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 90^o, 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 90^o. 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 = 90^o, 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 = 41.8^o

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