You are given a rectangular parallelepiped ABCDA1B1C1D1. AB = 1, AD = 1, AA1 = 2. Find the distance

| June 9, 2021 | education

You are given a rectangular parallelepiped ABCDA1B1C1D1. AB = 1, AD = 1, AA1 = 2. Find the distance from point D1 to line AC.

Since the parallelepiped is rectangular, all of its faces are rectangles.

By condition, AB = 1 cm, AD = 1 cm, then a square lies at the base of the parallelepiped.

By the Pythagorean theorem, we determine the length of the diagonal ВD.

BD ^ 2 = AB ^ 2 + AD ^ 2 = 1 + 1.

ВD = √2 cm.

The diagonals of the square intersect at right angles and are divided, at the point of intersection, in half, then OD = BD / 2 = √2 / 2 cm.

OD is the projection of the inclined OD1 on the ABCD plane, and since OD is perpendicular to the AC, then OD1 is perpendicular to the AC.

In a right-angled triangle ODD1, according to the Pythagorean theorem, OD1 ^ 2 = OD ^ 2 + DD1 ^ 2 = 2/4 + 4 = 18/4 = 9/2.

OD1 = 3 / √2 cm.

Answer: From point D1 to straight AC 3 / √2 cm.



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