Question

数学题目，请你们帮帮忙，谢谢你们，超级谢谢！！

Answer 1

解设c=(x，y)则x^2+y^2=20
ac=2x+4y，bc=4x+2y
cos<a，c>=ac/|a||c|=(2x+4y)/20
同理，得cos<b，c>=(2y+4y)/20
<a，c>=﹤b，c>则
2x+4y=2y+4x，则x=y
则x=y=±√10
则c(√10，√10)或(-√10，-√10)

Answer 2

(I)
(AB-BC).(AB+AC)=-1
AB.AB +AB.(AC-BC) -BC.AC =-1

|AB|^2 +AB.AB -BC.AC =-1
2|AB|^2 -BC.AC =-1
2|AB|^2 -BC.(AB+BC) =-1
2|AB|^2 -BC.AB- |BC|^2 =-1
2|AB|^2 +|BC||AB|cosB- |BC|^2 =-1
2+2cosB -4=-1
cosB = 1/2
B=π/3
(II)
tan(B+θ)=-3√3
(tanB+tanθ)/(1-tanB.tanθ)=-3√3
(√3 +tanθ)/(1-√3.tanθ)=-3√3
√3 +tanθ =-3√3 +9tanθ
8tanθ = 4√3
tanθ = √3/2