cos( véc tơ a, véc tơ b) có bằng cos( đường thẳng a, đường thẳng b) không 
Được , với điều kiện đó là vtpt nha bạn
Tiện mk cho luôn CT nè : 
Góc giữa hai đt : $\Delta :A_{1}x+B_{1}y+C_{1}=0$ và : $\Delta '=A_{2}x+B_{2}y+C_{2}=0$ :
 $cos(\Delta ,\Delta ')=|cos(\overrightarrow{n_{1}},\overrightarrow{n_{2}})|=\frac{|A_{1}A_{2}+B_{1}B_{2}|}{\sqrt{A^{2}_{1}+B^{2}_{1}}.\sqrt{A^{2}_{2}+B^{2}_{2}}}$
Trong đó : $\overrightarrow{n_{1}};\overrightarrow{n_{2}}$ là hai vtpt của 2 đt $\Delta và \Delta '$
À ý mình là giả sử vầy nha cos(AB,BD)=cos(vt AB,vt BD)=(vt AB x vt BD)/( AB x BD) đc k pn –  NTMT15399 07-04-16 09:23 PM

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