Question

已知向量a=(cosa,sina) b=(cosb,sinb) 且a不等于正负b,那么a+b与a-b的夹角的大小为

Answer 1

a+b=(cosa+cosb,sina+sinb)

a-b=(cosa-cosb,sina-sinb)

cos(a+b,a-b)=(a+b)*(a-b)/|a+b|*|a-b|

=[(cosa+cosb)(cosa-cosb)+(sina+sinb)(sina-sinb)]/√[(cosa+cosb)^2+(sina+sinb)^2]*√[(cosa-cosb)^2+(sina-sinb)^2]

=(cos^2 a -cos^2 b +sin^2 a -sin^2 b)/ √[(cosa+cosb)^2+(sina+sinb)^2]*√[(cosa-cosb)^2+(sina-sinb)^2]

=(1-1)/√[(cosa+cosb)^2+(sina+sinb)^2]*√[(cosa-cosb)^2+(sina-sinb)^2]

=0

所以夹角为90度