Gọi $O=AC\cap BD$, trong (SBD) gọi $K=B^{\prime }{D}^{\prime }\cap SO$, trong (SAC) gọi $C^{\prime }=AK\cap SC\iff SC\cap \left(A{B}^{\prime }{D}^{\prime }\right)=C^{\prime }$

Ta có:

$\begin{array}{c}\{\begin{array}{c}BC\perp AB\\ BC\perp SA\end{array}\Rightarrow BC\perp \left(SAC\right)\Rightarrow BC\perp A{B}^{\prime }\\ \{\begin{array}{c}A{B}^{\prime }\perp BC\\ A{B}^{\prime }\perp SB\end{array}\Rightarrow A{B}^{\prime }\perp \left(SBC\right)\Rightarrow A{B}^{\prime }\perp SC\end{array}$

Tương tự ta chứng minh được $A{D}^{\prime }\perp SC\Rightarrow SC\perp \left(A{B}^{\prime }{C}^{\prime }{D}^{\prime }\right)\Rightarrow A{C}^{\prime }\perp SC$

Xét tam giác vuông  SAB ta có $\frac{S{B}^{\prime }}{SB}=\frac{S{A}^2}{S{B}^2}=\frac{S{A}^2}{S{A}^2+A{B}^2}=\frac{4a^2}{4a^2+a^2}=\frac{4}{5}$

Xét tam giác vuông SAD ta có $\frac{S{D}^{\prime }}{SD}=\frac{S{A}^2}{S{D}^2}=\frac{S{A}^2}{S{A}^2+A{D}^2}=\frac{4a^2}{4a^2+a^2}=\frac{4}{5}$

Ta có: $A{C}^2=A{B}^2+B{C}^2=2a^2$

Xét tam giác vuông SAC ta có $\frac{S{C}^{\prime }}{SC}=\frac{S{A}^2}{S{C}^2}=\frac{S{A}^2}{S{A}^2+A{C}^2}=\frac{4a^2}{4a^2+2a^2}=\frac{2}{3}$

Ta có

$\begin{array}{c}\frac{V_{S.A{B}^{\prime }{C}^{\prime }}}{V_{S.ABC}}=\frac{S{B}^{\prime }}{SB}.\frac{S{C}^{\prime }}{SC}=\frac{4}{5}.\frac{2}{3}=\frac{8}{15}\Rightarrow V_{S.A{B}^{\prime }{C}^{\prime }}=\frac{8}{15}{V}_{S.ABC}=\frac{4}{15}{V}_{S.ABCD}\\ \frac{V_{S.A{C}^{\prime }{D}^{\prime }}}{V_{S.ACD}}=\frac{S{C}^{\prime }}{SC}.\frac{S{D}^{\prime }}{SD}=\frac{2}{3}.\frac{4}{5}=\frac{8}{15}\Rightarrow V_{S.A{C}^{\prime }{D}^{\prime }}=\frac{8}{15}{V}_{S.ACD}=\frac{4}{15}{V}_{S.ABCD}\\ \Rightarrow V_{S.A{B}^{\prime }{C}^{\prime }{D}^{\prime }}=\frac{8}{15}{V}_{S.ABCD}\end{array}$

Mà $V_{S.ABCD}=\frac{1}{3}SA.S_{ABCD}=\frac{1}{3}2a.a^2=\frac{2}{3}{a}^3\Rightarrow V_{S.A{B}^{\prime }{C}^{\prime }{D}^{\prime }}=\frac{8}{15}.\frac{1}{3}{a}^3=\frac{16a^3}{45}$

11,3 gio

Có AC // A′C′AC // A′C′ ⇒A′C′ //(AB′C)⇒A′C′ //AB′C
⇒d(A′C′;CB′)=d(A′C′;(ACB′))⇒dA′C′;CB′=dA′C′;ACB′ =d(C′;(ACB′))=dC′;ACB′ =d(B;(ACB′))=dB;ACB′
Vậy khoảng cách giữa hai đường thẳng CB′CB′ và A′C′A′C′ là 22√112211 .

a^3/3căn 2

a/2

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