Question

已知e1,e2,是夹角为60°的两个单位向量,设向量a=2*e1+e2,b=-3*e1+2*e2,求a与b的夹角

Answer 1

e1*e2=/e1//e2/cos60°=0.5

a1*a2=(2*e1+e2)(-3*e1+2*e2)

=-6e1^2+e1*e2+2e2^2

=-6+0.5+2=-3.5

/a/^2=a^2 ==(2*e1+e2)=(2*e1+e2)

/b/^2=b^2 )=(-3*e1+2*e2)(-3*e1+2*e2)

a*b=/a/*/b/*cosa

cosa=a*b/ (/a/*/b/)

a与b的夹角a=arccos[a*b/ (/a/*/b/)]

具体自己算一下