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a) Ta có $M\in mp(ABC)\Leftrightarrow \overrightarrow{AB},\overrightarrow{AC},\overrightarrow{AM} $ đồng phẳng $\Leftrightarrow [\overrightarrow{AB},\overrightarrow{AC} ].\overrightarrow{AM}=0 (*) $
$\overrightarrow{AB}=(2;-1;-4), \overrightarrow{AC}=(1;0;-3), \overrightarrow{AM}=(x; y-1;z-3) $
$[\overrightarrow{AB},\overrightarrow{AC} ]=\left( {\left| \begin{array}{l}
- 1\\
0
\end{array} \right.\,\,\,\,\left. \begin{array}{l}
4\\
- 3
\end{array} \right|;\left| \begin{array}{l}
- 4\\
3
\end{array} \right.\,\,\,\,\,\left. \begin{array}{l}
2\\
1
\end{array} \right|;\left| \begin{array}{l}
2\\
1
\end{array} \right.\,\,\,\,\,\left. \begin{array}{l}
- 1\\
0
\end{array} \right|} \right) = (3;2;1)$
$(*)\Leftrightarrow 3x+2y-2+z-3=0\Leftrightarrow 3x+2y+z-5=0$
b) Ta có $\left\{ \begin{array}{l}
\overrightarrow {AH} \bot \overrightarrow {BC} \\
\overrightarrow {BH} \bot \overrightarrow {AC} \\
H \in mp(ABC)
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
\overrightarrow {AH} \bot \overrightarrow {BC} = 0\\
\overrightarrow {BH} \bot \overrightarrow {AC} = 0\\
\overrightarrow {AB} ,\,\overrightarrow {AC} ,\,\overrightarrow {AH} đồng phẳng
\end{array} \right. (**)$
Với $\overrightarrow{BH}=(x-2;y;z+1) $
$(**) \Leftrightarrow \left\{ \begin{array}{l}
- x + y - 1 + z - 3 = 0\\
x - 2 - 3z - 3 = 0\\
3x + 2y + z - 5 = 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x - y - z = - 4\\
x - 3z = 5\\
3x + 2y + z = 5
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x = - 1\\
y = 5\\
z = - 2
\end{array} \right.$
Vậy $H(-1;5;-2)$
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Đăng bài 29-05-12 10:52 AM
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