定义平面向量之间的一种运算" $\odot$"如下，对任意的 $\stackrel{\rightharpoonup }{a}=(m,n)$， $\stackrel{\rightharpoonup }{b}=(p,q)$，令 $\stackrel{\rightharpoonup }{a}\odot \stackrel{\rightharpoonup }{b}=mq-np$，下面说法错误的是（   ）

A．	若 $\stackrel{\rightharpoonup }{a}$与 $\stackrel{\rightharpoonup }{b}$共线，则 $\stackrel{\rightharpoonup }{a}\odot \stackrel{\rightharpoonup }{b}=0$	B．	$\stackrel{\rightharpoonup }{a}\odot \stackrel{\rightharpoonup }{b}=\stackrel{\rightharpoonup }{b}\odot \stackrel{\rightharpoonup }{a}$
C．	对任意的 $\lambda \in R$，有 $\left(\lambda \stackrel{\rightharpoonup }{a}\right)\odot \stackrel{\rightharpoonup }{b}=\lambda (\stackrel{\rightharpoonup }{a}\odot \stackrel{\rightharpoonup }{b})$	D．	$(\stackrel{\rightharpoonup }{a}\odot \stackrel{\rightharpoonup }{b}{)}^2+(\stackrel{\rightharpoonup }{a}·\stackrel{\rightharpoonup }{b}{)}^2=\left|\stackrel{\rightharpoonup }{a}\right|·{\left|\stackrel{\rightharpoonup }{b}\right|}^2$