Question

已知a,b夹角为θ,且a=(3,4),a+2b=(5,8),则cosθ=

Answer 1

∵a=(3,4),a+2b=(5,8)

∴2b=a+2b-a=(5,8)-(3,4)=(2,4)

∴b=(1,2)

又a*b=|a|*|b|*cosθ

∴ cosθ=a*b/(|a|*|b|)

=(3×1+4×2) /[√(3²+4²)*√(1²+2²)]

=11√5/25