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 _________

Answer

To solve the problem, we need to analyze the given vectors and the condition provided.

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

Where \(a\) is a constant and \(\omega = \frac{\pi}{6} \, \text{rads}^{-1}\).

### Step 2: Find the magnitudes of \(\vec{A} + \vec{B}\) and \(\vec{A} - \vec{B}\)
The sum of the vectors is:
\[ \vec{A} + \vec{B} = a \hat{i} + a (\cos(\omega t) \hat{i} + \sin(\omega t) \hat{j}) = a(1 + \cos(\omega t)) \hat{i} + a \sin(\omega t) \hat{j} \]

The difference of the vectors is:
\[ \vec{A} - \vec{B} = a \hat{i} - a (\cos(\omega t) \hat{i} + \sin(\omega t) \hat{j}) = a(1 - \cos(\omega t)) \hat{i} - a \sin(\omega t) \hat{j} \]

### Step 3: Calculate the magnitudes
The magnitude of \(\vec{A} + \vec{B}\) is:
\[ |\vec{A} + \vec{B}| = \sqrt{(a(1 + \cos(\omega t)))^2 + (a \sin(\omega t))^2} \]
\[ = a \sqrt{(1 + \cos(\omega t))^2 + \sin^2(\omega t)} \]
Using the identity \((1 + \cos(\omega t))^2 + \sin^2(\omega t) = 1 + 2\cos(\omega t) + \cos^2(\omega t) + \sin^2(\omega t) = 2 + 2\cos(\omega t) = 2(1 + \cos(\omega t))\):
\[ |\vec{A} + \vec{B}| = a \sqrt{2(1 + \cos(\omega t))} \]

The magnitude of \(\vec{A} - \vec{B}\) is:
\[ |\vec{A} - \vec{B}| = \sqrt{(a(1 - \cos(\omega t)))^2 + (-a \sin(\omega t))^2} \]
\[ = a \sqrt{(1 - \cos(\omega t))^2 + \sin^2(\omega t)} \]
Using the identity \((1 - \cos(\omega t))^2 + \sin^2(\omega t) = 1 - 2\cos(\omega t) + \cos^2(\omega t) + \sin^2(\omega t) = 2 - 2\cos(\omega t) = 2(1 - \cos(\omega t))\):
\[ |\vec{A} - \vec{B}| = a \sqrt{2(1 - \cos(\omega t))} \]

### Step 4: Set up the equation
According to the problem, we have:
\[ |\vec{A} + \vec{B}| = \sqrt{3} |\vec{A} - \vec{B}| \]
Substituting the magnitudes we found:
\[ a \sqrt{2(1 + \cos(\omega t))} = \sqrt{3} a \sqrt{2(1 - \cos(\omega t))} \]
Dividing both sides by \(a\) (assuming \(a \neq 0\)):
\[ \sqrt{2(1 + \cos(\omega t))} = \sqrt{3} \sqrt{2(1 - \cos(\omega t))} \]
Squaring both sides:
\[ 2(1 + \cos(\omega t)) = 3 \cdot 2(1 - \cos(\omega t)) \]
\[ 1 + \cos(\omega t) = 3(1 - \cos(\omega t)) \]
\[ 1 + \cos(\omega t) = 3 - 3\cos(\omega t) \]
\[ 4\cos(\omega t) = 2 \]
\[ \cos(\omega t) = \frac{1}{2} \]

### Step 5: Solve for \(\omega t\)
The angle whose cosine is \(\frac{1}{2}\) is:
\[ \omega t = \frac{\pi}{3} + 2n\pi \quad \text{or} \quad \omega t = -\frac{\pi}{3} + 2n\pi \quad \text{for } n \in \mathbb{Z} \]

### Step 6: Substitute \(\omega\)
Since \(\omega = \frac{\pi}{6}\):
\[ \frac{\pi}{6} t = \frac{\pi}{3} \]
Solving for \(t\):
\[ t = \frac{\pi/3}{\pi/6} = 2 \text{ seconds} \]

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