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 = 90^o

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 = 35.26^o

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