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Câu hỏi: Trong không gian với hệ tọa độ Oxyz cho ba điểm $A\left( 0;1;0 \right),B\left( 2;2;2 \right),C\left( -2;3;1 \right)$ và đường thẳng $d:\dfrac{x-1}{2}=\dfrac{y+2}{-1}=\dfrac{z-3}{2}$. Tìm điểm M thuộc d để thể tích V của tứ diện MABC bằng 3.
A. $M\left( -\dfrac{15}{2};\dfrac{9}{4};-\dfrac{11}{2} \right);\ M\left( -\dfrac{3}{2};-\dfrac{3}{4};\dfrac{1}{2} \right).$
B. $M\left( -\dfrac{3}{5};\dfrac{3}{4};\dfrac{1}{2} \right);\ M\left( -\dfrac{15}{2};\dfrac{9}{4};\dfrac{11}{2} \right).$
C. $M\left( \dfrac{3}{2};-\dfrac{3}{4};\dfrac{1}{2} \right);\ M\left( \dfrac{15}{2};\dfrac{9}{4};\dfrac{11}{2} \right).$
D. $M\left( \dfrac{3}{5};-\dfrac{3}{4};\dfrac{1}{2} \right);\ M\left( \dfrac{15}{2};\dfrac{9}{4};\dfrac{11}{2} \right).$
A. $M\left( -\dfrac{15}{2};\dfrac{9}{4};-\dfrac{11}{2} \right);\ M\left( -\dfrac{3}{2};-\dfrac{3}{4};\dfrac{1}{2} \right).$
B. $M\left( -\dfrac{3}{5};\dfrac{3}{4};\dfrac{1}{2} \right);\ M\left( -\dfrac{15}{2};\dfrac{9}{4};\dfrac{11}{2} \right).$
C. $M\left( \dfrac{3}{2};-\dfrac{3}{4};\dfrac{1}{2} \right);\ M\left( \dfrac{15}{2};\dfrac{9}{4};\dfrac{11}{2} \right).$
D. $M\left( \dfrac{3}{5};-\dfrac{3}{4};\dfrac{1}{2} \right);\ M\left( \dfrac{15}{2};\dfrac{9}{4};\dfrac{11}{2} \right).$
Cách 1: Ta có $\overrightarrow{AB}=\left( 2;1;2 \right);\ \overrightarrow{AC}=\left( -2;2;1 \right)$.
Do $\left[ \overrightarrow{AB},\overrightarrow{AC} \right]=\left( -3;-6;6 \right)$ nên ${{S}_{\Delta ABC}}=\dfrac{1}{2}\left| \left[ \overrightarrow{AB},\overrightarrow{AC} \right] \right|=\dfrac{9}{2}.$
Gọi $\overrightarrow{n}$ là một vectơ pháp tuyến của mặt phẳng $\left( ABC \right)$ thì $\overrightarrow{n}=\left( 1;2;-2 \right)$ cùng phương với $\left[ \overrightarrow{AB},\overrightarrow{AC} \right]\Rightarrow $ Phương trình mặt phẳng $\left( ABC \right)$ là $x+2y-2z-2=0$.
Gọi $\begin{aligned}
& M\left( 1+2t;-2-t;3+2t \right)\in d \\
& \Rightarrow d\left( M,\left( ABC \right) \right)=\dfrac{\left| \left( 1+2t \right)+2\left( -2-t \right)-2\left( 3+2t \right)-2 \right|}{\sqrt{{{1}^{2}}+{{2}^{2}}+{{\left( -2 \right)}^{2}}}}=\dfrac{\left| 4t+11 \right|}{3}. \\
\end{aligned}$
Do thể tích V của tứ diện MABC bằng 3 nên:
$\dfrac{1}{3}.d\left( M,\left( ABC \right) \right).{{S}_{\Delta ABC}}=\dfrac{1}{3}.\dfrac{\left| 4t+11 \right|}{3}.\dfrac{9}{2}=3\Leftrightarrow \left| 4t+11 \right|=6\Leftrightarrow \left[ \begin{aligned}
& t=-\dfrac{5}{4}\Rightarrow M\left( -\dfrac{3}{2};-\dfrac{3}{4};\dfrac{1}{2} \right) \\
& t=-\dfrac{17}{4}\Rightarrow M\left( -\dfrac{15}{2};\dfrac{9}{4};-\dfrac{11}{2} \right) \\
\end{aligned} \right..$
Cách 2: Ta có $\overrightarrow{AB}=\left( 2;1;2 \right);\ \overrightarrow{AC}=\left( -2;2;1 \right)\Rightarrow \left[ \overrightarrow{AB},\overrightarrow{AC} \right]=\left( -3;-6;6 \right)$.
Gọi $\begin{aligned}
& M\left( 1+2t;-2-t;3+2t \right)\in d\Rightarrow \overrightarrow{AM}=\left( 1+2t;-3-t;3+2t \right) \\
& \Rightarrow \left[ \overrightarrow{AB},\overrightarrow{AC} \right].\overrightarrow{AM}=-3\left( 1+2t \right)-6\left( -3-t \right)+6\left( 3+2t \right)=12t+33. \\
\end{aligned}$
Vì ${{V}_{MABC}}=\dfrac{1}{6}\left| \left[ \overrightarrow{AB},\overrightarrow{AC} \right].\overrightarrow{AM} \right|=3$ nên $\left| 12t+33 \right|=18\Leftrightarrow \left[ \begin{aligned}
& t=-\dfrac{5}{4}\Rightarrow M\left( -\dfrac{3}{2};-\dfrac{3}{4};\dfrac{1}{2} \right) \\
& t=-\dfrac{17}{4}\Rightarrow M\left( -\dfrac{15}{2};\dfrac{9}{4};-\dfrac{11}{2} \right) \\
\end{aligned} \right..$
Do $\left[ \overrightarrow{AB},\overrightarrow{AC} \right]=\left( -3;-6;6 \right)$ nên ${{S}_{\Delta ABC}}=\dfrac{1}{2}\left| \left[ \overrightarrow{AB},\overrightarrow{AC} \right] \right|=\dfrac{9}{2}.$
Gọi $\overrightarrow{n}$ là một vectơ pháp tuyến của mặt phẳng $\left( ABC \right)$ thì $\overrightarrow{n}=\left( 1;2;-2 \right)$ cùng phương với $\left[ \overrightarrow{AB},\overrightarrow{AC} \right]\Rightarrow $ Phương trình mặt phẳng $\left( ABC \right)$ là $x+2y-2z-2=0$.
Gọi $\begin{aligned}
& M\left( 1+2t;-2-t;3+2t \right)\in d \\
& \Rightarrow d\left( M,\left( ABC \right) \right)=\dfrac{\left| \left( 1+2t \right)+2\left( -2-t \right)-2\left( 3+2t \right)-2 \right|}{\sqrt{{{1}^{2}}+{{2}^{2}}+{{\left( -2 \right)}^{2}}}}=\dfrac{\left| 4t+11 \right|}{3}. \\
\end{aligned}$
Do thể tích V của tứ diện MABC bằng 3 nên:
$\dfrac{1}{3}.d\left( M,\left( ABC \right) \right).{{S}_{\Delta ABC}}=\dfrac{1}{3}.\dfrac{\left| 4t+11 \right|}{3}.\dfrac{9}{2}=3\Leftrightarrow \left| 4t+11 \right|=6\Leftrightarrow \left[ \begin{aligned}
& t=-\dfrac{5}{4}\Rightarrow M\left( -\dfrac{3}{2};-\dfrac{3}{4};\dfrac{1}{2} \right) \\
& t=-\dfrac{17}{4}\Rightarrow M\left( -\dfrac{15}{2};\dfrac{9}{4};-\dfrac{11}{2} \right) \\
\end{aligned} \right..$
Cách 2: Ta có $\overrightarrow{AB}=\left( 2;1;2 \right);\ \overrightarrow{AC}=\left( -2;2;1 \right)\Rightarrow \left[ \overrightarrow{AB},\overrightarrow{AC} \right]=\left( -3;-6;6 \right)$.
Gọi $\begin{aligned}
& M\left( 1+2t;-2-t;3+2t \right)\in d\Rightarrow \overrightarrow{AM}=\left( 1+2t;-3-t;3+2t \right) \\
& \Rightarrow \left[ \overrightarrow{AB},\overrightarrow{AC} \right].\overrightarrow{AM}=-3\left( 1+2t \right)-6\left( -3-t \right)+6\left( 3+2t \right)=12t+33. \\
\end{aligned}$
Vì ${{V}_{MABC}}=\dfrac{1}{6}\left| \left[ \overrightarrow{AB},\overrightarrow{AC} \right].\overrightarrow{AM} \right|=3$ nên $\left| 12t+33 \right|=18\Leftrightarrow \left[ \begin{aligned}
& t=-\dfrac{5}{4}\Rightarrow M\left( -\dfrac{3}{2};-\dfrac{3}{4};\dfrac{1}{2} \right) \\
& t=-\dfrac{17}{4}\Rightarrow M\left( -\dfrac{15}{2};\dfrac{9}{4};-\dfrac{11}{2} \right) \\
\end{aligned} \right..$
Đáp án A.