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Câu hỏi: Viết phương trình đường thẳng d qua $M\left( 1;-2;3 \right)$ và vuông góc với hai đường thẳng ${{d}_{1}}:\dfrac{x}{1}=\dfrac{y-1}{-1}=\dfrac{z+1}{3},{{d}_{2}}:\left\{ \begin{aligned}
& x=1-t \\
& y=2+t \\
& z=1+3t \\
\end{aligned} \right.\left( t\in \mathbb{R} \right).$
A. $\left\{ \begin{aligned}
& x=1+t \\
& y=-2+t \\
& z=3 \\
\end{aligned} \right.\left( t\in \mathbb{R} \right)$
B. $\left\{ \begin{aligned}
& x=1+3t \\
& y=-2+t \\
& z=3+t \\
\end{aligned} \right.\left( t\in \mathbb{R} \right)$
C. $\left\{ \begin{aligned}
& x=1+t \\
& y=1-2t \\
& z=3t \\
\end{aligned} \right.\left( t\in \mathbb{R} \right)$
D. $\left\{ \begin{aligned}
& x=1 \\
& y=-2+t \\
& z=3+t \\
\end{aligned} \right.\left( t\in \mathbb{R} \right)$
& x=1-t \\
& y=2+t \\
& z=1+3t \\
\end{aligned} \right.\left( t\in \mathbb{R} \right).$
A. $\left\{ \begin{aligned}
& x=1+t \\
& y=-2+t \\
& z=3 \\
\end{aligned} \right.\left( t\in \mathbb{R} \right)$
B. $\left\{ \begin{aligned}
& x=1+3t \\
& y=-2+t \\
& z=3+t \\
\end{aligned} \right.\left( t\in \mathbb{R} \right)$
C. $\left\{ \begin{aligned}
& x=1+t \\
& y=1-2t \\
& z=3t \\
\end{aligned} \right.\left( t\in \mathbb{R} \right)$
D. $\left\{ \begin{aligned}
& x=1 \\
& y=-2+t \\
& z=3+t \\
\end{aligned} \right.\left( t\in \mathbb{R} \right)$
$\overrightarrow{{{u}_{{{d}_{1}}}}}=\left( 1;-1;3 \right);\overrightarrow{{{u}_{{{d}_{2}}}}}=\left( -1;1;3 \right)$
$\begin{aligned}
& \Rightarrow \overrightarrow{{{u}_{d}}}=\left[ \overrightarrow{{{u}_{{{d}_{1}}}}};\overrightarrow{{{u}_{{{d}_{2}}}}} \right]=\left( -6;-6;0 \right)=-6\left( 1;1;0 \right) \\
& \Rightarrow \left( d \right):\left\{ \begin{aligned}
& x=1+t \\
& y=-2+t \\
& z=3 \\
\end{aligned} \right.\left( t\in \mathbb{R} \right). \\
\end{aligned}$
$\begin{aligned}
& \Rightarrow \overrightarrow{{{u}_{d}}}=\left[ \overrightarrow{{{u}_{{{d}_{1}}}}};\overrightarrow{{{u}_{{{d}_{2}}}}} \right]=\left( -6;-6;0 \right)=-6\left( 1;1;0 \right) \\
& \Rightarrow \left( d \right):\left\{ \begin{aligned}
& x=1+t \\
& y=-2+t \\
& z=3 \\
\end{aligned} \right.\left( t\in \mathbb{R} \right). \\
\end{aligned}$
Đáp án A.