In a regular triangular prism ABCA1B1C1 with a side of 3 √3, find the distance between straight lines AA1 and BC.

Lines AA1 and BC intersect, then the distance between them will be the perpendicular drawn from point A to the segment BC.

Point A is the apex of triangle ABC, and BC is its opposite side. Then AH is the height of the equilateral triangle ABC, which means its median.

Then ВН = СН = ВС / 2 = 3 * √3 / 2.

AH ^ 2 = AC ^ 2 – CH ^ 2 = 27 – 27/4 = 81/4.

AH = 9/2 = 4.5 cm.

Answer: The distance between the lines is 4.5 cm.