Google
Câu hỏi: Cho hình chóp $S.ABC$ có $SA=SB=SC$, $\widehat{ASC}=120{}^\circ $, $\widehat{BSC}=60{}^\circ $, $\widehat{ASB}=90{}^\circ $. Tính cosin của góc giữa hai đường thẳng $SB$ và $AC$.
A. $0$.
B. $\dfrac{\sqrt{3}}{6}$.
C. $\dfrac{-\sqrt{3}}{6}$.
D. $\dfrac{\sqrt{3}}{3}$.
Gọi $M$ là trung điểm $AC$ $\Rightarrow SM\bot AC$ ; $AC=2MC=2.SC.\dfrac{\sqrt{3}}{2}=SC\sqrt{3}$.
Do $\widehat{ASB}=90{}^\circ $ $\Rightarrow \overrightarrow{SB}.\overrightarrow{SA}=0$.
Có $cos\left( SB;AC \right)=\left| \cos \left( \overrightarrow{SB};\overrightarrow{AC} \right) \right|=\dfrac{\left| \overrightarrow{SB}.\overrightarrow{AC} \right|}{\left| \overrightarrow{SB} \right|\left| \overrightarrow{AC} \right|}=\dfrac{\left| \overrightarrow{SB}.\left( \overrightarrow{SC}-\overrightarrow{SA} \right) \right|}{SB.AC}$ $=\dfrac{\left| \overrightarrow{SB}.\overrightarrow{SC}-\overrightarrow{SB}\overrightarrow{.SA} \right|}{SB.AC}=\dfrac{\left| \overrightarrow{SB}.\overrightarrow{SC} \right|}{SB.AC}$
$=\dfrac{\left| SB.SC.cos\left( \overrightarrow{SB};\overrightarrow{SC} \right) \right|}{SB.AC}$ $=\dfrac{SB.SC.cos60{}^\circ }{SB.SC\sqrt{3}}=\dfrac{1}{2\sqrt{3}}=\dfrac{\sqrt{3}}{6}$.
A. $0$.
B. $\dfrac{\sqrt{3}}{6}$.
C. $\dfrac{-\sqrt{3}}{6}$.
D. $\dfrac{\sqrt{3}}{3}$.
Do $\widehat{ASB}=90{}^\circ $ $\Rightarrow \overrightarrow{SB}.\overrightarrow{SA}=0$.
Có $cos\left( SB;AC \right)=\left| \cos \left( \overrightarrow{SB};\overrightarrow{AC} \right) \right|=\dfrac{\left| \overrightarrow{SB}.\overrightarrow{AC} \right|}{\left| \overrightarrow{SB} \right|\left| \overrightarrow{AC} \right|}=\dfrac{\left| \overrightarrow{SB}.\left( \overrightarrow{SC}-\overrightarrow{SA} \right) \right|}{SB.AC}$ $=\dfrac{\left| \overrightarrow{SB}.\overrightarrow{SC}-\overrightarrow{SB}\overrightarrow{.SA} \right|}{SB.AC}=\dfrac{\left| \overrightarrow{SB}.\overrightarrow{SC} \right|}{SB.AC}$
$=\dfrac{\left| SB.SC.cos\left( \overrightarrow{SB};\overrightarrow{SC} \right) \right|}{SB.AC}$ $=\dfrac{SB.SC.cos60{}^\circ }{SB.SC\sqrt{3}}=\dfrac{1}{2\sqrt{3}}=\dfrac{\sqrt{3}}{6}$.
Đáp án B.