AB=e1+e2,BC=2e1+8e2,CD=3(e1-e2)
(1)
AB = e1+e2
OB-OA = e1+e2 (1)
BC = 2e1+8e2
OC-OB = 2e1+8e2 (2)
CD=3(e1-e2)
OD-OC = 3(e1-e2) (3)
(2)+(3)
OD -OB = 5e1+5e2
= 5(OB-OA) ( from (1))
6OB = OD+5OA
OB = (OD+5OA)/6
=> A,B,D 三點共線 such that
|AB|/| BD| = 1:5
(2)
ke1+e2 和 e1+ke2 共線
=> ke1+e2 = m(e1+ke2) ( m 是不等于0 的常數(shù))
=> ke1+ ke2 = me1 + mke2
=> k = m and k=mk (e1與e2不共線)
=> k = k^2
=>k(k-1) =0
=> k= 1
or k=0 ( rejected)
ie k=1 ,ke1+e2 和 e1+ke2 共線