A cube of side 1 unit is placed so that the origin coincides with one of its…

Concept

For an axis-aligned unit cube, the three coordinate axes run along the three cube edges that meet at the origin, so the cube occupies the region where every coordinate lies between 0 and 1 and each vertex has coordinates that are each either 0 or 1. Two vertices lie at opposite ends of the cube's space (body) diagonal exactly when each coordinate of one is the complement of the other: every 0 becomes 1 and every 1 becomes 0. Such a pair is separated by the longest distance inside the cube, the space diagonal of length √3.

Here the origin is the vertex where the three positive axis directions start, taken along the cube's edges. Calling it a "top" vertex only names its position along one axis; it does not move the cube into negative coordinates and does not change the complement rule used to find the diagonally opposite vertex. So all eight vertices still have 0/1 coordinates, and the answer is found the same way whichever vertex you label "top" or "bottom".

Applying it here

1. Write the coordinates of the given vertex: (1, 0, 1).

2. Complement each coordinate independently: 1 → 0, 0 → 1, 1 → 0.

3. Read off the resulting vertex: (0, 1, 0).

Cross-check

Measure the straight-line distance from (1, 0, 1) to (0, 1, 0): √((1−0)2 + (0−1)2 + (1−0)2) = √3. That equals the cube's full space diagonal, so these two vertices are joined by the longest diagonal of the cube. For comparison, (0, 0, 0) lies only √2 away (a face diagonal) and (1, 1, 1) lies only 1 unit away (an edge), so the diagonal-opposite of (1, 0, 1) is the vertex (0, 1, 0).