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Câu hỏi: Trong không gian $Oxyz,$ cho đường thẳng $\left( \Delta \right):\dfrac{x-1}{2}=\dfrac{y}{2}=\dfrac{z+2}{3}.$ Phương trình đường thẳng $\left( d \right)$ qua $O\left( 0;0;0 \right)$ cắt $\left( \Delta \right)$ tại $A$ và $\left( d \right)\bot \left( \Delta \right)$ là
A. $3x-6y+2z=0$.
B. $x=y=\dfrac{3z}{-4}$.
C. $\dfrac{x}{3}=\dfrac{y}{30}=\dfrac{z}{-4}$.
D. $\dfrac{x}{25}=\dfrac{y}{8}=\dfrac{z}{-22}$.
A. $3x-6y+2z=0$.
B. $x=y=\dfrac{3z}{-4}$.
C. $\dfrac{x}{3}=\dfrac{y}{30}=\dfrac{z}{-4}$.
D. $\dfrac{x}{25}=\dfrac{y}{8}=\dfrac{z}{-22}$.
Do $A\in \left( \Delta \right)$ nên $A\left( 1+2t;2t;-2+3t \right)$ và $\overrightarrow{OA}=\left( 1+2t;2t;-2+3t \right)$
Vì $\left( d \right)\bot \left( \Delta \right)$ nên $\overrightarrow{OA}\bot {{\overrightarrow{u}}_{\left( \Delta \right)}}=\left( 2;2;3 \right)$ hay $2\left( 1+2t \right)+2\left( 2t \right)+3\left( -2+3t \right)=0\Leftrightarrow t=\dfrac{4}{17}$
Suy ra $\overrightarrow{OA}=\left( \dfrac{25}{17};\dfrac{8}{17};-\dfrac{22}{17} \right)=\dfrac{1}{17}\left( 25;8;-22 \right)$ và $\left( d \right):\dfrac{x}{25}=\dfrac{y}{8}=\dfrac{z}{-22}.$
Vì $\left( d \right)\bot \left( \Delta \right)$ nên $\overrightarrow{OA}\bot {{\overrightarrow{u}}_{\left( \Delta \right)}}=\left( 2;2;3 \right)$ hay $2\left( 1+2t \right)+2\left( 2t \right)+3\left( -2+3t \right)=0\Leftrightarrow t=\dfrac{4}{17}$
Suy ra $\overrightarrow{OA}=\left( \dfrac{25}{17};\dfrac{8}{17};-\dfrac{22}{17} \right)=\dfrac{1}{17}\left( 25;8;-22 \right)$ và $\left( d \right):\dfrac{x}{25}=\dfrac{y}{8}=\dfrac{z}{-22}.$
Đáp án D.