Question

已知向量a=(1,2),b=(cosα,sinα),设m=a+tb（t为实数）,若a⊥b且a-b与m的夹角为π/4,则t

Answer 1

a⊥b,则a*b＝0

|a-b|^2＝(a-b)*(a-b)＝|a|^2+|b|^2＝5+1＝6,|a-b|＝√6

|a+tb|^2＝(a+tb)*(a+tb)＝|a|^2+t^2×|b|^2＝5+t^2,|a+tb|＝√(5+t^2)

(a-b)*m＝(a-b)*(a+tb)＝|a|^2-t|b|^2＝5-t

a-b与m的夹角为π/4,则cos(π/4)＝[(a-b)*(a+tb)]/|a-b|*|a+tb|]＝(5-t)/[√6*(5+t^2),t＝(-5±3√5)/2