Cho hình lập phương ABCD.A'B'C'D' . Góc giữa hai mặt phẳng

Đáp án D

Gọi $H=B^{\text{'}}C\cap B{C}^{\text{'}},K=A{D}^{\text{'}}\cap A^{\text{'}}D$ .

Khi đó $\left(AB{C}^{\text{'}}{D}^{\text{'}}\right)\cap \left(A^{\text{'}}{B}^{\text{'}}CD\right)=HK$ .

Có$\left\{\begin{array}{l}{D}^{\text{'}}{C}^{\text{'}}\perp B^{\text{'}}{C}^{\text{'}}\\ D^{\text{'}}{C}^{\text{'}}\perp C{C}^{\text{'}}\end{array}\right.\Rightarrow D^{\text{'}}{C}^{\text{'}}\perp \left(BC{C}^{\text{'}}{B}^{\text{'}}\right)\Rightarrow \left\{\begin{array}{l}{D}^{\text{'}}{C}^{\text{'}}\perp B^{\text{'}}C\\ D^{\text{'}}{C}^{\text{'}}\perp B{C}^{\text{'}}\end{array}\right.$ .

Mà  song song nhau nên $\left\{\begin{array}{l}HK\perp B^{\text{'}}C\\ HK\perp B{C}^{\text{'}}\end{array}\right.$

Ta có:$\left\{\begin{array}{l}\left(AB{C}^{\text{'}}{D}^{\text{'}}\right)\cap \left(A^{\text{'}}{B}^{\text{'}}CD\right)=HK\\ HK\perp B{C}^{\text{'}},B{C}^{\text{'}}\subset \left(AB{C}^{\text{'}}{D}^{\text{'}}\right)\\ HK\perp B^{\text{'}}C,B^{\text{'}}C\subset \left(A^{\text{'}}{B}^{\text{'}}CD\right)\end{array}\right.\iff \left(\left(AB{C}^{\text{'}}{D}^{\text{'}}\right),\left(A^{\text{'}}{B}^{\text{'}}CD\right)\right)=\left(B{C}^{\text{'}},B^{\text{'}}C\right)=90°$