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Find the angle betweeen a plane and a line segment

Question:

EFGH-ABCD is a cube. GE is a diagonal in plane EFGH. Connect ED and FC. What is the degree measure of the angle between the plane EFCD and the line segment GE?

Solution:

Connect GB which intersects FC at point K, then connect EK.

Because EF is perpendicular to FG and EF is perpendicular to FB, so EF is perpendicular to the plane FGCB.

Because GB is on the plane FGCB, so EF is perpendicular to GB

Because K in the intersection point of two diagonals FC and BG in square BCGF, so EF is perpendicular to GK.

Because BCGF is a square, so its two diagonals are perpendicular and bisect each other, so GB is perpendicular to FC.

Because K in the midpoint of GB, so GK is perpendicular to FC.

Now we have, GK is perpendicular to EF and GK is perpendicular to FC.

Because EF and FC are two intersected lines and two intersected lines make a plane, so GK is perpendicular to the plane EFCD.

So the angle GEK is the angle between the line segment EG and the plane EFCD.

In right triangle GEK, from sine definition,

sin(angle GEK) = (opposite side of the angle GEK)/(hypotenuse of the right triangle GEK)

= GK/GE = [(square root of 2)/2]a ÷ (square root of 2)a = 1/2

So angle GEK = 30^o

Therefore, the angle between the line segment EG and the plane EFCD is 30^o.