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Câu hỏi: Trong không gian $Oxyz$, cho hai đường thẳng ${{d}_{1}}:\dfrac{x-1}{1}=\dfrac{y+1}{-1}=\dfrac{z}{2}$, ${{d}_{2}}:\dfrac{x}{1}=\dfrac{y-1}{2}=\dfrac{z}{1}$. Đường thẳng $d$ đi qua điểm $A\left( 5;-3;5 \right)$ cắt ${{d}_{1}}$, ${{d}_{2}}$ tại hai điểm $B$ và $C$. Độ dài đoạn thẳng $BC$ bằng
A. $3\sqrt{2}$
B. $\sqrt{19}$
C. $2\sqrt{5}$
D. $2\sqrt{3}$
A. $3\sqrt{2}$
B. $\sqrt{19}$
C. $2\sqrt{5}$
D. $2\sqrt{3}$
Gọi $B\left( b+1;-1-b;2b \right)\in {{d}_{1}};C\left( c;2c+1;c \right)\in {{d}_{2}}$.
$\overrightarrow{AB}\left( b-4;2-b;2b-5 \right);\overrightarrow{AC}\left( c-5;2c+4;c-5 \right)$
$A,B,C$ thẳng hàng $\Leftrightarrow \overrightarrow{AB}=k\overrightarrow{AC}\Leftrightarrow \left\{ \begin{matrix}
b-4=k\left( c-5 \right) \\
2-b=k\left( 2c+4 \right) \\
2b-5=k\left( c-5 \right) \\
\end{matrix} \right.\Leftrightarrow \left\{ \begin{matrix}
b-kc+5k=4 \\
-b-2kc-4k=-2 \\
2b-kc+5k=5 \\
\end{matrix}\Leftrightarrow \left\{ \begin{matrix}
b=1 \\
kc=-\dfrac{1}{2} \\
k=\dfrac{1}{2} \\
\end{matrix} \right. \right.$
$c=-1\Rightarrow B\left( 2,-2,2 \right),C\left( -1,-1,-1 \right)\Rightarrow BC=\sqrt{19}$
$\overrightarrow{AB}\left( b-4;2-b;2b-5 \right);\overrightarrow{AC}\left( c-5;2c+4;c-5 \right)$
$A,B,C$ thẳng hàng $\Leftrightarrow \overrightarrow{AB}=k\overrightarrow{AC}\Leftrightarrow \left\{ \begin{matrix}
b-4=k\left( c-5 \right) \\
2-b=k\left( 2c+4 \right) \\
2b-5=k\left( c-5 \right) \\
\end{matrix} \right.\Leftrightarrow \left\{ \begin{matrix}
b-kc+5k=4 \\
-b-2kc-4k=-2 \\
2b-kc+5k=5 \\
\end{matrix}\Leftrightarrow \left\{ \begin{matrix}
b=1 \\
kc=-\dfrac{1}{2} \\
k=\dfrac{1}{2} \\
\end{matrix} \right. \right.$
$c=-1\Rightarrow B\left( 2,-2,2 \right),C\left( -1,-1,-1 \right)\Rightarrow BC=\sqrt{19}$
Đáp án B.