Shifting of Line Segment

Question

                      

$L_1$ and $L_2$ are skew lines in $3$-dimensional space. $A_1 C \mathbin{\bot} L_1,L_2$. If line segment $AB$ is shifted parallelly, such that $A$ moves along $L_1$ and reaches $A_1$ while $B$ reaches $B_1$, then prove that $B_1 C \mathbin\bot A_1 C$.

It seems really intuitive but I wasn't able to get it. Can anyone prove it using only simple geometry (lines, congruency or similarity) without vectors?

Answer 1

Here's a vector solution, but coordinate-free, and really simple . . .

The advantages? It works in $n$ dimensions, for all $n \ge 2$, and requires minimal visualization.

From $A_1 C \mathbin{\bot} L_1,L_2$, we get $$(\mathbf{A_1}-\mathbf{C})\cdot(\mathbf{A}-\mathbf{A_1})=0\tag{eq1}$$ $$(\mathbf{A_1}-\mathbf{C})\cdot(\mathbf{B}-\mathbf{C})=0\tag{eq2}$$ Since line segment $AB$ moves in parallel, ending up as line segment $A_1B_1$, it follows that $$\mathbf{B}-\mathbf{A}=\mathbf{B_1}-\mathbf{A_1}\tag{eq3}$$ The goal is to prove $B_1 C \mathbin\bot A_1 C$, or equivalently $$(\mathbf{B_1}-\mathbf{C})\cdot(\mathbf{A_1}-\mathbf{C})=0\tag{g}$$ Now here's the intuition . . .

The $3$ equations $(\text{eq}1),\,(\text{eq}2),\,(\text{eq}3)$, represent all the information in the problem, effectively replacing the diagram. Hence, either those $3$ equations identically imply equation $(\text{g})$, or else the implication doesn't hold, in which case, the claim of the problem is false.

With that intuition, elementary algebra (of vectors) should suffice . . .

Subtracting $(\text{eq}1)$ from $(\text{eq}2)$, we get \begin{align*} & (\mathbf{A_1}-\mathbf{C}) \cdot \bigl((\mathbf{B}-\mathbf{C}) - (\mathbf{A}-\mathbf{A_1}) \bigr) =0 \\[4pt] \implies\;& (\mathbf{A_1}-\mathbf{C}) \cdot \bigl((\mathbf{B}-\mathbf{A}) - (\mathbf{C}-\mathbf{A_1}) \bigr) =0 \\[4pt] \implies\;& (\mathbf{A_1}-\mathbf{C}) \cdot \bigl((\mathbf{B_1}-\mathbf{A_1}) - (\mathbf{C}-\mathbf{A_1}) \bigr) =0 &&\text{[by $(\text{eq}3)$]} \\[4pt] \implies\;& (\mathbf{A_1}-\mathbf{C}) \cdot (\mathbf{B_1}-\mathbf{C}) =0 \\[4pt] \end{align*} which proves equation $(\text{g})$, as required.

Answer 2

Line $BC$ lies on plane $\alpha$, perpendicular to line $A_1C$ at $C$, and line $AA_1$ is parallel to $\alpha$.

Quadrilateral $ABB_1A_1$ is a parallelogram by construction, thus $BB_1$ is parallel to $AA_1$ and lies then on $\alpha$. It follows that point $B_1$ belongs to $\alpha$, as well as line $CB_1$. By the definition of perpendicular plane, we have then $A_1C\perp B_1C$.