Câu hỏi 31

Ta có

\(\begin{array}{l}\overrightarrow {SB} .\overrightarrow {AC}  = \left( {\overrightarrow {SC}  + \overrightarrow {CB} } \right).\overrightarrow {AC} \\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \overrightarrow {SC} .\overrightarrow {AC}  + \overrightarrow {CB} .\overrightarrow {AC} \end{array}\)

+) \(\overrightarrow {SC} .\overrightarrow {AC}  = SC.AC.cos\angle \left( {\overrightarrow {SC} ;\overrightarrow {AC} } \right)\).

Xét \(\Delta SAC\) ta có \(SA = AC = SC = a \Rightarrow \Delta SAC\) đều \( \Rightarrow \angle \left( {\overrightarrow {SC} ;\overrightarrow {AC} } \right) = {60^0}\).

\( \Rightarrow \overrightarrow {SC} .\overrightarrow {AC}  = a.a.\cos {60^0} = \frac{1}{2}{a^2}\).

+) \(\overrightarrow {CB} .\overrightarrow {AC}  = CB.AC.\cos \angle \left( {\overrightarrow {CB} ;\overrightarrow {AC} } \right)\).

Xét tam giác \(ABC\) có \(AB = AC = a,\,\,BC = a\sqrt 2 \) \( \Rightarrow \Delta ABC\) vuông cân tại \(A\).

\( \Rightarrow \angle \left( {\overrightarrow {CB} ;\overrightarrow {AC} } \right) = {180^0} - {45^0} = {135^0}\).

\( \Rightarrow \overrightarrow {CB} .\overrightarrow {AC}  = a\sqrt 2 .a.\cos {135^0} =  - {a^2}\).

\(\begin{array}{l} \Rightarrow \overrightarrow {SB} .\overrightarrow {AC}  = \frac{1}{2}{a^2} - {a^2} =  - \frac{1}{2}{a^2}\\ \Rightarrow SB.AC.\cos \angle \left( {\overrightarrow {SB} ;\overrightarrow {AC} } \right) =  - \frac{1}{2}{a^2}\\ \Leftrightarrow a.a.\cos \angle \left( {\overrightarrow {SB} ;\overrightarrow {AC} } \right) =  - \frac{1}{2}{a^2}\\ \Leftrightarrow \cos \angle \left( {\overrightarrow {SB} ;\overrightarrow {AC} } \right) =  - \frac{1}{2}\\ \Leftrightarrow \angle \left( {\overrightarrow {SB} ;\overrightarrow {AC} } \right) = {120^0}\end{array}\).

Chọn D.