1. Cho hình chóp $S.ABCD.$ Gọi $G$ là trọng tâm tam giác $ABC,\,\overrightarrow{a}=\overrightarrow{BC},\,\overrightarrow{b}=\overrightarrow{SB},\,\overrightarrow{c}=\overrightarrow{SG}.$ Phân tích $\overrightarrow{SA}$ theo $\overrightarrow{a},\,\overrightarrow{b},\,\overrightarrow{c}.$

2. Cho tứ diện $ABCD;\,\overrightarrow{b}=\overrightarrow{AB},\,\overrightarrow{c}=\overrightarrow{AC},\,\overrightarrow{d}=\overrightarrow{AD}.$
      a) Biểu diễn $\overrightarrow{BC},\,\overrightarrow{CD},\,\overrightarrow{DB}$ theo $\overrightarrow{b},\,\overrightarrow{c},\,\overrightarrow{d}.$
      b) $M$ là trung điểm $BC$. Biểu diễn $\overrightarrow{DM}$ theo $\overrightarrow{b},\,\overrightarrow{c},\,\overrightarrow{d}.$
      c) $G$ là trọng tâm tam giác $BCD.$ Biểu diễn $\overrightarrow{AG}$ theo $\overrightarrow{b},\,\overrightarrow{c},\,\overrightarrow{d}.$
2 bài này cô mình đều cho! –  kellyhoang297 11-03-13 09:02 PM
2/
a/$\overrightarrow{BC}=\overrightarrow{BA}+\overrightarrow{AC}=\overrightarrow{c}-\overrightarrow{b}$
$\overrightarrow{CD}=\overrightarrow{CA}+\overrightarrow{AD}=\overrightarrow{d}-\overrightarrow{c}$
$\overrightarrow{DB}=\overrightarrow{DA}+\overrightarrow{AB}=\overrightarrow{b}-\overrightarrow{d}$
b/
$\overrightarrow{DM}=\overrightarrow{DC}+\overrightarrow{CM}=\overrightarrow{DC}+\frac{1}{2}\overrightarrow{CB}$
$=(\overrightarrow{DA}+\overrightarrow{AC})+\frac{1}{2}(\overrightarrow{CA}+\overrightarrow{AB})$ 
$=\frac{1}{2}\overrightarrow{b} +\frac{1}{2}\overrightarrow{c}-\overrightarrow{d}$ 
c/
$3\overrightarrow{AG}=(\overrightarrow{AB}+\overrightarrow{BG})+(\overrightarrow{AC}+\overrightarrow{CG})+(\overrightarrow{AD}+\overrightarrow{DG})$ 
$\Leftrightarrow 3\overrightarrow{AG}=\overrightarrow{b}+\overrightarrow{c}+\overrightarrow{d}$ 
$\Rightarrow \overrightarrow{AG}=\frac{1}{3}(\overrightarrow{b}+\overrightarrow{c}+\overrightarrow{d})$  
nếu phát hiện lỗi hay thắc mắc bạn hãy để lại tin nhắn, mình sẽ giải quyết ngay khi rảnh. Thanks :) –  gunbk92 12-03-13 12:20 AM
1/
$\overrightarrow{SA}+\overrightarrow{SB}+\overrightarrow{SC}=(\overrightarrow{SG}+\overrightarrow{GA})+(\overrightarrow{SG}+\overrightarrow{GB})+(\overrightarrow{SG}+\overrightarrow{GC})$ 
$\Leftrightarrow \overrightarrow{SA}+\overrightarrow{SB}+(\overrightarrow{SB}+\overrightarrow{BC})=3\overrightarrow{SG}$
$\Leftrightarrow \overrightarrow{SA}=3\overrightarrow{c}-2\overrightarrow{b}-\overrightarrow{a}$ 

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