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Câu hỏi: Trong không gian $Oxyz,$ cho điểm $M\left( 3;-4;-5 \right)$ và các đường thẳng ${{d}_{1}}:\dfrac{x+4}{-5}=\dfrac{y-4}{2}=\dfrac{z-2}{3};{{d}_{2}}:\dfrac{x-1}{-1}=\dfrac{y-2}{3}=\dfrac{z+5}{-2}.$ Đường thẳng $d$ đi qua $M$ và cắt ${{d}_{1}},{{d}_{2}}$ lần lượt tại $A,B$. Diện tích tam giác $OAB$ bằng
A. $5\sqrt{3}$
B. $\dfrac{3\sqrt{5}}{2}$
C. $3\sqrt{5}$
D. $\dfrac{5\sqrt{3}}{2}$
A. $5\sqrt{3}$
B. $\dfrac{3\sqrt{5}}{2}$
C. $3\sqrt{5}$
D. $\dfrac{5\sqrt{3}}{2}$
Phương pháp:
- Tham số hóa tọa độ điểm $A,B.$
- Sử dụng điều kiện $M,A,B$ thẳng hàng tìm tọa độ điểm $A,B.$
- Sử dụng công thức ${{S}_{OAB}}=\dfrac{1}{2}\left| \left[ \overrightarrow{OA},\overrightarrow{OB} \right] \right|.$
Cách giải:
Gọi $A\left( -4-5a;4+2a;2+3a \right)\in {{d}_{1}},B\left( 1-b;2+3b;-5-2b \right)\in {{d}_{2}}.$
Vì $M,A,B\in d$ nên chúng thẳng hàng $\Rightarrow \overrightarrow{MA},\overrightarrow{MB}$ cùng phương.
Ta có: $\left\{ \begin{aligned}
& \overrightarrow{MA}=\left( 5a+7;-2a-8;-3a-7 \right) \\
& \overrightarrow{MB}=\left( b+2;-3b-6;2b \right) \\
\end{aligned} \right.$
$\Rightarrow \dfrac{5a+7}{b+2}=\dfrac{-2a-8}{-3b-6}=\dfrac{-3a-7}{2b}$
$\Leftrightarrow \left\{ \begin{aligned}
& 15ab+30a+21b+42=2ab+8b+4a+16 \\
& 10ab+14b=-3ab-6a-7b-14 \\
\end{aligned} \right.$
$\Leftrightarrow \left\{ \begin{aligned}
& 13ab+26a+13b+26=0 \\
& 13ab+6a+21b+14=0 \\
\end{aligned} \right.$
$\Leftrightarrow \left\{ \begin{aligned}
& 20a-8b+12=0 \\
& ab+2a+b+2=0 \\
\end{aligned} \right.$
$\Leftrightarrow \left\{ \begin{aligned}
& 5a-2b+3=0 \\
& ab+2a+b+2=0 \\
\end{aligned} \right.$
$\Leftrightarrow \left\{ \begin{aligned}
& b=\dfrac{5a+3}{2} \\
& a.\dfrac{5a+3}{2}+2a+\dfrac{5a+3}{2}+2=0 \\
\end{aligned} \right.$
$\Leftrightarrow \left\{ \begin{aligned}
& b=\dfrac{5a+3}{2} \\
& 5{{a}^{2}}+3a+4a+5a+3+4=0 \\
\end{aligned} \right.$
$\Leftrightarrow \left\{ \begin{aligned}
& b=\dfrac{5a+3}{2} \\
& 5{{a}^{2}}+12a+7=0 \\
\end{aligned} \right.$
$\Leftrightarrow \left[ \begin{aligned}
& a=-1,b=-1 \\
& a=-\dfrac{7}{5},b=-2 \\
\end{aligned} \right.$
$\Rightarrow \left[ \begin{aligned}
& A\left( 1;2;-1 \right),B\left( 2;-1;-3 \right) \\
& A\left( 3;\dfrac{6}{5};-\dfrac{11}{5} \right),B\left( 3;-4;-1 \right) \\
\end{aligned} \right.$
TH1: $A\left( 1;2;-1 \right),B\left( 2;-1;-3 \right)\Rightarrow \overrightarrow{OA}\left( 1;2;-1 \right),\overrightarrow{OB}\left( 2;-1;-3 \right)$
$\Rightarrow {{S}_{OAB}}=\dfrac{1}{2}\left| \left[ \overrightarrow{OA},\overrightarrow{OB} \right] \right|=\dfrac{1}{2}\sqrt{{{3}^{2}}+{{6}^{2}}+{{0}^{2}}}=\dfrac{3\sqrt{5}}{2}$
TH2: $A\left( 3;\dfrac{6}{5};-\dfrac{11}{5} \right),B\left( 3;-4;-1 \right)\Rightarrow \overrightarrow{OA}\left( 3;\dfrac{6}{5};-\dfrac{11}{5} \right),\overrightarrow{OB}\left( 3;-4;-1 \right)$
$\Rightarrow {{S}_{OAB}}=\dfrac{1}{2}\left| \left[ \overrightarrow{OA},\overrightarrow{OB} \right] \right|=\dfrac{1}{2}.\sqrt{{{10}^{2}}+3,{{6}^{2}}+15,{{6}^{2}}}=\dfrac{\sqrt{8908}}{10}$.
- Tham số hóa tọa độ điểm $A,B.$
- Sử dụng điều kiện $M,A,B$ thẳng hàng tìm tọa độ điểm $A,B.$
- Sử dụng công thức ${{S}_{OAB}}=\dfrac{1}{2}\left| \left[ \overrightarrow{OA},\overrightarrow{OB} \right] \right|.$
Cách giải:
Gọi $A\left( -4-5a;4+2a;2+3a \right)\in {{d}_{1}},B\left( 1-b;2+3b;-5-2b \right)\in {{d}_{2}}.$
Vì $M,A,B\in d$ nên chúng thẳng hàng $\Rightarrow \overrightarrow{MA},\overrightarrow{MB}$ cùng phương.
Ta có: $\left\{ \begin{aligned}
& \overrightarrow{MA}=\left( 5a+7;-2a-8;-3a-7 \right) \\
& \overrightarrow{MB}=\left( b+2;-3b-6;2b \right) \\
\end{aligned} \right.$
$\Rightarrow \dfrac{5a+7}{b+2}=\dfrac{-2a-8}{-3b-6}=\dfrac{-3a-7}{2b}$
$\Leftrightarrow \left\{ \begin{aligned}
& 15ab+30a+21b+42=2ab+8b+4a+16 \\
& 10ab+14b=-3ab-6a-7b-14 \\
\end{aligned} \right.$
$\Leftrightarrow \left\{ \begin{aligned}
& 13ab+26a+13b+26=0 \\
& 13ab+6a+21b+14=0 \\
\end{aligned} \right.$
$\Leftrightarrow \left\{ \begin{aligned}
& 20a-8b+12=0 \\
& ab+2a+b+2=0 \\
\end{aligned} \right.$
$\Leftrightarrow \left\{ \begin{aligned}
& 5a-2b+3=0 \\
& ab+2a+b+2=0 \\
\end{aligned} \right.$
$\Leftrightarrow \left\{ \begin{aligned}
& b=\dfrac{5a+3}{2} \\
& a.\dfrac{5a+3}{2}+2a+\dfrac{5a+3}{2}+2=0 \\
\end{aligned} \right.$
$\Leftrightarrow \left\{ \begin{aligned}
& b=\dfrac{5a+3}{2} \\
& 5{{a}^{2}}+3a+4a+5a+3+4=0 \\
\end{aligned} \right.$
$\Leftrightarrow \left\{ \begin{aligned}
& b=\dfrac{5a+3}{2} \\
& 5{{a}^{2}}+12a+7=0 \\
\end{aligned} \right.$
$\Leftrightarrow \left[ \begin{aligned}
& a=-1,b=-1 \\
& a=-\dfrac{7}{5},b=-2 \\
\end{aligned} \right.$
$\Rightarrow \left[ \begin{aligned}
& A\left( 1;2;-1 \right),B\left( 2;-1;-3 \right) \\
& A\left( 3;\dfrac{6}{5};-\dfrac{11}{5} \right),B\left( 3;-4;-1 \right) \\
\end{aligned} \right.$
TH1: $A\left( 1;2;-1 \right),B\left( 2;-1;-3 \right)\Rightarrow \overrightarrow{OA}\left( 1;2;-1 \right),\overrightarrow{OB}\left( 2;-1;-3 \right)$
$\Rightarrow {{S}_{OAB}}=\dfrac{1}{2}\left| \left[ \overrightarrow{OA},\overrightarrow{OB} \right] \right|=\dfrac{1}{2}\sqrt{{{3}^{2}}+{{6}^{2}}+{{0}^{2}}}=\dfrac{3\sqrt{5}}{2}$
TH2: $A\left( 3;\dfrac{6}{5};-\dfrac{11}{5} \right),B\left( 3;-4;-1 \right)\Rightarrow \overrightarrow{OA}\left( 3;\dfrac{6}{5};-\dfrac{11}{5} \right),\overrightarrow{OB}\left( 3;-4;-1 \right)$
$\Rightarrow {{S}_{OAB}}=\dfrac{1}{2}\left| \left[ \overrightarrow{OA},\overrightarrow{OB} \right] \right|=\dfrac{1}{2}.\sqrt{{{10}^{2}}+3,{{6}^{2}}+15,{{6}^{2}}}=\dfrac{\sqrt{8908}}{10}$.
Đáp án B.