a) Sai. $overset{rightarrow}{AB} + overset{rightarrow}{CD} = (overset{rightarrow}{AP} + overset{rightarrow}{PQ} + overset{rightarrow}{QB}) + (overset{rightarrow}{CP} + overset{rightarrow}{PQ} + overset{rightarrow}{QD}) = (overset{rightarrow}{AP} + overset{rightarrow}{CP}) + 2overset{rightarrow}{PQ} + (overset{rightarrow}{QB} + overset{rightarrow}{QD}) = 2overset{rightarrow}{PQ}$
b) Sai: Ta có: $left. overset{rightarrow}{AB}(overset{rightarrow}{AB} – overset{rightarrow}{CA}) = overset{rightarrow}{AB} cdot overset{rightarrow}{AB} + overset{rightarrow}{AB} cdot overset{rightarrow}{AC} = {overset{rightarrow}{AB}}^{2} + middle| overset{rightarrow}{AB} middle| cdot middle| overset{rightarrow}{AC} middle| cdot cos(overset{rightarrow}{AB},overset{rightarrow}{AC}) right.$
$= AB^{2} + AB cdot AC cdot cos(widehat{BAC}) = a^{2} + a cdot a cdot cos 60^{{^circ}} = a^{2} + dfrac{a^{2}}{2} = dfrac{3a^{2}}{2}.$
c) Đúng. Ta có $overset{rightarrow}{GA} + overset{rightarrow}{GB} = 2overset{rightarrow}{GM};overset{rightarrow}{GC} + overset{rightarrow}{GD} = 2overset{rightarrow}{GN};overset{rightarrow}{GM} + overset{rightarrow}{GN} = overset{rightarrow}{0}$
Suy ra: $overset{rightarrow}{GA} + overset{rightarrow}{GB} + overset{rightarrow}{GC} + overset{rightarrow}{GD} = overset{rightarrow}{0}$ hay $overset{rightarrow}{GA} + overset{rightarrow}{GB} + overset{rightarrow}{GC} = – overset{rightarrow}{GD}$.
d) Góc giữa $overset{rightarrow}{AC}$ và $overset{rightarrow}{MN}$ là: $60^{{^circ}}$
d) Sai:Ta có $cos(overset{rightarrow}{AC},overset{rightarrow}{MN}) = dfrac{overset{rightarrow}{AC} cdot overset{rightarrow}{MN}}{left| overset{rightarrow}{AC} middle| cdot middle| overset{rightarrow}{MN} right|}$.
$left. overset{rightarrow}{MN} = dfrac{1}{2}(overset{rightarrow}{AC} + overset{rightarrow}{BD})Rightarrow middle| overset{rightarrow}{MN} middle| {}_{2} = {overset{rightarrow}{MN}}^{2} = dfrac{1}{4}left( {{overset{rightarrow}{AC}}^{2} + 2overset{rightarrow}{AC} cdot overset{rightarrow}{BD} + {overset{rightarrow}{BD}}^{2}} right) = dfrac{1}{4}left( {2a^{2} + 2overset{rightarrow}{AC} cdot overset{rightarrow}{BD}} right) right.$
Mà $left. overset{rightarrow}{AC} cdot overset{rightarrow}{BD} = overset{rightarrow}{AC} cdot (overset{rightarrow}{AD} – overset{rightarrow}{AB}) = overset{rightarrow}{AC} cdot overset{rightarrow}{AD} – overset{rightarrow}{AC} cdot overset{rightarrow}{AB} = middle| overset{rightarrow}{AC} middle| cdot middle| overset{rightarrow}{AD} middle| cos 60^{{^circ}} – middle| overset{rightarrow}{AC} middle| cdot middle| overset{rightarrow}{AB} middle| cdot cos 60^{{^circ}} = 0 right.$
Vậy $left. {overset{rightarrow}{MN}}^{2} = dfrac{1}{4}.2a^{2} = dfrac{a^{2}}{2}Rightarrowleft| overset{rightarrow}{MN} right| = dfrac{a}{sqrt{2}} right.$.
Ta có $overset{rightarrow}{AC} cdot overset{rightarrow}{MN} = dfrac{1}{2}overset{rightarrow}{AC} cdot (overset{rightarrow}{AC} + overset{rightarrow}{BD}) = dfrac{1}{2}left( {{overset{rightarrow}{AC}}^{2} + overset{rightarrow}{AC} cdot overset{rightarrow}{BD}} right) = dfrac{a^{2}}{2}$.
$left. cos(overset{rightarrow}{AC},overset{rightarrow}{MN}) = dfrac{overset{rightarrow}{AC} cdot overset{rightarrow}{MN}}{left| overset{rightarrow}{AC} middle| cdot middle| overset{rightarrow}{MN} right|} = dfrac{dfrac{a^{2}}{2}}{a cdot dfrac{a}{sqrt{2}}} = dfrac{sqrt{2}}{2}Rightarrow(overset{rightarrow}{AC},overset{rightarrow}{MN}) = 45^{{^circ}} right.$