Giải Câu 5 Bài 2: Hai đường thẳng vuông góc.
Đặt $SA=SB=SC=a$, \(\widehat{ABC}= \widehat{BSC}=\widehat{CSA}=\alpha \)
- \(\overrightarrow{SA}.\overrightarrow{BC}=\overrightarrow{SA}.(\overrightarrow{SC}-\overrightarrow{SB})\)
\(=\overrightarrow{SA}.\overrightarrow{SC}-\overrightarrow{SA}.\overrightarrow{SB}\)
\(= SA.SC.\cos\widehat{ASC} - SA.SB.\cos\widehat{ASB} = a.a.cos\alpha -a.a.cos\alpha =0\).
Vậy \(SA ⊥ BC\).
- \(\overrightarrow{SB}.\overrightarrow{AC}=\overrightarrow{SB}.(\overrightarrow{SC}-\overrightarrow{SA})\)
\(=\overrightarrow{SB}.\overrightarrow{SC}-\overrightarrow{SB}.\overrightarrow{SA}\)
\(= SB.SC.\cos\widehat{BSC} - SB.SA.\cos\widehat{ASB} = a.a.cos\alpha -a.a.cos\alpha =0\).
Vậy \(SB ⊥ AC\).
- \(\overrightarrow{SC}.\overrightarrow{AB}=\overrightarrow{SC}.(\overrightarrow{SB}-\overrightarrow{SA})\)
\(=\overrightarrow{SC}.\overrightarrow{SB}-\overrightarrow{SC}.\overrightarrow{SA}\)
\(= SC.SB.\cos\widehat{BSC} - SC.SA.\cos\widehat{ASC} = a.a.cos\alpha -a.a.cos\alpha =0\).
Vậy \(SC ⊥ AB\).