Given the points A (1,3), B (4,-1), then the vector The unit vector of AB in the same direction is ______. Given points A (1,3), B (4,-1), then the vector The unit vector of AB in the same direction is ______. Given the point A (1,3), B (4,-1), then the vector The unit vector of AB in the same direction is ______.

Given the points A (1,3), B (4,-1), then the vector The unit vector of AB in the same direction is ______. Given points A (1,3), B (4,-1), then the vector The unit vector of AB in the same direction is ______. Given the point A (1,3), B (4,-1), then the vector The unit vector of AB in the same direction is ______.

Point A (1,3), B (4,-1),
∴

AB=(3,-4), available |

AB |=
32+(−4)2=5,
Therefore, the vector

The unit vector of AB in the same direction is:

E =1
|

AB |•

AB =1
5(3,-4)=(3
5,−4
5)
Therefore, the answer is:(3
5,−4
5)

Point A (1,3), B (4,-1),
∴

AB=(3,-4), available |

AB |=
32+(−4)2=5,
Therefore, the vector

The unit vector of AB in the same direction is:

E =1
|

AB |•

AB=1
5(3,-4)=(3
5,−4
5)
Therefore, the answer is:(3
5,−4
5)

Add two vectors in the same direction, and take the direction of the sum vector

Same direction as two addition vectors

Do two collinear vectors include zero vectors

Your question is:
Let a and b be collinear vectors:
1, Two zero vectors, then collinear;
2. If there is exactly a zero vector, it is collinear;
3.Both are not zero vectors, any one can be used as base vector when collinear;
B=λa

The vectors a, b are known to be two non-collinear nonzero vectors, and t is a constant. If the module of vector a is equal to the module of vector b and the angle between vector a and vector b is 60°, then what is the minimum value of the module of t (vector a-t*vector b)?

A*b=|a||b|cos60=1/2||a|^2
|A-tb|= root number [a^2-2ta*b+t^2b^2]= root number (a^2-t*a^2+t^2*a^2)= root number [a^2[(t-1/2)^2+3/4]]
Therefore, when t=1/2, the minimum value is: root number 3/2|a|

A*b=|a||b|cos60=1/2|^2
|A-tb|= root number [a^2-2ta*b+t^2b^2]= root number (a^2-t*a^2+t^2*a^2)= root number [a^2[(t-1/2)^2+3/4]]
Therefore, when t=1/2, the minimum value is: root number 3/2|a|

Given point A (4,3) B (2,5) c (3,4), find the coordinates of the product of vector AB and vector BC

AB=(2-4,5-3)=(-2,2)
BC =(3-2,4-5)=(1,-1)
AB*BC=-2*1+2*(-1)=-4
AB×BC =(0,0,0)

Given two points A (4,1) B (7,-3), the unit vector in the same direction as vector AB is

Vector AB is (3,-4)
The modulus of AB is 5
The module of unit vector is only 1
So unit vectors are (3/5,-4/5)
The method is to find the original vector and divide it by the module of the original vector.