Cho hình chóp S.ABC có SA = SB = SC và góc ASB = góc BSC = góc CSA. Chứng minh rằng vecto SA. vecto BC = vecto SB . vecto AC = vecto SC. vecto AB = vecto 0

Ta có: \(\overrightarrow {SA} .\overrightarrow {BC} = \overrightarrow {SA} \left( {\overrightarrow {SC} - \overrightarrow {SB} } \right)\)

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

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

                       = 0.

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

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

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

                       = 0.

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

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

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

                       = 0.

Vậy \(\overrightarrow {SA} .\overrightarrow {BC} = \overrightarrow {SB} .\overrightarrow {AC} = \overrightarrow {SC} .\overrightarrow {AB} = 0\).