Question

（1）如果向量AB=向量e1+向量e2,向量BC=2向量e1+8向量e2,向量CD=3（向量e1-e2）求证：A,B,D

Answer 1

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 三点共线 such that

|AB|/| BD| = 1:5

(2)

ke1+e2 和 e1+ke2 共线

=> ke1+e2 = m(e1+ke2) ( m 是不等于0 的常数)

=> ke1+ ke2 = me1 + mke2

=> k = m and k=mk (e1与e2不共线)

=> k = k^2

=>k(k-1) =0

=> k= 1

or k=0 ( rejected)

ie k=1 ,ke1+e2 和 e1+ke2 共线