Tính khoảng cách từ các điểm $M_0(2;3;1), M_1(1;-1;1)$ đến đường thẳng: $\frac{x+2}{1}=\frac{y-1}{2}=\frac{z+1}{-2}   $
Áp dụng công thức $d({M_1},\Delta ) = \frac{{|{\rm{[}}\overrightarrow u {\rm{,}}\overrightarrow {{M_0}{M_1}} {\rm{]}}|}}{{|\overrightarrow u |}}$
Trong đó: $M_0(x_0;y_0;z_0)\in \Delta $
Ta có: $M(-2;1;-1)\in \Delta:\frac{x+2}{1}=\frac{y-1}{2}=\frac{z+1}{-2}    $ và $M_0(2;3;1)$
Áp dụng công thức trên ta có:  $d({M_0},\Delta ) = \frac{{|{\rm{[}}\overrightarrow u {\rm{,}}\overrightarrow {M{M_0}} {\rm{]}}|}}{{|\overrightarrow u |}}\,\,  (1)$
Ta tính: $\overrightarrow{u}=(1;2;-2), \overrightarrow{MM_0}=(4;2;2)  $
$\begin{array}{l}
|{\rm{[}}\overrightarrow u {\rm{,}}\overrightarrow {M{M_0}} {\rm{]}}| = (8; - 10; - 6) \Rightarrow |{\rm{[}}\overrightarrow u {\rm{,}}\overrightarrow {M{M_0}} {\rm{]}}| = 10\sqrt 2 \\
|\overrightarrow u | = \sqrt {{1^1} + {2^2} + {{( - 2)}^2}}  = 3
\end{array}$
Thay vào $(1)$ ta có: $d({M_0},\Delta ) = \frac{{10\sqrt 2 }}{3}$
Ta tính: $d({M_1},\Delta ) = \frac{{|{\rm{[}}\overrightarrow u {\rm{,}}\overrightarrow {M{M_1}} {\rm{]}}|}}{{|\overrightarrow u |}}$
Ta có: $\overrightarrow{u}=(1;2;-2),  \overrightarrow{MM_1}=(3;-2;2)  $
nên:   $\begin{array}{l}
|{\rm{[}}\overrightarrow u {\rm{,}}\overrightarrow {M{M_1}} {\rm{]}}| = (0; - 8; - 8) \Rightarrow |{\rm{[}}\overrightarrow u {\rm{,}}\overrightarrow {M{M_1}} {\rm{]}}| = 8\sqrt 2 \\
|\overrightarrow u | = \sqrt {{1^1} + {2^2} + {{( - 2)}^2}}  = 3\\
 \Rightarrow d({M_1},\Delta ) = \frac{{8\sqrt 2 }}{3}
\end{array}$

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