Rectangle ABCD and right triangle DCK lie in different planes. Vertex K is projected to point B

Rectangle ABCD and right triangle DCK lie in different planes. Vertex K is projected to point B. BK = 4cm, AB = 4√2, AD = 4. Calculate the distance between lines AB and KC.

Since ABCD is a rectangle, BC = AD = 4 cm.

Point K is projected to point B, then BK is perpendicular to square ABCD, and triangle BSC is rectangular and isosceles, BC = BK = 4 cm.

Let us define the length of the hypotenuse СK.

CK ^ 2 = BK ^ 2 + BC ^ 2 = 16 + 16 = 32.

СK = 4 * √2 cm.

The plane BCK is perpendicular to the plane of the square, then the height BH of the triangle BCK is the desired distance.

KH = CH = KC / 2 = 4 * √2 / 2 = 2 * √2 cm.

Then BH ^ 2 = BK ^ 2 – KH ^ 2 = 16 – 8 = 8.

BH = 2 * √2 cm.

Answer: The distance between the lines is 2 * √2 cm.