ABCD is a regular tetrahedron, A is the origin and B lies on x-axis. ABC lies in the xy-plane `|vec(AB)|=2` Under these conditions, the number of possible tetrahedrons is :

Answer

To solve the problem step by step, let's analyze the conditions given for the regular tetrahedron ABCD.

1. **Understanding the Tetrahedron**:
- A tetrahedron has four vertices. In this case, we have vertices A, B, C, and D.
- A is the origin (0, 0, 0).
- B lies on the x-axis, which means its coordinates can be represented as (x, 0, 0).

2. **Finding the Position of B**:
- The modulus of the vector AB is given as |AB| = 2. Since A is at the origin, we can say:
\[ |AB| = |B - A| = |B| = 2 \]
- Therefore, B can be at (2, 0, 0) or (-2, 0, 0). This gives us **2 possible positions for B**.

3. **Position of C**:
- The triangle ABC lies in the xy-plane. Therefore, the z-coordinate of C must be 0.
- The distance from B to C must also be equal to 2 (since all sides of a regular tetrahedron are equal).
- If B is at (2, 0, 0), C can be at (2, y, 0) or (2, -y, 0) such that:
\[ |BC| = 2 \Rightarrow \sqrt{(2 - 2)^2 + (y - 0)^2 + (0 - 0)^2} = 2 \Rightarrow |y| = 2 \]
- Thus, C can be at (2, 2, 0) or (2, -2, 0).
- If B is at (-2, 0, 0), similar reasoning applies, and C can be at (-2, 2, 0) or (-2, -2, 0).
- Hence, we have **2 possible positions for C** regardless of the position of B.

4. **Position of D**:
- The vertex D can be positioned above or below the xy-plane. Thus, D can have a z-coordinate of either +z or -z.
- The distance from A to D must also be equal to 2. Therefore, D can be at (0, 0, 2) or (0, 0, -2).
- This gives us **2 possible positions for D**.

5. **Calculating Total Combinations**:
- Now we can combine the possibilities:
- 1 position for A (the origin)
- 2 positions for B
- 2 positions for C
- 2 positions for D
- The total number of possible tetrahedrons is given by:
\[ \text{Total Tetrahedrons} = 1 \times 2 \times 2 \times 2 = 8 \]

Thus, the number of possible tetrahedrons is **8**.