# Angle between two planes in a cube

Find the cosine of the angle between the planes $$(A,B,C,D)$$ and $$(M,N,K)$$ in the cube $$ABCDA_1B_1C_1D_1$$ where $$M,N$$ and $$K$$ are the midpoints of $$BB_1,A_1B_1$$ and $$B_1C_1$$, respectively.

As we can see on the diagram, the intersection line of $$(A,B,C,D)$$ and $$(M,N,K)$$ isn't inside the cube and I don't see what characterizes it. Can you give me a hint?

• Think about the angle between the normals to the two planes. What is the relationship between that angle and the angle you need? Apr 20, 2022 at 17:18
• Alternatively, due to symmetry you can show that it's the angle between $BD$ and $MP$, where $P$ is the midpoint between $K$ and $N$. Apr 20, 2022 at 17:22
• @Andrei, thank you for the response. Can you clarify your answer a little more for me? I am not familiar with coordinate geometry.
– Hipo
Apr 20, 2022 at 17:25
• The angle you need is also the angle between diagonal $DB_1$ and edge $DD_1$. Apr 20, 2022 at 17:40
• @Intelligentipauca, thank you for the response. What is the explanation behind this?
– Hipo
Apr 20, 2022 at 17:43

This can be done quite easily using coordinate geometry. The plane (ABCD) has an equation $$z = 0$$ and the equation of plane $$MNK$$ is

$$[1, 1, 1] \cdot [r - r_0] = 0$$

where $$r_0 = \dfrac{1}{3} (M + N + k)$$

The angle between the normals to the two planes is

$$\theta = \cos^{-1} \dfrac{[0, 0, 1] \cdot [1,1,1]}{\sqrt{3}} = \cos^{-1}\left(\dfrac{1}{\sqrt{3}}\right)$$

And this is an acute angle.

The angle between two planes can be defined in two equivalent ways: the angle between normals (like in the other answer), or the angle between two lines that are perpendicular on the intersection of the two planes. This second procedure is a little longer.

In your figure, the intersection of the two planes is a line in the $$ABCD$$ plane, perpendicular to $$BD$$. Let's call $$T$$ the point where $$BD$$ meets the intersection between planes. Then, due to symmetry, $$TM$$ will intersect $$KN$$ at $$P$$, where $$P$$ is the midpoint between $$K$$ and $$N$$. Note that $$\triangle PMB_1$$ and $$\triangle TMB$$ are congruent (right angle triangles, opposite angles are the same, and $$BM=B_1M$$). Then the angle between planes is equal to the angle $$\angle B_1PM$$. If the side of the cube is $$a$$, then $$B_1M=\frac a2=B_1N=B_1K$$. Then in the isosceles right angle triangle $$B_1KN$$ the height $$B_1P=\frac a{2\sqrt 2}$$. From Pythagoras' theorem in $$\triangle B_1PM$$ you get $$PM=\frac {a\sqrt 3}{2\sqrt 2}$$ and then $$\cos\angle B_1PM=\frac{\frac a{2\sqrt 2}}{\frac {a\sqrt 3}{2\sqrt 2}}=\frac 1{\sqrt 3}$$

• Thanks! Can I ask you why the two planes intersect in a line perpendicular to BD$? – Hipo Apr 20, 2022 at 17:50 •$NK||AC$from the definition of$N$and$K$. Two lines in different (intersecting) planes can be parallel only if they are parallel to the intersection line. So the intersection line is parallel to$AC$, which in turn is perpendicular to$BD\$ (diagonals of a square) Apr 20, 2022 at 17:58