Let OABC be a tetrahedron whose edges are of unit length. If `vec OA = vec a` , `vec OB = vec b`, and `vec OC` = `alpha(vec a + vec b) + beta(vec a xx vec b), `then `(alpha beta)^2 = p/q`(where p & q are relatively prime to each other),then the value of `[q/(2p)]` is

Answer

To solve the problem, we need to find the values of \(\alpha\) and \(\beta\) for the tetrahedron \(OABC\) with unit edges. We know that:

- \(\vec{OA} = \vec{a}\)
- \(\vec{OB} = \vec{b}\)
- \(\vec{OC} = \alpha(\vec{a} + \vec{b}) + \beta(\vec{a} \times \vec{b})\)

### Step 1: Understand the Geometry of the Tetrahedron
In a regular tetrahedron, all edges are of equal length (unit length in this case). Therefore, we have:

\[ |\vec{a}| = |\vec{b}| = |\vec{c}| = 1 \]

### Step 2: Calculate the Height of the Tetrahedron
The height \(CD\) from point \(C\) to the base triangle \(OAB\) can be calculated using the formula for the height of a regular tetrahedron:

\[ h = \frac{\sqrt{6}}{3} \cdot \text{side length} \]

For a unit tetrahedron, the height is:

\[ h = \frac{\sqrt{6}}{3} \]

### Step 3: Determine the Relationship Between Vectors
Since \(CD\) is perpendicular to both \(\vec{a}\) and \(\vec{b}\), we can express it in terms of the cross product:

\[ \vec{CD} = k(\vec{a} \times \vec{b}) \]

where \(k\) is some scalar. Taking magnitudes, we have:

\[ |\vec{CD}| = |k| |\vec{a} \times \vec{b}| \]

The magnitude of \(\vec{a} \times \vec{b}\) is given by:

\[ |\vec{a} \times \vec{b}| = |\vec{a}||\vec{b}|\sin(60^\circ) = 1 \cdot 1 \cdot \frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{2} \]

Thus,

\[ \frac{\sqrt{6}}{3} = |k| \cdot \frac{\sqrt{3}}{2} \]

Solving for \(k\):

\[ |k| = \frac{\sqrt{6}}{3} \cdot \frac{2}{\sqrt{3}} = \frac{2\sqrt{2}}{3} \]

Let \(k = \beta\), so:

\[ \beta = \frac{2\sqrt{2}}{3} \]

### Step 4: Find \(\alpha\)
The centroid \(D\) of triangle \(OAB\) can be calculated as:

\[ \vec{OD} = \frac{\vec{a} + \vec{b}}{3} \]

The magnitude of \(\vec{OD}\) is:

\[ |\vec{OD}| = \left|\frac{\vec{a} + \vec{b}}{3}\right| = \frac{1}{3} |\vec{a} + \vec{b}| \]

Using the cosine rule in triangle \(OAB\):

\[ |\vec{a} + \vec{b}|^2 = |\vec{a}|^2 + |\vec{b}|^2 + 2|\vec{a}||\vec{b}|\cos(60^\circ) = 1 + 1 + 2 \cdot \frac{1}{2} = 3 \]

Thus,

\[ |\vec{a} + \vec{b}| = \sqrt{3} \]

Then,

\[ |\vec{OD}| = \frac{1}{3} \sqrt{3} \]

Now, we equate this to the expression for \(\vec{OC}\):

\[ |\alpha(\vec{a} + \vec{b}) + \beta(\vec{a} \times \vec{b})| = 1 \]

Since the magnitude of \(\vec{OC}\) must also equal 1, we can express this as:

\[ |\alpha| \cdot |\vec{a} + \vec{b}| + |\beta| \cdot |\vec{a} \times \vec{b}| = 1 \]

Substituting the values we have:

\[ |\alpha| \cdot \sqrt{3} + |\beta| \cdot \frac{\sqrt{3}}{2} = 1 \]

Substituting \(\beta\):

\[ |\alpha| \cdot \sqrt{3} + \frac{2\sqrt{2}}{3} \cdot \frac{\sqrt{3}}{2} = 1 \]

This simplifies to:

\[ |\alpha| \cdot \sqrt{3} + \frac{\sqrt{6}}{3} = 1 \]

Solving for \(\alpha\):

\[ |\alpha| \cdot \sqrt{3} = 1 - \frac{\sqrt{6}}{3} \]

\[ |\alpha| = \frac{1 - \frac{\sqrt{6}}{3}}{\sqrt{3}} = \frac{3 - \sqrt{6}}{3\sqrt{3}} \]

### Step 5: Calculate \((\alpha \beta)^2\)
Now we can find \((\alpha \beta)^2\):

\[ \alpha \beta = \left(\frac{3 - \sqrt{6}}{3\sqrt{3}}\right) \cdot \left(\frac{2\sqrt{2}}{3}\right) \]

Calculating this gives:

\[ \alpha \beta = \frac{(3 - \sqrt{6}) \cdot 2\sqrt{2}}{9\sqrt{3}} \]

Squaring this:

\[ (\alpha \beta)^2 = \frac{(3 - \sqrt{6})^2 \cdot 8}{81 \cdot 3} = \frac{(9 - 6\sqrt{6} + 6) \cdot 8}{243} = \frac{(15 - 6\sqrt{6}) \cdot 8}{243} \]

This results in:

\[ (\alpha \beta)^2 = \frac{120 - 48\sqrt{6}}{243} \]

### Step 6: Identify \(p\) and \(q\)
To express this in the form \(\frac{p}{q}\), we need to simplify and identify \(p\) and \(q\).

Assuming \(p = 120 - 48\sqrt{6}\) and \(q = 243\), we need to ensure they are coprime.

### Step 7: Calculate \(\frac{q}{2p}\)
Finally, we need to find \(\frac{q}{2p}\):

\[ \frac{q}{2p} = \frac{243}{2(120 - 48\sqrt{6})} \]

This gives us the final answer.

### Final Answer
The value of \(\left[\frac{q}{2p}\right]\) is calculated based on the values of \(p\) and \(q\) obtained above.