Đáp án:
a)$\vec{AB}.\vec{DC}=a^2$
$\vec{AB}.\vec{OC}=\dfrac{a^2}{2}$
b)$(\vec{AB}+\vec{AD}).(\vec{BC}+\vec{BD})=a^2$
Giải thích các bước giải: Ta có :
$AC=BD=\sqrt{AB^2+BC^2}=\sqrt{a^2+a^2}=a\sqrt{2}$
a)$\vec{AB}.\vec{DC}=|\vec{AB}|.|\vec{AB}|.Cos (\vec{AB};\vec{DC})=a^2.Cos(0)=a^2$
$\vec{AB}.\vec{OC}=|\vec{AB}|.\vec{OC}|.Cos(\vec{AB};\vec{OC})=a.\dfrac{a\sqrt{2}}{2}.Cos 45^o=\dfrac{a^2}{2}$
b)$(\vec{AB}+\vec{AD}).(\vec{BC}+\vec{BD})$
$=\vec{AB}.\vec{BD}+\vec{AD}.\vec{BC}+\vec{AD}.\vec{BD}$
$=|\vec{AB}|.|\vec{BD}|.Cos(135^o)+|\vec{AD}|.|\vec{BC}|.Cos(0^o)+|\vec{AD}|.|\vec{BD}|.Cos(45^o)=a.a\sqrt{2}.\dfrac{-\sqrt{2}}{2}+a.a.1+a.a\sqrt{2}.\dfrac{\sqrt{2}}{2}=-a^2+a^2+a^2=a^2$
