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# Find the angle formed by a line segment and a plane

Question:

In the figure shown, ABCD-A1B1C1D1 is a cube. Each side of the cube is a. Find the degree measure of the angle between the line segment DD1 and the plane ACB1.

Solution:

In the figure shown, AB1 is a diagonal of the square ABB1A1 in the plane ABB1A1. CB1 is a diagonal of the square BCC1B1 in the plane BCC1B1. AC is a diagonal of the square ABCD in the plane ABCD. Connect BD, BD intersects AC at point O, then connect B1O.

Because line segment DD1 parallel to line segment BB1, so the angle between DD1 and the plane ACB1 is equal to the angle between BB1 and the plane ACB1. Because ABCD is a square, AC and BD are two diagonals, so AC and BD bisect and perpendicular to each other at the point O. Point O is the midpoint of AC and AC lies on the plane ACB1. Point B1 is a point on the plane ACB1. So, line segment B1O lies on the plane ACB1. So, the angle between BB1 and the plane ACB1 is the angle between BB1 and B1O. Let the angle BB1O is the angle theta, the angle theta is the angle between DD1 and the plane ACB1. Because line segment BB1 is perpendicular to the plane ABCD and BD is a line on the plane ABCD, so B1B is perpendicular to BD. Because O is the midpoint of BD, so, B1B is perpendicular to BO. so, the triangle B1BO is a right triangle.

In right triangle B1BO,
tan theta = BO/B1B
Because ABCD is a square with side a,
so, the angle BCD = 90o
so triangle BCD is a right triangle.
In right triangle BCD,
BD2 = BC2 + CD2 = a2 + a2 = 2a2
BD = (square root of 2) a
Because O is the midpoint of BD, so
BO = BD/2 = [(square root of 2)/2] a
B1B = a
tan theta = BO/B1B = [(square root of 2)/2] a/a = (square root of 2)/2 = 0.7071
theta = arc tan 0.7071 = 35.26o

Therefore, the angle between the line segment DD1 and the plane ACB1 is 35.26 degrees. Please watch the video for more details.