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Câu hỏi: Trong không gian với hệ tọa độ Oxyz, cho ba đường thẳng ${{d}_{1}}:\dfrac{x-1}{2}=\dfrac{y}{3}=\dfrac{z+1}{-3},$ ${{d}_{2}}:\dfrac{x+2}{1}=\dfrac{y-1}{-2}=\dfrac{z}{2},{{d}_{3}}:\dfrac{x+3}{-3}=\dfrac{y-2}{-4}=\dfrac{z+5}{8}.$ Đường thẳng song song với d3 cắt d1 và d2 có phương trình là
A. $\dfrac{x-1}{-3}=\dfrac{y}{-4}=\dfrac{z+1}{8}.$
B. $\dfrac{x-1}{-3}=\dfrac{y}{-4}=\dfrac{z-1}{8}.$
C. $\dfrac{x+1}{-3}=\dfrac{y-3}{-4}=\dfrac{z}{8}.$
D. $\dfrac{x-1}{-3}=\dfrac{y-3}{-4}=\dfrac{z}{8}.$
A. $\dfrac{x-1}{-3}=\dfrac{y}{-4}=\dfrac{z+1}{8}.$
B. $\dfrac{x-1}{-3}=\dfrac{y}{-4}=\dfrac{z-1}{8}.$
C. $\dfrac{x+1}{-3}=\dfrac{y-3}{-4}=\dfrac{z}{8}.$
D. $\dfrac{x-1}{-3}=\dfrac{y-3}{-4}=\dfrac{z}{8}.$
Gọi $A\left( 1+2t;3t;-1-3t \right)\in {{d}_{1}} v\grave{a} B\left( -2+u;1-2u;2u \right)\in {{d}_{2}} $
Ta có: $=\left( -3+u-2t;1-2u-3t;2u+1+3t \right)$
Mặt khác $\overrightarrow{{{u}_{{{d}_{3}}}}}=\left( -3;-4;8 \right),AB//{{d}_{3}}\Rightarrow \overrightarrow{AB}=k.\overrightarrow{{{u}_{{{d}_{3}}}}}\Rightarrow \dfrac{-3+u-2t}{-3}=\dfrac{1-2u-3t}{-4}=\dfrac{2u+1+3t}{8}$
$\Leftrightarrow \left\{ \begin{aligned}
& 10u+t=15 \\
& -14u+7t=-21 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& t=0 \\
& u=\dfrac{3}{2} \\
\end{aligned} \right.\Rightarrow A\left( 1;0;-1 \right). $ Suy ra $ AB:\dfrac{x-1}{-3}=\dfrac{y}{-4}=\dfrac{z+1}{8}.$
Ta có: $=\left( -3+u-2t;1-2u-3t;2u+1+3t \right)$
Mặt khác $\overrightarrow{{{u}_{{{d}_{3}}}}}=\left( -3;-4;8 \right),AB//{{d}_{3}}\Rightarrow \overrightarrow{AB}=k.\overrightarrow{{{u}_{{{d}_{3}}}}}\Rightarrow \dfrac{-3+u-2t}{-3}=\dfrac{1-2u-3t}{-4}=\dfrac{2u+1+3t}{8}$
$\Leftrightarrow \left\{ \begin{aligned}
& 10u+t=15 \\
& -14u+7t=-21 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& t=0 \\
& u=\dfrac{3}{2} \\
\end{aligned} \right.\Rightarrow A\left( 1;0;-1 \right). $ Suy ra $ AB:\dfrac{x-1}{-3}=\dfrac{y}{-4}=\dfrac{z+1}{8}.$
Đáp án A.