In a regular triangular prism ABCA1B1C1, all edges of which are equal to 3, find the distance between straight lines AA1 and BC1.
| April 1, 2021 | education
The BC1 segment belongs to the BB1C1C plane, which is the lateral face of the prism. Then the distance between straight lines АА1 and ВС1 is the shortest distance between the edge АА1 and the face В1С1С.
At the base of the prism, draw a perpendicular from the vertex A to the edge BC, which will be the desired distance.
The ABC triangle is equilateral with a side of 3 cm, then its height AH = BC * √3 / 2 = 3 * √3 / 2 cm.
Answer: The distance between lines AA1 and BC1 is 3 * √3 / 2 cm.
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