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Câu hỏi: Trong không gian với hệ tọa độ Oxyz, viết phương trình mặt phẳng $\left( P \right)$ chứa đường thẳng $d:\dfrac{x-1}{2}=\dfrac{y}{1}=\dfrac{z+1}{3}$ và vuông góc với mặt phẳng $\left( Q \right):2x+y-z+0$.
A. $x+2y+z=0$
B. $x-2y-1=0$
C. $x+2y-1=0$
D. $x-2y+z=0$
A. $x+2y+z=0$
B. $x-2y-1=0$
C. $x+2y-1=0$
D. $x-2y+z=0$
Cách 1: $A\left( 1;0;-1 \right);B\left( 3;1;2 \right)\in d:\dfrac{x-1}{2}=\dfrac{y}{1}=\dfrac{z+1}{3}$
$\begin{aligned}
& \Rightarrow \left( P \right):a\left( x-1 \right)+b\left( y-0 \right)+c\left( z+1 \right)=0 \\
& \Rightarrow a\left( 3-1 \right)+b\left( 1-0 \right)+c\left( 2+1 \right)=0\Rightarrow b=-2a-3c \\
& \Rightarrow \left( P \right):a\left( x-1 \right)-\left( 2a+3c \right)y+c\left( x+1 \right)=0 \\
& \left( Q \right):2x+y-z=0 \\
& \left( P \right)\bot \left( Q \right)\Leftrightarrow 2a-\left( 2a+3c \right)-c=0\Leftrightarrow c=0 \\
& \Rightarrow \left( P \right):x-1-2y=0 \\
\end{aligned}$
Cách 2: $\overrightarrow{{{n}_{\left( P \right)}}}=\left[ \overrightarrow{{{n}_{\left( Q \right)}}},\overrightarrow{{{u}_{d}}} \right]=\left( -4;8;0 \right)$.
$\begin{aligned}
& \Rightarrow \left( P \right):a\left( x-1 \right)+b\left( y-0 \right)+c\left( z+1 \right)=0 \\
& \Rightarrow a\left( 3-1 \right)+b\left( 1-0 \right)+c\left( 2+1 \right)=0\Rightarrow b=-2a-3c \\
& \Rightarrow \left( P \right):a\left( x-1 \right)-\left( 2a+3c \right)y+c\left( x+1 \right)=0 \\
& \left( Q \right):2x+y-z=0 \\
& \left( P \right)\bot \left( Q \right)\Leftrightarrow 2a-\left( 2a+3c \right)-c=0\Leftrightarrow c=0 \\
& \Rightarrow \left( P \right):x-1-2y=0 \\
\end{aligned}$
Cách 2: $\overrightarrow{{{n}_{\left( P \right)}}}=\left[ \overrightarrow{{{n}_{\left( Q \right)}}},\overrightarrow{{{u}_{d}}} \right]=\left( -4;8;0 \right)$.
Đáp án B.