12.7. The base of the pyramid $V A B C$ is the isosceles triangle $A B C$, where $A B=A C=6 \sqrt{2} \mathrm{~cm}$ and $B C=4 \sqrt{6} \mathrm{~cm}$. The lateral edges of the pyramid are $\sqrt{51} \mathrm{~cm}$. Determine the distance between the lines $A B$ and $V C$.
Solution. Let $O$ be the projection of vertex $V$ onto the plane $(A B C)$. Since the lateral edges are congruent, $O$ is the center of the circumcircle of triangle $A B C$.

![](https://cdn.mathpix.com/cropped/2024_06_06_0cfd4fdde813dd8be1cdg-2.jpg?height=916&width=1282&top_left_y=214&top_left_x=411)

Let $K \in B C$ such that $A K \perp B C$. Then $A K=4 \sqrt{3} \mathrm{~cm}$.

Let $\alpha=m(\angle A B C)$. Then $\sin \alpha=\frac{2 \sqrt{3}}{3 \sqrt{2}}, \cos \alpha=\frac{\sqrt{6}}{3 \sqrt{2}}$.

By the Law of Sines, $\frac{A C}{\sin \alpha}=2 O C$, which implies $O C=3 \sqrt{3} \mathrm{~cm}$.

We construct $B B_{1}\left\|V C, A A_{1}\right\| V C$ and $A A_{1}=B B_{1}=V C$. Then $V C \|\left(A B B_{1}\right)$.

The distance $d$ between $A B$ and $V C$ is the distance between $V C$ and $A B B_{1}$.

We construct the parallelepiped $A B C D A_{1} B_{1} V D_{1}$. Then the distance between $V C$ and $A B B_{1}$ is the height of the parallelepiped $A B C D A_{1} B_{1} V D_{1}$ drawn from vertex $V$ to the plane $\left(A B B_{1}\right)$.

$$
\mathcal{V}_{A B C D A_{1} B_{1} V D_{1}}=\mathcal{A}_{A B C D} \cdot V O, \mathcal{V}_{A B C D A_{1} B_{1} V D_{1}}=\mathcal{A}_{B B_{1} A_{1} A} \cdot d
$$

implies

$$
\begin{gathered}
d=\frac{\mathcal{A}_{A B C D} \cdot V O}{\mathcal{A}_{B B_{1} A_{1} A}} \\
\mathcal{A}_{A B C D}=A B \cdot B C \cdot \sin \alpha=48 \sqrt{2}\left(\mathrm{~cm}^{2}\right)
\end{gathered}
$$

From triangle $V O C$, we calculate $V O=2 \sqrt{6} \mathrm{~cm}$.

To calculate $\mathcal{A}_{B B_{1} A_{1} A}$, consider $O_{1} \in(A B C), B_{1} O_{1} \perp(A B C)$. Then $B_{1} V O O_{1}$ is a rectangle, and $O_{1} O C B$ is a parallelogram, and $O_{1} B=O C=3 \sqrt{3} \mathrm{~cm}$.

Consider point $P \in(A B)$ such that $O_{1} P \perp A B$. Then $B_{1} P \perp A B$.

Then from $\triangle O_{1} P B, O_{1} P=O_{1} B \cdot \sin \beta$, where $\beta=m\left(\angle O_{1} B P\right)=\left(180^{\circ}-\alpha-m(\angle O C K)\right.$. In $\triangle O K C$, we calculate $O K=A K-O A=\sqrt{3}(\mathrm{~cm}), \sin (\angle O C K)=\frac{1}{3}$ and $\cos (\angle O C K)=\frac{2 \sqrt{2}}{3}$.

Then $\sin \beta=\sin (\alpha+m(\angle O C K))=\frac{2 \sqrt{3}}{3 \sqrt{2}} \cdot \frac{2 \sqrt{2}}{3}+\frac{\sqrt{6}}{3 \sqrt{2}} \cdot \frac{1}{3}=\frac{5 \sqrt{3}}{9}$.

We obtain $O_{1} P=O_{1} B \cdot \sin \beta=5$. In the right triangle $B_{1} O_{1} P$, we calculate

$$
B_{1} P=\sqrt{B_{1} O_{1}^{2}+O_{1} P^{2}}=7(\mathrm{~cm})
$$

Then $\mathcal{A}_{B B_{1} A_{1} A}=A B \cdot B_{1} P=42 \sqrt{2}\left(\mathrm{~cm}^{2}\right)$,
and

$$
d=\frac{48 \sqrt{2} \cdot 2 \sqrt{6}}{42 \sqrt{2}}=\frac{16 \sqrt{6}}{7}(\mathrm{~cm})
$$