Prove that the points `O(0,0,0), A(2.0,0), B(1,sqrt3,0) and C(1,1/sqrt3,(2sqrt2)/sqrt3)` are the vertices of a regular tetrahedron.,

Answer

To prove that the points \( O(0,0,0) \), \( A(2,0,0) \), \( B(1,\sqrt{3},0) \), and \( C\left(1,\frac{1}{\sqrt{3}},\frac{2\sqrt{2}}{\sqrt{3}}\right) \) are the vertices of a regular tetrahedron, we need to show that the distances between all pairs of these points are equal.

### Step 1: Calculate the distance \( OA \)
Using the distance formula:
\[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2} \]
For points \( O(0,0,0) \) and \( A(2,0,0) \):
\[ OA = \sqrt{(2 - 0)^2 + (0 - 0)^2 + (0 - 0)^2} = \sqrt{2^2} = \sqrt{4} = 2 \]

### Step 2: Calculate the distance \( OB \)
For points \( O(0,0,0) \) and \( B(1,\sqrt{3},0) \):
\[ OB = \sqrt{(1 - 0)^2 + (\sqrt{3} - 0)^2 + (0 - 0)^2} = \sqrt{1^2 + (\sqrt{3})^2} = \sqrt{1 + 3} = \sqrt{4} = 2 \]

### Step 3: Calculate the distance \( OC \)
For points \( O(0,0,0) \) and \( C\left(1,\frac{1}{\sqrt{3}},\frac{2\sqrt{2}}{\sqrt{3}}\right) \):
\[ OC = \sqrt{\left(1 - 0\right)^2 + \left(\frac{1}{\sqrt{3}} - 0\right)^2 + \left(\frac{2\sqrt{2}}{\sqrt{3}} - 0\right)^2} \]
Calculating each term:
\[ = \sqrt{1^2 + \left(\frac{1}{\sqrt{3}}\right)^2 + \left(\frac{2\sqrt{2}}{\sqrt{3}}\right)^2} = \sqrt{1 + \frac{1}{3} + \frac{8}{3}} = \sqrt{1 + 3} = \sqrt{4} = 2 \]

### Step 4: Calculate the distance \( AB \)
For points \( A(2,0,0) \) and \( B(1,\sqrt{3},0) \):
\[ AB = \sqrt{(1 - 2)^2 + (\sqrt{3} - 0)^2 + (0 - 0)^2} = \sqrt{(-1)^2 + (\sqrt{3})^2} = \sqrt{1 + 3} = \sqrt{4} = 2 \]

### Step 5: Calculate the distance \( AC \)
For points \( A(2,0,0) \) and \( C\left(1,\frac{1}{\sqrt{3}},\frac{2\sqrt{2}}{\sqrt{3}}\right) \):
\[ AC = \sqrt{(1 - 2)^2 + \left(\frac{1}{\sqrt{3}} - 0\right)^2 + \left(\frac{2\sqrt{2}}{\sqrt{3}} - 0\right)^2} \]
Calculating:
\[ = \sqrt{(-1)^2 + \left(\frac{1}{\sqrt{3}}\right)^2 + \left(\frac{2\sqrt{2}}{\sqrt{3}}\right)^2} = \sqrt{1 + \frac{1}{3} + \frac{8}{3}} = \sqrt{1 + 3} = \sqrt{4} = 2 \]

### Step 6: Calculate the distance \( BC \)
For points \( B(1,\sqrt{3},0) \) and \( C\left(1,\frac{1}{\sqrt{3}},\frac{2\sqrt{2}}{\sqrt{3}}\right) \):
\[ BC = \sqrt{(1 - 1)^2 + \left(\frac{1}{\sqrt{3}} - \sqrt{3}\right)^2 + \left(\frac{2\sqrt{2}}{\sqrt{3}} - 0\right)^2} \]
Calculating:
\[ = \sqrt{0 + \left(\frac{1 - 3}{\sqrt{3}}\right)^2 + \left(\frac{2\sqrt{2}}{\sqrt{3}}\right)^2} = \sqrt{0 + \left(-\frac{2}{\sqrt{3}}\right)^2 + \left(\frac{2\sqrt{2}}{\sqrt{3}}\right)^2} \]
\[ = \sqrt{\frac{4}{3} + \frac{8}{3}} = \sqrt{4} = 2 \]

### Conclusion
Since all distances \( OA, OB, OC, AB, AC, \) and \( BC \) are equal to \( 2 \), we conclude that the points \( O, A, B, \) and \( C \) are the vertices of a regular tetrahedron.