2 + 2\cos(\omega t) = 3(2 - 2\cos(\omega t)) \]
\[ 2 + 2\cos(\omega t) = 6 - 6\cos(\omega t) \]
Combining like terms:
\[ 8\cos(\omega t) = 4 \implies \cos(\omega t) = \frac{1}{2} \]

### Step 7: Solve for \(\omega t\)
The cosine function equals \(\frac{1}{2}\) at:
\[ \omega t = \frac{\pi}{3} + 2n\pi \quad \text{or} \quad \omega t = -\frac{\pi}{3} + 2n\pi \quad (n \in \mathbb{Z}) \]
Using \(\omega = \frac{\pi}{6}\):
\[ \frac{\pi}{6} t = \frac{\pi}{3} \implies t = 2 \, \text{seconds} \]

### Final Answer
Thus, the value of \(\tau\) at which the condition is satisfied for the first time is:
\[ \boxed{2} \]