已知矩形ABCD中，AB=3，BC=4，e1=（向量AB）/（向量AB模），e2=（向量AD）/（向量AD模）.

Question

(1)若向量AC=xe1+ye2,求想，x,y的值
(2)求向量AC与向量BD的夹角的余弦值

Answer 1

(1)设A（0,0）则B(3,0)；D(0,4)；C(3,4)
e1=(1,0);e2=(0,1);AC=(3,4)=3e1+4e2
x=3,y=4
(2)设夹角为M BD=(-3,4)
AC*BD=|AC|*|BD|*cosM
cosM=(3,4)*(-3,4)/5/5=7/25

Answer 2

x=3
y=4

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