已知单位向量e1,e2的夹角为π/3,求向量a=e1+e2,b=e2-2e1的夹角

Question

已知单位向量e1,e2的夹角为π/3,求向量a=e1+e2,b=e2-2e1的夹角

Answer 1

∵单位向量e1,e2的夹角为π/3
∴|e1|=|e2|=1 ,
e1●e2=1*1*cosπ/3=1/2
a=e1+e2,b=e2-2e1
a●b=(e1+e2)●(e2-2e1)
=-e1●e2+|e2|-2|e1|
=-1/2+1-2
=-3/2
|a|²=|e1+e2|²ji
=|e1|²+|e2|²+2e1●e2
=1+1+1=3
∴|a|=√3
|b|²=|e2-2e1|²
=|e2|²|+4|e1|²-4e1●e2
=5-4×1/2
=3
∴|b|=√3
∴cos=(a●b)/(|a||b|)=(-3/2)/(√3²)=-1/2
∴=2π/3
即a,b夹角为2π/3