Let $d_1, d_2, d_3, d_4$ be pairwise skew straight lines. Assuming that $d_{12} ⊥ d_{34}$ and $d_{13} ⊥ d_{24}$, show that $d_{14} ⊥ d_{23}$, where $d_{ik}$ is the common perpendicular of the lines $d_i$ and $d_k$.
Can somebody help me, please? I don't know how to imagine these 4 skew lines in space.
I will first prove the vector identity $$(a\times b)\cdot(c\times d)=(a\cdot c)(b\cdot d)-(a\cdot d)(b\cdot c),\tag{1}$$ where $a,b,c,d\in\Bbb{R}^3$. First recall that $$x\times (y\times z)=(x\cdot z)y-(x\cdot y)z=(z\times y)\times x\tag{2}$$ and \begin{align}\det\begin{pmatrix}\vert&\vert&\vert\\x&y&z\\\vert&\vert&\vert\end{pmatrix}&=x\cdot(y\times z)=y\cdot(z\times x)=z\cdot (x\times y)\\&=(x\times y)\cdot z=(y\times z)\cdot x=(z\times x)\cdot y.\tag{3}\end{align}
To prove $(1)$, we note that $$(a\times b)\cdot(c\times d)=\big(b\times(c\times d)\big)\cdot a$$ by $(3)$. That is, $$(a\times b)\cdot(c\times d)=\big((b\cdot d)c-(b\cdot c)d\big)\cdot a=(a\cdot c)(b\cdot d)-(a\cdot d)(b\cdot c),$$ by $(2)$.
To prove $(2)$, we observe that it holds trivially when $y$ and $z$ are not linearly independent. Assume now that $y$ and $z$ are linearly independent. Then, $x\times(y\times z)$ is perpendicular to the normal vector $y\times z$ of the subspace of $\Bbb{R}^3$ spanned by $y$ and $z$. Therefore, $x\times(y\times z)$ lies in the span of $y$ and $z$. Suppose that $$x\times (y\times z)=\alpha(x) y+\beta(x) z$$ for some linear functionals $\alpha,\beta:\Bbb{R}^3\to\Bbb{R}$. By the Riesz representation theorem, $\alpha(x)=x\cdot u$ and $\beta(x)=x\cdot v$ for some $u,v\in\Bbb{R}^3$. If $x\perp y$, then we see that $x\times (y\times z)$ is in the span of $y$. Thus $\beta(x)=0$ when $x\cdot y=0$. This means $v=\mu y$ for some $\mu\in\Bbb R$. Similarly, $u=\lambda z$ for some $\lambda\in\Bbb R$. Since $x$ is perpendicular to $x\times(y\times z)$, we obtain $$0=x\cdot\big(x\times (y\times z)\big)=\alpha(x)(x\cdot y)+\beta(x)(x\cdot z)=\lambda(x\cdot z)(x\cdot y)+\mu(x\cdot y)(x\cdot z).$$ This shows that $\mu=-\lambda$. However, by $(3)$ we have $$z\cdot \big(y\times (y\times z)\big)=(y\times z)\cdot (z\times y)=-|y\times z|^2,$$ while $$z\cdot\big(y\times (y\times z)\big)=\lambda z\cdot\big((y\cdot z)y-(y\cdot y)z\big)=\lambda \big((y\cdot z)^2-|y|^2|z|^2\big).$$ Since $|y\times z|^2=|y|^2|z|^2-(y\cdot z)^2$, we conclude that $\lambda=1$. This proves $(2)$.
For $(3)$, we just expand the determinant and compare it with the expansion of, for example, $x\cdot (y\times z)$. Using the Einstein notation can make things even simpler.
Back to the problem, let $v_1$, $v_2$, $v_3$, and $v_4$ denote vectors parallel to $d_1$, $d_2$, $d_3$, and $d_4$, respectively. Therefore, $d_{ij}$ is parallel to $v_i\times v_j$. We need to show that if $$(v_1\times v_2)\cdot (v_3\times v_4)=0\tag{4}$$ and $$(v_1\times v_3)\cdot (v_2\times v_4)=0,\tag{5}$$ then $$(v_1\times v_4)\cdot (v_2\times v_3)=0.\tag{6}$$ By $(1)$, $(4)$ is equivalent to $$(v_1\cdot v_3)(v_2\cdot v_4)=(v_1\cdot v_4)(v_2\cdot v_3).\tag{7}$$ By $(1)$, $(5)$ is equivalent to $$(v_1\cdot v_2)(v_3\cdot v_4)=(v_1\cdot v_4)(v_2\cdot v_3).\tag{8}$$ From $(7)$ and $(8)$, we get $$(v_1\cdot v_2)(v_3\cdot v_4)=(v_1\cdot v_3)(v_2\cdot v_4).\tag{9}$$ By $(1)$, $(6)$ is equivalent to $(9)$.