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# How to find the angle formed by a line outside a plane and a plane?

Question:

ABCD-A1B1C1D1 is a cube. The side of the cube is 2. E is the midpoint of A1B1. Find the cosine of the angle formed by the line segment EC and the plane ABCD.

Solution:

Because ABCD-A1B1C1D1 is a cube, so, A1B1 = D1C1 = 2. Because E is the midpoint of A1B1, so, A1E = B1E = (1/2)A1B1 = 2/2 = 1

In the figure above, point F is the projection of the point E on the plane ABCD. So, AF = BF = (1/2) AB = 2/2 = 1.

Connect FC, FC is the projection of the line segment EC on the plane ABCD. So, the angle formed by EC and the plane ABCD is the angle formed by EC and FC. Let the angle ECF be the angle theta.

Because EF parallel to A1A, and A1A is perpendicular to the plane ABCD, so, EF is perpendicular to plane ABCD. Because FC lies on plane ABCD, so, EF is perpendicular to FC.

the angle ECF is the angle theta
In right triangle EFC,
cos theta = FC/EC

Because ABCD is a square, F is the midpoint of AB and the angle FBC is 90o, so, triangle FBC is a right triangle.

In right triangle FBC,
FC2 = FB2 + BC2
= 12 + 22 = 5
FC = square root of 5.

Because EF is perpendicular to FC, so, the angle EFC is 90o. So, the triangle EFC is a right triangle.

In right triangle EFC,
EC2 = EF2 + FC2

Because E is the midpoint of A1B1, so, A1E = half of A1B1 = 2/2 = 1. Since AF = AB/2 = 2/2 = 1. Because A1E = AF and and A1E is parallel to AF and angle A1AF = 90o, so A1EFA is a rectangle. So, EF = A1A = 2.

In right triangle EFC,
EC2 = EF2 + FC2
= 22 + (square root of 5)2 = 4 + 5 = 9
EC = 3
cos theta = FC/EC = (square root of 5)/3 = 0.7454
angle theta = arc cos 0.7454 = 41.8o

Therefore, the angle between segment EC and the plane ABCD is 41.8o. Please watch the video for more details.