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Câu hỏi: Trong không gian với hệ trục tọa độ Oxyz cho $M\left( 1;-2;1 \right), N\left( 0;1;3 \right)$. Phương trình đường thẳng qua hai điểm M, N là
A. $\dfrac{x+1}{-1}=\dfrac{y-2}{3}=\dfrac{x+1}{2}.$
B. $\dfrac{x+1}{1}=\dfrac{y-3}{-2}=\dfrac{z-2}{1}.$
C. $\dfrac{x}{-1}=\dfrac{1-y}{3}=\dfrac{z-3}{2}.$
D. $\dfrac{x}{-1}=\dfrac{1-y}{-3}=\dfrac{z-3}{2}.$
A. $\dfrac{x+1}{-1}=\dfrac{y-2}{3}=\dfrac{x+1}{2}.$
B. $\dfrac{x+1}{1}=\dfrac{y-3}{-2}=\dfrac{z-2}{1}.$
C. $\dfrac{x}{-1}=\dfrac{1-y}{3}=\dfrac{z-3}{2}.$
D. $\dfrac{x}{-1}=\dfrac{1-y}{-3}=\dfrac{z-3}{2}.$
Ta có $\overrightarrow{MN}=\left( -1;3;2 \right)\Rightarrow d:\dfrac{x}{-1}=\dfrac{y-1}{3}=\dfrac{z-3}{2}$ hay $d:\dfrac{x}{-1}=\dfrac{1-y}{-3}=\dfrac{z-3}{2}.$
Đáp án D.