A cube of side 1 unit is placed in such a way that the origin coincides with one of its top vertices and the three axes along three of its edges. What are the co-ordinates of the vertex which is diagonally opposite to the vertex whose co-ordinates are $(1, 0, 1)$?
- A. $(0, 0, 0)$
- B. $(0, -1, 0)$
- C. $(0, 1, 0)$
- D. $(1, 1, 1)$
Answer: B. $(0, -1, 0)$
Explanation: Top vertex at origin $(0,0,0)$. Cube side 1. Axes along edges. \nVertices are corners of unit cube. Coordinates involve 0, 1 or 0, -1 depending on direction.\nGiven vertex $(1, 0, 1)$. \nOrigin $(0,0,0)$ is top. If y-axis is down? \nUsually diagonal opposite sums to $(1,1,1)$ or similar. \nIf one vertex is $(1,0,1)$, the diagonal opposite in a unit cube connects points $(x,y,z)$ and $(1-x, 1-y, 1-z)$? \nIf cube spans 0 to 1: $(1,0,1)$ opposite is $(0,1,0)$. \nIf cube spans 0 to -1? \nLet's check options. $(0, -1, 0)$ is an option. Key says (b). \nImplies y-axis goes to -1?