Angle between two planes in a cube

Question

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?

Answer 1

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.

Answer 2

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 $$\cos\angle B_1PM=\frac{\frac a{2\sqrt 2}}{\frac {a\sqrt 3}{2\sqrt 2}}=\frac 1{\sqrt 3}$$