In a rectangular parallelepiped ABCDA1B1C1D1: AB = 2cm, AD = 3cm, AC1 = 7cm.

| August 12, 2021 | education

In a rectangular parallelepiped ABCDA1B1C1D1: AB = 2cm, AD = 3cm, AC1 = 7cm. Find the distance between lines AB and B1C1.

Straight lines AB and B1C1 are intersecting lines, then the distance between them will be the segment BB1, which is their common perpendicular.

The first way.

The square of the diagonal of a parallelepiped is equal to the sum of the squares of the lengths of its three dimensions.

AC1 ^ 2 = AB ^ 2 + AD ^ 2 + BB1 ^ 2.

Then: BB1 ^ 2 = AC1 ^ 2 – AB ^ 2 – AD ^ 2 = 49 – 4 – 9 = 36.

BB1 = 6 cm.

Second way.

Since the diagonal of the parallelepiped is equal, then B1D = AC1 = 7 cm.

We construct a diagonal BD, then BD ^ 2 = AB ^ 2 + AD ^ 2 = 4 + 9 = 13.

From the right-angled triangle BB1D we determine the length of the leg BB1.

BB1 ^ 2 = DB1 ^ 2 – BD ^ 2 = 49 – 13 = 36.

BB1 = 6 cm.

Answer: Between straight lines AB and B1C1 6 cm.



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