苏棉
春芽
共回答了13个问题采纳率:76.9% 举报
Ia+bI=√(a+b)²
=√[IaI²+2IaIIbIcos120°+IbI²]
=√[4-2*2*2*(1/2)+4]
=2
Ia﹣bI=√(a-b)²
=√[IaI²-2IaIIbIcos120°+IbI²]
=√[4+2*2*2*(1/2)+4]
=2√3
设向量a+向量b与向量a﹣向量b的夹角为x
所以(a+b)(a-b)=IaI²-IbI²
=Ia+bI*Ia-bIcosx
即2*2√3cosx=4-4=0
所以cosx=0
x=90°
1年前
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