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# How to find the angle between two lines in a cube?

Question:

In the figure shown, ABCD-A1B1C1D1 is a cube with side length a. M is the midpoint of DD1. Find the angle between CM and BA1.

Solution:

Connect CD1. Because this is a cube, CD1 parallel to BA1. So, the angle between CM and BA1 is the angle between CM and CD1. CM and CD1 lie on the plane CC1D1D.

Because this is a cube, CDD1C1 is a square. The angle CDM is right angle.

In right triangle CDM,
CD = a, DM = (a/2), because M is the midpoint of DD1
CM2 = CD2 + DM2
= a2 + (a/2)2
= a2 + a2/4
= (5/4)a2
so, CM = (square root of 5) a/2

Because this is a cube, CDD1C1 is a square. The angle CC1D1 is right angle.

In right triangle CC1D1,
CC1 = C1D1 = a
CD12 = CC12 + C1D12
= a2 + a2
= 2a2
so, CD1 = (square root of 2) a
MD1 = (a/2)

Now, we know the three sides of the triangle CMD1. We can use the cosine law to find the angle MCD1 which is the angle between the line segments CM and CD1.

Let the angle MCD1 be the angle theta
In triangle MCD1,
cos theta = (CM2 + CD12 - MD12 ÷ [2 × CM × CD1]
= [(5/4)a2 + 2a2 - (a/2)2] ÷ [2 × (square root of 5) a/2 × (square root of 2) a]
= [a2(5/4 + 2 - 1/4)] ÷ [(square root of 10) a2]
= (5/4 + 8/4 - 1/4) ÷ (square root of 10)
= 3 ÷ (square root of 10)
= 3 × (square root of 10) ÷ 10
= 0.9487
angle theta = arc cos 0.9487 = 18.43 degrees

Therefore, the angle between segment CM and segment CD1 is 18.43 degrees. Because CD1 parallel to BA1, so the angle between segment CM and segment BA1 is equal to the angle between segment CM and segment CD1. Please watch the video for more details.