Known A , B , C Is three vectors in the same plane, where A =(1,2) (1) If | C |=2 5 , And C ∥ A , Seeking C Coordinates of; (2) If | B |= 5 2 , And A +2 B And A - B Vertical, seeking A And B θ. Known A , B , C Is three vectors in the same plane, where A =(1,2) (1) If | C |=2 5 , And C ∥ A , Please C Coordinates of; (2) If | B |= 5 2 , And A +2 B And A - B Vertical, seeking A And B θ. Known A , B , C Is three vectors in the same plane, where A =(1,2) (1) If | C |=2 5 , And C ∥ A , Seeking C Coordinates of; (2) If | B |= 5 2 , And A +2 B And A - B Vertical, seek A And B θ.

Known A , B , C Is three vectors in the same plane, where A =(1,2) (1) If | C |=2 5 , And C ∥ A , Seeking C Coordinates of; (2) If | B |= 5 2 , And A +2 B And A - B Vertical, seeking A And B θ. Known A , B , C Is three vectors in the same plane, where A =(1,2) (1) If | C |=2 5 , And C ∥ A , Please C Coordinates of; (2) If | B |= 5 2 , And A +2 B And A - B Vertical, seeking A And B θ. Known A , B , C Is three vectors in the same plane, where A =(1,2) (1) If | C |=2 5 , And C ∥ A , Seeking C Coordinates of; (2) If | B |= 5 2 , And A +2 B And A - B Vertical, seek A And B θ.

(1) Because

A
,

B
,

C
Is three vectors in the same plane, where

A
=(1,2),
If |

C
|=2
5
, And

C
∥

A
, Can be set

C
=λ•

A
=(λ,2λ), Then |

C
|=
λ2+(2λ)2
=2
5
,
λ=±2,

C
=(2,4), Or

C
=(-2,4).
(2)∵|

B
|=
5

2
, And

A
+2

B
And

A
-

B
Vertical,(

A
+2

B
)•(

A
-

B
)=

A
2+

A
•

B
-2

B
2=0,
Simplified available

A
•

B
=-
5
2
, I.e.
5
×
5

2
×Cos θ=-
5
2
, Cosθ=-1,
Gu

A
And

B
θ=π.

How to judge three vectors in the same plane

If three vectors are mixed product, the result is 0, then it is coplanar.

Given abc is three vectors in the same plane, where a =(1,2)1. If |b|= root 5/2, and a+2b is perpendicular to 2a-b, find the angle α between a and b

A+2b perpendicular to 2a-b
Therefore (a+2b)(2a-b)=0
3Ab=2b^2-2a^2=2*(5/4)-2*5=-7.5
Therefore ab=-2.5
So cosα=-2.5/(√5 5/2)=-1
So α=π

A+2b perpendicular to 2a-b
Therefore (a+2b)(2a-b)=0
Get 3ab=2b^2-2a^2=2*(5/4)-2*5=-7.5
Therefore ab=-2.5
So cosα=-2.5/(√5 5/2)=-1
So α=π

Given A (-1,-5) and vector a =(2.3), what is the coordinate of point B if vector A B =3a?

(5,4)
0B-OA=AB=3a=(6,9)
So Xb=6-1=5. Yb=9-5=4

Given that a vector‖b vector, a=(2,3), b=(-4, m) and c modulus=5, the angle between c and a is 60°, then the value of (a+b)*c is? Given a vector ‖b vector, a=(2,3), b=(-4, m) and c modulus=5, the angle between c and a is 60°, then the value of (a+b)*c is?

Because a vector‖b vector
So a=xb
A=(2,3), b=(-4, m)
X=-4/2=-2
M=3*-2=-6
B=(-4,-6)
(A+b)*c=|(a+b)*|c|cos60°
(A+b)*c=root 13*5*1/2=5root 13/2

Because a vector‖b vector
So a=xb
A=(2,3), b=(-4, m)
X=-4/2=-2
M=3*-2=-6
B=(-4,-6)
(A+b)*c=|(a+b) c|cos60°
(A+b)*c=root 13*5*1/2=5root 13/2

Given the vector a=(3,0), b=(k,5), and the angle between a and b is 3 divisions/4, what is the value of k

The cosine of the included angle of the vector is equal to the product of the vector divided by the product of the vector modulus
Product of vector=3×k+0×5=3k
Product of modulus=3(k2+25)
Cos3π/4=(3k)÷[3√(k2+25)]=-√2/2
I.e. k/[3√(k2+25)]=-√2/2 visible k