Set vector groups a1, a2... An is linearly related and a1 is not equal to zero proves that there is a vector ak (2 Don't paste the problem in Baidu, that is not very detailed, anyway I do n' t understand... Let the vector group a1, a2... An is linearly related and a1 is not equal to zero proves that there is a vector ak (2 Don't paste the problem in Baidu, that is not very detailed, anyway I do n' t understand... Let the vector group a1, a2... An is linearly related, and a1 is not equal to zero proves that there is a vector ak (2 Don't paste the problem in Baidu, that is not very detailed, anyway I do n' t understand...

Set vector groups a1, a2... An is linearly related and a1 is not equal to zero proves that there is a vector ak (2 Don't paste the problem in Baidu, that is not very detailed, anyway I do n' t understand... Let the vector group a1, a2... An is linearly related and a1 is not equal to zero proves that there is a vector ak (2 Don't paste the problem in Baidu, that is not very detailed, anyway I do n' t understand... Let the vector group a1, a2... An is linearly related, and a1 is not equal to zero proves that there is a vector ak (2 Don't paste the problem in Baidu, that is not very detailed, anyway I do n' t understand...

Vector group a1, a2... Am linear correlation,
There is λ1,λ2,...λm is not all zero, which is satisfied,
λ1A1+λ2a2+. am=0.1
1)λ2,..λM is not all zero,
Otherwise,λ2=.=λm=0,λ1=0, substitute 1:
λ1 A1=0, a1=0, then λ1=0, contradictory.
2)λK=0,(2≤k≤m),
According to formula 1:λkak=-(λ1a1+λ2a2+. mam)
Ak =-(1/λk)(λ1a1+λ2a2+. am)
A k can be expressed linearly by a 1, a 2, am.
The above with your problem a little error!
K

If plane vector B and vector A =(−1,2) is 180°, and B=3 5, Then B =() A.(-3,6) B.(3,-6) C.(6,-3) D.(-6,3)

Set

B=λ

A =(-λ,2λ)(λ<0),
∵|

B |=3
5,
∴(-λ)2+(2λ)2=45
∴λ2=9
∵λ<0,∴λ=-3
∴

B=(3,-6)
Therefore, B.

Given vector a=(cosα, sinα), vector b=(cosβ, sinβ),|a-b|=2 root number 5/5 1. Find the value of cos (α-β)? 2. If 0<α<π/2,-π<β<0, sinβ=-5/13, find the value of sinα?

Analysis: a-b|=2√5/5,
A^2-2a.b+b^2=4/5
Also a^2=│a│^2=(cosα)^2+(sinα)^2=1
B^2=│b│^2=(cosβ)^2+(sinβ)^2=1,
A.b=3/5
Cos (α-β)= cos α cos β+ sin α sin β= a.b =3/5
∵-π<β<0,0<α<π/2,
0<α-β<3π/2,且Cos(α-β)=3/5>0
Then 0<α-β<π/2,-π/2<β<0
Sinβ=-5/13, cosβ=12/13
12 Cosa-5 sina =39/5
Simultaneous (cosα)^2+(sinα)^2=1,
Solve sinα=(3√46+15)/65

Known vector A =(cosθ, sinθ), vector B =( 3,−1), Then |2 A− B|, the minimum values are ___.

2

A-

B=(2 cosθ-
3,2Sin 1),|2

A-

B |=
(2 Cosθ-
3)2+(2Sin 1)2=
8+4Sinθ-4
3 Cosθ=
8+8S in (θ-π
3),
4 Max,0 min
Therefore, the answer is:4,0.

Given vector a=(cosα, sinα), b=(root number 3,1),(0,π), and a⊥b, then α is equal to () Given vector a=(cosα, sinα), b=(root 3,1),(0,π), and a⊥b, then α is equal to ()

A▪b=sin Root 3 cosα=2 s in (/3)=0
So /3=π, so α=2π/3

Given vector a=(cos⊙, sin⊙) vector b=(root 3,-1) The maximum value of /2a-b vector / is?

Vector a=(cos⊙, sin⊙) Vector b=(root 3,-1)
Vector (2a-b)=(2 cos 3,2 sin 1),
|2A-b vector |=√[(2cos⊙-√3)^2+(2sin⊙+1)^2]
=√[12+4*(Sin ⊙-√3 cos⊙)]
=2√[3+2(Sin 1/2-cos 3/2)]
=2√[3+2*Sin (⊙-∏/3)],
Only when sin (⊙-∏/3)=1,|2a vector-b vector| has the maximum value,
|2A-b vector |max=2√(3+2)=2√5.