In a regular triangular prism ABCA1B1C1, all edges of which are equal to 1, find the distance from point

In a regular triangular prism ABCA1B1C1, all edges of which are equal to 1, find the distance from point B to straight line A1C1.

The distance from point B to straight line A1C1 will be a perpendicular drawn from point B to straight line A1C1, that is, segment BK.

The projection of the BK segment onto the ABC plane is the BH segment, which is perpendicular to the AC side, then AC is the height of the equilateral triangle ABC, and therefore its median.

Then the height HH = AC * √3 / 2 = 1 * √3 / 2 = √3 / 2 cm.

Segment КН = АА1 = 1 cm.

Then in a right-angled triangle BHK, by the Pythagorean theorem, KB ^ 2 = KH ^ 2 + BH ^ 2 = 1 + 3/4 = 7/4.

KB = √7 / 2 cm.

Answer: From point B to straight line A1C1 is equal to √7 / 2 cm.