Question

[Nguyễn Kim Trí]

Cho hình tứ diện ABCD có AB=c, CD=c', AC=b, BD=b', BC=a, AD=a'. Tính góc giữa các vecto  BC và DA

Answer 1

[๖Fly༉Donutღღ]

Tao có: \(\overrightarrow{BC}.\overrightarrow{AD}=\overrightarrow{BC}\left(\overrightarrow{DC}+\overrightarrow{CA}\right)=\overrightarrow{CB}.\overrightarrow{CD}-\overrightarrow{CB}.\overrightarrow{CA}\)

\(=\frac{1}{2}\left(CB^2+CD^2-BD^2\right)-\frac{1}{2}\left(CB^2+CA^2-AB^2\right)\)

\(=\frac{1}{2}\left(AB^2+CD^2-BD^2-CA^2\right)\)

\(\Rightarrow\cos\left(\overrightarrow{BC},\overrightarrow{DA}\right)=\frac{1}{2}.\frac{c^2+c'^2-b^2-b'^2}{2aa'}\)