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Câu hỏi: Trong không gian Oxyz, cho mặt phẳng $\left( P \right):x+y+z-3=0$ và đường thẳng $d:\dfrac{x}{1}=\dfrac{y+1}{2}=\dfrac{z-2}{-1}.$ Đường thẳng d' đối xứng với d qua mặt phẳng (P) có phương trình là:
A. $\dfrac{x-1}{1}=\dfrac{y-1}{-2}=\dfrac{z-1}{7}.$
B. $\dfrac{x+1}{1}=\dfrac{y+1}{-2}=\dfrac{z+1}{7}.$
C. $\dfrac{x-1}{1}=\dfrac{y-1}{2}=\dfrac{z-1}{7}.$
D. $\dfrac{x+1}{1}=\dfrac{y+1}{2}=\dfrac{z+1}{7}.$
A. $\dfrac{x-1}{1}=\dfrac{y-1}{-2}=\dfrac{z-1}{7}.$
B. $\dfrac{x+1}{1}=\dfrac{y+1}{-2}=\dfrac{z+1}{7}.$
C. $\dfrac{x-1}{1}=\dfrac{y-1}{2}=\dfrac{z-1}{7}.$
D. $\dfrac{x+1}{1}=\dfrac{y+1}{2}=\dfrac{z+1}{7}.$
$I=d\cap \left( P \right)\Rightarrow I\left( 1;1;1 \right).$ Tìm ${A}'?$
AH qua A có $\overrightarrow{{{u}_{AH}}}=\overrightarrow{{{n}_{p}}}=\left( 1;1;1 \right)\Rightarrow AH:\left\{ \begin{aligned}
& x=t \\
& y=-1+t \\
& z=2+t \\
\end{aligned} \right.$
Suy ra $H\left( t;t-1;t+2 \right).$
Mà $H\in \left( P \right)\Rightarrow H\left( \dfrac{2}{3};\dfrac{-1}{3};\dfrac{8}{3} \right).$
Ta có: ${A}'\left( \dfrac{4}{3};\dfrac{1}{3};\dfrac{10}{3} \right)\Rightarrow \overrightarrow{I{A}'}=\left( \dfrac{1}{3};\dfrac{-2}{3};\dfrac{7}{3} \right)\Rightarrow {d}':\dfrac{x-1}{1}=\dfrac{y-1}{-2}=\dfrac{z-1}{7}.$
AH qua A có $\overrightarrow{{{u}_{AH}}}=\overrightarrow{{{n}_{p}}}=\left( 1;1;1 \right)\Rightarrow AH:\left\{ \begin{aligned}
& x=t \\
& y=-1+t \\
& z=2+t \\
\end{aligned} \right.$
Suy ra $H\left( t;t-1;t+2 \right).$
Mà $H\in \left( P \right)\Rightarrow H\left( \dfrac{2}{3};\dfrac{-1}{3};\dfrac{8}{3} \right).$
Ta có: ${A}'\left( \dfrac{4}{3};\dfrac{1}{3};\dfrac{10}{3} \right)\Rightarrow \overrightarrow{I{A}'}=\left( \dfrac{1}{3};\dfrac{-2}{3};\dfrac{7}{3} \right)\Rightarrow {d}':\dfrac{x-1}{1}=\dfrac{y-1}{-2}=\dfrac{z-1}{7}.$
Đáp án A.