Vector

(a) AB = OB - OA = λ( u + v ) - ( 4u - v )
AB = (λ- 4 ) u + (λ+ 1 ) v


AC = OC - OA = u + 2v - ( 4u - v )
AC = -3u +3v


(b) B為AC上的一點
AB // AC
AB = kAC where k is a constant
(λ- 4 ) u + (λ+ 1 ) v = k ( -3u + 3v )
λ- 4 = -3k . . . . (1)
λ+ 1 = 3k . . . . (2)
(1)/(2): (λ- 4 ) / ( λ+ 1 ) = -1
4 -λ= λ+ 1
2λ= 3
λ= 3/2

(a) AB=OB-OA=λ(u+v)-(4u-v)=(λ-4)u+(λ+1)v

AC=OC-OA=(u+2v)-(4u-v)=-3u+3v

(b)

B為AC上的一點

AB//AC=>(λ-4)u+(λ+1)v=m(-3u+3v)

=>λ-4=-(λ+1)=>λ=3/2