Find the projection of the inclined AM on the plane α, if the distance from point M to the plane

| February 16, 2021 | education

Find the projection of the inclined AM on the plane α, if the distance from point M to the plane α is 15 cm, and the length of the inclined one is 25 cm.

1. The distance from point M to plane α is the perpendicular to the plane. Let the base of the perpendicular be T. K. Then, by condition, the perpendicular MK = 15 cm, the oblique projection is the projection AK.

2. MK is perpendicular to the plane α, which means that MK is perpendicular to any straight line in the plane α, that is, MK is perpendicular to AK (since AK belongs to the plane α). This means that the inclined AM, the perpendicular MK and the projection of the AK form a right-angled triangle, the angle of the AKM is a straight line.

3. In the triangle AKM: angle AKM – straight line, MK = 15 cm (leg), AM = 25 cm (hypotenuse). By the Pythagorean theorem, we find the leg AK:

AM ^ 2 = AK ^ 2 + MK ^ 2, hence

AK ^ 2 = AM ^ 2 – MK ^ 2 = 25 ^ 2 – 15 ^ 2 = 625 – 225 = 400,

AK = √400 = √ (100 * 4) = √100 * √4 = 10 * 2 = 20 (cm).

Answer: the projection of the oblique AM is the straight line AK = 20 cm.



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