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Câu hỏi: Cho hình chóp $S.ABC$ có $SA=BC=2a$. Gọi $M$, $N$ lần lượt là trung điểm của $AB$ và $SC$, $MN=a\sqrt{3}$. Tính số đo góc giữa hai đường thẳng $SA$ và $BC$.
A. ${{120}^{{}^\circ }}$.
B. ${{150}^{{}^\circ }}$.
C. ${{30}^{{}^\circ }}$.
D. ${{60}^{{}^\circ }}$.
A. ${{120}^{{}^\circ }}$.
B. ${{150}^{{}^\circ }}$.
C. ${{30}^{{}^\circ }}$.
D. ${{60}^{{}^\circ }}$.
Ta có: $\overrightarrow{MN}=\overrightarrow{MA}+\overrightarrow{AS}+\overrightarrow{SN}$.
$\overrightarrow{MN}=\overrightarrow{MB}+\overrightarrow{BC}+\overrightarrow{CN}$
$\Rightarrow 2\overrightarrow{MN}=\overrightarrow{AS}+\overrightarrow{BC}$
$\Rightarrow 4{{\left( \overrightarrow{MN} \right)}^{2}}={{\left( \overrightarrow{AS}+\overrightarrow{BC} \right)}^{2}}$
$\Leftrightarrow 4{{\overrightarrow{MN}}^{2}}={{\overrightarrow{AS}}^{2}}+{{\overrightarrow{BC}}^{2}}+2\overrightarrow{AS}.\overrightarrow{BC}$
$\Leftrightarrow 4M{{N}^{2}}=A{{S}^{2}}+B{{C}^{2}}+2.AS.BC.\cos \left( \overrightarrow{AS},\overrightarrow{BC} \right)$
$\Leftrightarrow \cos \left( \overrightarrow{AS},\overrightarrow{BC} \right)=\dfrac{4M{{N}^{2}}-A{{S}^{2}}-B{{C}^{2}}}{2.AS.BC}=\dfrac{1}{2}$
$\Rightarrow \left( \overrightarrow{AS},\overrightarrow{BC} \right)={{60}^{{}^\circ }}$.
Vậy $\left( AS,BC \right)={{60}^{{}^\circ }}$.
$\overrightarrow{MN}=\overrightarrow{MB}+\overrightarrow{BC}+\overrightarrow{CN}$
$\Rightarrow 2\overrightarrow{MN}=\overrightarrow{AS}+\overrightarrow{BC}$
$\Rightarrow 4{{\left( \overrightarrow{MN} \right)}^{2}}={{\left( \overrightarrow{AS}+\overrightarrow{BC} \right)}^{2}}$
$\Leftrightarrow 4{{\overrightarrow{MN}}^{2}}={{\overrightarrow{AS}}^{2}}+{{\overrightarrow{BC}}^{2}}+2\overrightarrow{AS}.\overrightarrow{BC}$
$\Leftrightarrow 4M{{N}^{2}}=A{{S}^{2}}+B{{C}^{2}}+2.AS.BC.\cos \left( \overrightarrow{AS},\overrightarrow{BC} \right)$
$\Leftrightarrow \cos \left( \overrightarrow{AS},\overrightarrow{BC} \right)=\dfrac{4M{{N}^{2}}-A{{S}^{2}}-B{{C}^{2}}}{2.AS.BC}=\dfrac{1}{2}$
$\Rightarrow \left( \overrightarrow{AS},\overrightarrow{BC} \right)={{60}^{{}^\circ }}$.
Vậy $\left( AS,BC \right)={{60}^{{}^\circ }}$.
Đáp án D.