Solid Geometry Advanced – Problem 5: Find , where is the angle between planes and

In the right triangular prism $ABC-A_1B_1C_1$, the lateral edges are perpendicular to the base plane. Assume $\angle ABC=90^\circ$ and $AB=BC=BB_1=2$. Let $M$ be the midpoint of $AB$.

Find $\cos\theta$, where $\theta$ is the angle between planes $MB_1C$ and $B_1CA_1$.

(1) Use the same coordinates as in the proof problem: $$B(0,0,0),\ A(2,0,0),\ C(0,2,0),\ B_1(0,0,2),\ A_1(2,0,2).$$ Then $M(1,0,0)$.

(2) A normal vector to plane $MB_1C$ is $$\mathbf n_1=(\overrightarrow{B_1M})\times(\overrightarrow{CM}).$$ Compute $$\overrightarrow{B_1M}=(-1,0,2),\quad \overrightarrow{CM}=(-1,2,0),$$ so $$\mathbf n_1=(-1,0,2)\times(-1,2,0)=(-4,-2,-2)\sim(2,1,1).$$ (3) A normal vector to plane $B_1CA_1$ is $$\mathbf n_2=(\overrightarrow{CB_1})\times(\overrightarrow{A_1B_1}).$$ Compute $$\overrightarrow{CB_1}=(0,2,-2),\quad \overrightarrow{A_1B_1}=(2,0,0),$$ so $$\mathbf n_2=(0,2,-2)\times(2,0,0)=(0,-4,-4)\sim(0,1,1).$$ (4) Therefore $$\cos\theta=\frac{|\mathbf n_1\cdot\mathbf n_2|}{|\mathbf n_1||\mathbf n_2|}=\frac{|(2,1,1)\cdot(0,1,1)|}{\sqrt6\cdot\sqrt2}=\frac{2}{\sqrt{12}}=\frac{1}{\sqrt3}.$$

The angle between the planes satisfies $\cos\theta=\dfrac{1}{\sqrt3}$.