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Câu hỏi: Trong không gian với hệ tọa độ Oxyz, tính góc giữa hai đường thẳng ${{d}_{1}}:\dfrac{x}{1}=\dfrac{y+1}{-1}=\dfrac{z-1}{2}$ và ${{d}_{2}}:\dfrac{x+1}{-1}=\dfrac{y}{1}=\dfrac{z-3}{1}$.
A. $45{}^\circ $
B. $30{}^\circ $
C. $60{}^\circ $
D. $90{}^\circ $
A. $45{}^\circ $
B. $30{}^\circ $
C. $60{}^\circ $
D. $90{}^\circ $
$\begin{aligned}
& cos\left( \widehat{{{d}_{1}};{{d}_{2}}} \right)=\left| cos\left( \overrightarrow{{{n}_{{{d}_{1}}}}};\overrightarrow{{{n}_{{{d}_{2}}}}} \right) \right| \\
& =\dfrac{\left| 1.\left( -1 \right)+\left( -1 \right).1+2.1 \right|}{\sqrt{{{1}^{2}}+{{\left( -1 \right)}^{2}}+{{2}^{2}}}\sqrt{{{\left( -1 \right)}^{2}}+{{1}^{2}}+{{1}^{2}}}}=0\Rightarrow \left( \widehat{{{d}_{1}};{{d}_{2}}} \right)=90{}^\circ \\
\end{aligned}$
& cos\left( \widehat{{{d}_{1}};{{d}_{2}}} \right)=\left| cos\left( \overrightarrow{{{n}_{{{d}_{1}}}}};\overrightarrow{{{n}_{{{d}_{2}}}}} \right) \right| \\
& =\dfrac{\left| 1.\left( -1 \right)+\left( -1 \right).1+2.1 \right|}{\sqrt{{{1}^{2}}+{{\left( -1 \right)}^{2}}+{{2}^{2}}}\sqrt{{{\left( -1 \right)}^{2}}+{{1}^{2}}+{{1}^{2}}}}=0\Rightarrow \left( \widehat{{{d}_{1}};{{d}_{2}}} \right)=90{}^\circ \\
\end{aligned}$
Đáp án D.