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Câu hỏi: Cho hình chóp $S.ABC$ có $SA=SB=SC=AB=AC=a$ và $BC=a\sqrt{2}$. Tính góc giữa hai đường thẳng $AB$ và $SC$.
A. $\widehat{\left( AB,SC \right)}={{30}^{0}}$.
B. $\widehat{\left( AB,SC \right)}={{90}^{0}}$.
C. $\widehat{\left( AB,SC \right)}={{60}^{0}}$.
D. $\widehat{\left( AB,SC \right)}={{45}^{0}}$.
Ta có: $\overrightarrow{AB}.\overrightarrow{SC}=AB. SC.\cos \left( \overrightarrow{AB},\overrightarrow{SC} \right)$
$\Rightarrow \cos \left( \overrightarrow{AB},\overrightarrow{SC} \right)=\dfrac{\overrightarrow{AB}.\overrightarrow{SC}}{AB. SC}=\dfrac{\left( \overrightarrow{SB}-\overrightarrow{SA} \right).\overrightarrow{SC}}{AB. SC}=\dfrac{\overrightarrow{SB}.\overrightarrow{SC}-\overrightarrow{SA}.\overrightarrow{SC}}{AB. SC}$
Mặt khác $SB=SC=a;BC=a\sqrt{2}\Rightarrow B{{C}^{2}}=S{{B}^{2}}+S{{C}^{2}}\Rightarrow \Delta SBC$ vuông tại $S$, tức $\overrightarrow{SB}.\overrightarrow{SC}=0$.
Lại có $SA=SC=AC=a\Rightarrow \Delta SAC$ đều, do đó
$\overrightarrow{SA}.\overrightarrow{SC}=SA.SC.\cos \left( \overrightarrow{SA},\overrightarrow{SC} \right)=a.a.\cos {{60}^{0}}=\dfrac{{{a}^{2}}}{2}$.
Vậy $\cos \left( \overrightarrow{AB},\overrightarrow{SC} \right)=\dfrac{0-\dfrac{{{a}^{2}}}{2}}{a.a}=-\dfrac{1}{2}\Rightarrow \widehat{\left( \overrightarrow{AB},\overrightarrow{SC} \right)}={{120}^{0}}.$ Do đó $\widehat{\left( AB,SC \right)}={{60}^{0}}$.
A. $\widehat{\left( AB,SC \right)}={{30}^{0}}$.
B. $\widehat{\left( AB,SC \right)}={{90}^{0}}$.
C. $\widehat{\left( AB,SC \right)}={{60}^{0}}$.
D. $\widehat{\left( AB,SC \right)}={{45}^{0}}$.
$\Rightarrow \cos \left( \overrightarrow{AB},\overrightarrow{SC} \right)=\dfrac{\overrightarrow{AB}.\overrightarrow{SC}}{AB. SC}=\dfrac{\left( \overrightarrow{SB}-\overrightarrow{SA} \right).\overrightarrow{SC}}{AB. SC}=\dfrac{\overrightarrow{SB}.\overrightarrow{SC}-\overrightarrow{SA}.\overrightarrow{SC}}{AB. SC}$
Mặt khác $SB=SC=a;BC=a\sqrt{2}\Rightarrow B{{C}^{2}}=S{{B}^{2}}+S{{C}^{2}}\Rightarrow \Delta SBC$ vuông tại $S$, tức $\overrightarrow{SB}.\overrightarrow{SC}=0$.
Lại có $SA=SC=AC=a\Rightarrow \Delta SAC$ đều, do đó
$\overrightarrow{SA}.\overrightarrow{SC}=SA.SC.\cos \left( \overrightarrow{SA},\overrightarrow{SC} \right)=a.a.\cos {{60}^{0}}=\dfrac{{{a}^{2}}}{2}$.
Vậy $\cos \left( \overrightarrow{AB},\overrightarrow{SC} \right)=\dfrac{0-\dfrac{{{a}^{2}}}{2}}{a.a}=-\dfrac{1}{2}\Rightarrow \widehat{\left( \overrightarrow{AB},\overrightarrow{SC} \right)}={{120}^{0}}.$ Do đó $\widehat{\left( AB,SC \right)}={{60}^{0}}$.
Đáp án C.