向量AB与AC的夹角为30度,向量AC的模为2,AB-AC的模为根号2,求向量AB与AB-AC的夹角.

题目

向量AB与AC的夹角为30度,向量AC的模为2,AB-AC的模为根号2,求向量AB与AB-AC的夹角.
老师说有2种情况,怎么也算不出来

答案

设x为AB与AB-AC的夹角
|AC|=2
|AB-AC|=√2
(AB-AC).(AB-AC)=2
|AB|^2-2|AC||AB|cos30°+|AC|^2 = 2
|AB|^2-2√3|AB|+4 = 2
|AB|^2-2√3|AB|+2 = 0
|AB| = √3 + 1 or √3 - 1
AB.(AB-AC) = |AB|^2 - |AB||AC|cos30°
|AB||AB-BC|cosx = |AB|^2 - |AB||AC|cos30°
√2cosx= |AB|-√3
cosx =√2/2 or -√2/2
x = 45° or 135°
AB与AB-AC的夹角=45° or 135°