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Câu hỏi: Phương trình đường thẳng song song với đường thẳng $d:\dfrac{x-1}{1}=\dfrac{y+2}{1}=\dfrac{z}{-1}$ và cắt hai đường thẳng ${{d}_{1}}:\dfrac{x+1}{2}=\dfrac{y+1}{1}=\dfrac{z-2}{-1}$ và ${{d}_{2}}:\dfrac{x-1}{-1}=\dfrac{y-2}{1}=\dfrac{z-3}{3}$ là
A. $\dfrac{x+1}{-1}=\dfrac{y+1}{-1}=\dfrac{z-2}{1}$
B. $\dfrac{x-1}{1}=\dfrac{y}{1}=\dfrac{z-1}{-1}$
C. $\dfrac{x-1}{1}=\dfrac{y-2}{1}=\dfrac{z-3}{-1}$
D. $\dfrac{x-1}{1}=\dfrac{y}{-1}=\dfrac{z-1}{1}$
A. $\dfrac{x+1}{-1}=\dfrac{y+1}{-1}=\dfrac{z-2}{1}$
B. $\dfrac{x-1}{1}=\dfrac{y}{1}=\dfrac{z-1}{-1}$
C. $\dfrac{x-1}{1}=\dfrac{y-2}{1}=\dfrac{z-3}{-1}$
D. $\dfrac{x-1}{1}=\dfrac{y}{-1}=\dfrac{z-1}{1}$
Gọi $A\left( -1+2t;-1+t;2-t \right)\in {{\text{d}}_{1}},\text{ B}\left( 1-u;2+u;3+3u \right)\in {{\text{d}}_{2}}$
Khi đó: $\overrightarrow{AB}\left( 2-u-2t;3+u-t;1+3u+t \right)$
Do đó $AB\text{ // d}\Rightarrow \dfrac{2-u-2t}{1}=\dfrac{3+u-t}{1}=\dfrac{1+3u+t}{1}\Leftrightarrow \left\{ \begin{aligned}
& t=1 \\
& u=-1 \\
\end{aligned} \right.\Rightarrow A\left( 1;0;1 \right)\Rightarrow \left( \Delta \right):\dfrac{x-1}{1}=\dfrac{y}{1}=\dfrac{z-1}{-1}$.
Khi đó: $\overrightarrow{AB}\left( 2-u-2t;3+u-t;1+3u+t \right)$
Do đó $AB\text{ // d}\Rightarrow \dfrac{2-u-2t}{1}=\dfrac{3+u-t}{1}=\dfrac{1+3u+t}{1}\Leftrightarrow \left\{ \begin{aligned}
& t=1 \\
& u=-1 \\
\end{aligned} \right.\Rightarrow A\left( 1;0;1 \right)\Rightarrow \left( \Delta \right):\dfrac{x-1}{1}=\dfrac{y}{1}=\dfrac{z-1}{-1}$.
Đáp án B.