a) Theo đề bài ta có :
- Đáy: \(ABCD\)
- Mặt trên: \(A'B'C'D'\)
Ta chọn hệ trục tọa độ như sau :
\(A\left(0;0;0\right);B\left(a;0;0\right);C\left(a;a;0\right);D\left(0;a;0\right);A'\left(0;0;a\right);C'\left(a;a;a\right)\)
\(\overrightarrow{A'B}=\left(a;0;-a\right)\)
\(\overrightarrow{A'D}=\left(0;a;-a\right)\)
\(\overrightarrow{n\left(A'BD\right)}=\left[\overrightarrow{A'B}.\overrightarrow{A'D}\right]=\left(a^2;a^2;a^2\right)=\left(1;1;1\right)\)
\(\overrightarrow{n\left(ABCD\right)}=\left(0;0;1\right)\)
\(cos\left(\widehat{\left(A'BD\right);\left(ABCD\right)}\right)=\dfrac{1.0+1.0+1.1}{\sqrt{1^2+1^2+1^2}.\sqrt{1^2}}=\dfrac{1}{\sqrt{3}}=\dfrac{\sqrt{3}}{3}\)
b) \(\overrightarrow{C'B}=\left(0;-a;-a\right)\)
\(\overrightarrow{C'D}=\left(-a;0;-a\right)\)
\(\overrightarrow{n\left(C'BD\right)}=\left[\overrightarrow{C'B}.\overrightarrow{C'D}\right]=\left(a^2;a^2;-a^2\right)=\left(1;1;-1\right)\)
\(\left[A';BD;C'\right]=\left(\widehat{\left(A'BD\right);\left(C'BD\right)}\right)=\theta\)
\(cos\theta=\dfrac{1.1+1.1+1.\left(-1\right)}{\sqrt{1^2+1^2+1^2}.\sqrt{1^2+1^2+\left(-1\right)^2}}=\dfrac{1}{3}\)
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