Câu 5. Cho hình chóp tam giác \(S. ABC\) có \(SA = SB = SC\) và có \(\widehat{ASB}= \widehat{BSC}=\widehat{CSA}. \) Chứng minh rằng \(SA ⊥ BC, SB ⊥ AC, SC ⊥ AB\).

Giải

(h. 3. 19)

\(\overrightarrow{SA}. \overrightarrow{BC}=\overrightarrow{SA}. (\overrightarrow{SC}-\overrightarrow{SB})\)

\(=\overrightarrow{SA}. \overrightarrow{SC}-\overrightarrow{SA}. \overrightarrow{SB}\)

\(= SA. SC. \cos\widehat{ASC} - SA. SB. \cos\widehat{ASB} = 0\).

Vậy \(SA ⊥ BC\).

\(\overrightarrow{SB}. \overrightarrow{AC}=\overrightarrow{SB}. (\overrightarrow{SC}-\overrightarrow{SA})\)

\(=\overrightarrow{SB}. \overrightarrow{SC}-\overrightarrow{SB}. \overrightarrow{SA}\)

\(= SB. SC. \cos\widehat{BSC} - SB. SA. \cos\widehat{ASB} = 0\).

Vậy \(SB ⊥ AC\).

\(\overrightarrow{SC}. \overrightarrow{AB}=\overrightarrow{SC}. (\overrightarrow{SB}-\overrightarrow{SA})\)

\(=\overrightarrow{SC}. \overrightarrow{SB}-\overrightarrow{SC}. \overrightarrow{SA}\)

\(= SC. SB. \cos\widehat{BSC} - SC. SA. \cos\widehat{ASC} = 0\).

Vậy \(SC ⊥ AB\).