(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三点共线(2)试确定实数k,使k向量e1+向量e2和e1+k向量e2共线
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![(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三点共线(2)试确定实数k,使k向量e1+向量e2和e1+k向量e2共线](/uploads/image/z/5174900-44-0.jpg?t=%EF%BC%881%EF%BC%89%E5%A6%82%E6%9E%9C%E5%90%91%E9%87%8FAB%3D%E5%90%91%E9%87%8Fe1%2B%E5%90%91%E9%87%8Fe2%2C%E5%90%91%E9%87%8FBC%3D2%E5%90%91%E9%87%8Fe1%2B8%E5%90%91%E9%87%8Fe2%2C%E5%90%91%E9%87%8FCD%3D3%EF%BC%88%E5%90%91%E9%87%8Fe1-e2%EF%BC%89%E6%B1%82%E8%AF%81%EF%BC%9AA%2CB%2CD%E4%B8%89%E7%82%B9%E5%85%B1%E7%BA%BF%EF%BC%882%EF%BC%89%E8%AF%95%E7%A1%AE%E5%AE%9A%E5%AE%9E%E6%95%B0k%2C%E4%BD%BFk%E5%90%91%E9%87%8Fe1%2B%E5%90%91%E9%87%8Fe2%E5%92%8Ce1%2Bk%E5%90%91%E9%87%8Fe2%E5%85%B1%E7%BA%BF)
(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三点共线(2)试确定实数k,使k向量e1+向量e2和e1+k向量e2共线
(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三点共线(2)试确定实数k,使k向量e1+向量e2和e1+k向量e2共线
(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三点共线(2)试确定实数k,使k向量e1+向量e2和e1+k向量e2共线
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
(1)∵AD=AB+BC+CD=6(e1+e2)=6AB
∴AB//AD
∵AB与AD有公共点A
∴A,B,D三点共线
(2)∵ke1+e2和e1+ke2共线,
∴存在λ使ke1+e2=λ(e1+ke...
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(1)∵AD=AB+BC+CD=6(e1+e2)=6AB
∴AB//AD
∵AB与AD有公共点A
∴A,B,D三点共线
(2)∵ke1+e2和e1+ke2共线,
∴存在λ使ke1+e2=λ(e1+ke2),
即(k-λ)e1+(1-λk)e2=0.
∵e1与e2为非零不共线向量,
∴k-λ=0且1-λk=0.
∴k=±1.
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