[Pythagorean Theorem in Space]

Hey

On the diagonals $D_1A, A_1B, B_1C, C_1D$ of the faces of the cube $ABCD A_1B_1C_1D_1$, points $M, N, P, Q$ are taken respectively, such that

$$
D_1M : D_1A = B N : B A_1 = B_1 P : B_1 C = D Q : D C_1 = \mu,
$$

and the lines $MN$ and $PQ$ are mutually perpendicular. Find $\mu$.
Let $A B=x, A D=y, A A 1=z, \overrightarrow{A B}=\vec{x}, \overrightarrow{A D}=\vec{y}, \overrightarrow{A A_{1}}=\vec{z}, x=y=z=a$. Then

$$
\begin{gathered}
\overrightarrow{M N}=\overrightarrow{M D_{1}}+\overrightarrow{D_{1} A_{1}}+\overrightarrow{A_{1} N}=\mu \overrightarrow{A D_{1}}+\overrightarrow{D_{1} A_{1}}+(1-\mu) \overrightarrow{A_{1} B}= \\
=\mu(\vec{y}+\vec{z})-\vec{y}+(1-\mu)(\vec{x}-\vec{z})=(1-\mu) \vec{x}+(\mu-1) \vec{y}+(2 \mu-1) \vec{z} \\
\overrightarrow{P Q}=\overrightarrow{P B_{1}}+\overrightarrow{B_{1} C_{1}}+\overrightarrow{C_{1} Q}=\mu \overrightarrow{C B_{1}}+\overrightarrow{B_{1} C_{1}}+(1-\mu) \overrightarrow{C_{1} D}= \\
=\mu(\overrightarrow{-y}+\vec{z})+\vec{y}+(1-\mu)\left(-\overrightarrow{x^{\prime}}-\vec{z}\right)=(\mu-1) \vec{x}+(1-\mu) \vec{y}+(2 \mu-1) \vec{z} .
\end{gathered}
$$

Since $M N \perp P Q$, we have:

$$
\begin{gathered}
\overrightarrow{M N} \cdot \overrightarrow{P Q}=((1-\mu) \vec{x}+(\mu-1) \vec{y}+(2 \mu-1) \vec{z})((\mu-1) \vec{x}+(1-\mu) \vec{y}+(2 \mu-1) \vec{z})= \\
=-(\mu-1) 2 a 2-(\mu-1) 2 a 2+(2 \mu-1) 2 a 2= \\
=(-(\mu-1) 2-(\mu-1) 2+(2 \mu-1) 2) a 2= \\
=(-2 \mu 2+4 \mu-2+1-4 \mu+4 \mu 2) a 2=(2 \mu 2-1) a 2=0
\end{gathered}
$$

from which we find that $|\mu|=\frac{1}{\sqrt{2}}$, and since $\mu>0$, then $\mu=\frac{1}{\sqrt{2}}$.