Solid Geometry Advanced – Problem 1: Prove that line is parallel to plane

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 $AB=BC=BB_1=2$. Point $M$ is the midpoint of $AB$, and point $N$ is the midpoint of segment $A_1C$.

Prove that line $MN$ is parallel to plane $BCC_1B_1$.

(1) Place coordinates: let the base plane be $z=0$. Set $$B(0,0,0),\ A(2,0,0),\ C(0,2,0),\ B_1(0,0,2),\ A_1(2,0,2),\ C_1(0,2,2).$$ Then $$M\left(1,0,0\right),\quad N=\frac{A_1+C}{2}=\left(1,1,1\right).$$ (2) Compute the direction vector $$\overrightarrow{MN}=N-M=(0,1,1).$$ (3) In plane $BCC_1B_1$, two non-parallel direction vectors are $$\overrightarrow{BC}=(0,2,0),\quad \overrightarrow{BB_1}=(0,0,2).$$ We have $$\overrightarrow{MN}=\tfrac12\overrightarrow{BC}+\tfrac12\overrightarrow{BB_1},$$ so $\overrightarrow{MN}$ is a linear combination of two directions in plane $BCC_1B_1$. Therefore the line $MN$ is parallel to plane $BCC_1B_1$. \]

Since $\overrightarrow{MN}$ lies in the direction span of $\overrightarrow{BC}$ and $\overrightarrow{BB_1}$, we conclude $MN\parallel$ plane $BCC_1B_1$.