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Câu hỏi: . Cho 2 đường thẳng ${{d}_{1}}:\dfrac{x}{1}=\dfrac{y}{-2}=\dfrac{z-1}{1}$ và ${{d}_{2}}:\dfrac{x-1}{2}=\dfrac{y}{1}=\dfrac{z+2}{-1}.$ Phương trình đường thẳng qua $A\left( 2;1;-1 \right)$ và vuông góc với cả ${{d}_{1}};{{d}_{2}}$ là
A. $\dfrac{x-2}{1}=\dfrac{y-1}{-2}=\dfrac{z+1}{3}.$
B. $\dfrac{x-2}{3}=\dfrac{y-1}{-3}=\dfrac{z+1}{1}.$
C. $\dfrac{x-2}{1}=\dfrac{y-1}{3}=\dfrac{z+1}{3}.$
D. $\dfrac{x-2}{1}=\dfrac{y-1}{3}=\dfrac{z+1}{5}.$
A. $\dfrac{x-2}{1}=\dfrac{y-1}{-2}=\dfrac{z+1}{3}.$
B. $\dfrac{x-2}{3}=\dfrac{y-1}{-3}=\dfrac{z+1}{1}.$
C. $\dfrac{x-2}{1}=\dfrac{y-1}{3}=\dfrac{z+1}{3}.$
D. $\dfrac{x-2}{1}=\dfrac{y-1}{3}=\dfrac{z+1}{5}.$
Gọi $d$ là đường thẳng cần tìm, gọi $\left\{ \begin{aligned}
& A=d\cap {{d}_{1}} \\
& B=d\cap {{d}_{2}} \\
\end{aligned} \right.$.
+ ${{d}_{1}}:\left\{ \begin{aligned}
& x=a \\
& y=-2a \\
& z=1+a \\
\end{aligned} \right.\Rightarrow A\left( a;-2a;a+1 \right) $; $ {{d}_{2}}:\left\{ \begin{aligned}
& x=1+2b \\
& y=b \\
& z=-2-b \\
\end{aligned} \right.\Rightarrow B\left( 2b+1;b;-b-2 \right)$
+ $d$ nhận $\overrightarrow{AB}=\left( 2b-a+1;2a+b;-a-b-3 \right)$ là một VTCP.
Mà $d\bot {{d}_{1}},d\bot {{d}_{2}}$ và $\overrightarrow{{{u}_{{{d}_{1}}}}}=\left( 1;-2;1 \right),\overrightarrow{{{u}_{{{d}_{2}}}}}=\left( 2;1;-1 \right)$ nên $\left\{ \begin{aligned}
& \overrightarrow{AB}.\overrightarrow{{{u}_{{{d}_{1}}}}}=0 \\
& \overrightarrow{AB}.\overrightarrow{{{u}_{{{d}_{2}}}}}=0 \\
\end{aligned} \right.$
$\Leftrightarrow \left\{ \begin{aligned}
& \left( 2b-a+1 \right)-2\left( 2a+b \right)-\left( -a-b-3 \right)=0 \\
& 2\left( 2b-a+1 \right)+\left( 2a+b \right)-\left( -a-b-3 \right)=0 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& -6a-b=2 \\
& a+6b=-5 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& a=-\dfrac{1}{5} \\
& b=-\dfrac{4}{5} \\
\end{aligned} \right.$
$\Rightarrow \overrightarrow{AB}=\left( -\dfrac{2}{5};-\dfrac{6}{5};-2 \right)\Rightarrow d$ nhận $\overrightarrow{u}=\left( 1;3;5 \right)$ là một VTCP.
Mà $d$ qua $A\left( 2;1;-1 \right)\Rightarrow d:\dfrac{x-2}{1}=\dfrac{y-1}{3}=\dfrac{z+1}{5}$.
& A=d\cap {{d}_{1}} \\
& B=d\cap {{d}_{2}} \\
\end{aligned} \right.$.
+ ${{d}_{1}}:\left\{ \begin{aligned}
& x=a \\
& y=-2a \\
& z=1+a \\
\end{aligned} \right.\Rightarrow A\left( a;-2a;a+1 \right) $; $ {{d}_{2}}:\left\{ \begin{aligned}
& x=1+2b \\
& y=b \\
& z=-2-b \\
\end{aligned} \right.\Rightarrow B\left( 2b+1;b;-b-2 \right)$
+ $d$ nhận $\overrightarrow{AB}=\left( 2b-a+1;2a+b;-a-b-3 \right)$ là một VTCP.
Mà $d\bot {{d}_{1}},d\bot {{d}_{2}}$ và $\overrightarrow{{{u}_{{{d}_{1}}}}}=\left( 1;-2;1 \right),\overrightarrow{{{u}_{{{d}_{2}}}}}=\left( 2;1;-1 \right)$ nên $\left\{ \begin{aligned}
& \overrightarrow{AB}.\overrightarrow{{{u}_{{{d}_{1}}}}}=0 \\
& \overrightarrow{AB}.\overrightarrow{{{u}_{{{d}_{2}}}}}=0 \\
\end{aligned} \right.$
$\Leftrightarrow \left\{ \begin{aligned}
& \left( 2b-a+1 \right)-2\left( 2a+b \right)-\left( -a-b-3 \right)=0 \\
& 2\left( 2b-a+1 \right)+\left( 2a+b \right)-\left( -a-b-3 \right)=0 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& -6a-b=2 \\
& a+6b=-5 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& a=-\dfrac{1}{5} \\
& b=-\dfrac{4}{5} \\
\end{aligned} \right.$
$\Rightarrow \overrightarrow{AB}=\left( -\dfrac{2}{5};-\dfrac{6}{5};-2 \right)\Rightarrow d$ nhận $\overrightarrow{u}=\left( 1;3;5 \right)$ là một VTCP.
Mà $d$ qua $A\left( 2;1;-1 \right)\Rightarrow d:\dfrac{x-2}{1}=\dfrac{y-1}{3}=\dfrac{z+1}{5}$.
Đáp án D.