1) Vector a, vector b satisfies |vector a|,|vector b|=10, then what is the value range of |vector a-vector b| 2) Given |vector a|=6,|vector b|=8,|vector a-vector b|=10, then |vector a+vector b| equals what 3) Given △ABC vertex A (-1,-1/2), B (2,3) and center of gravity coordinate G (1,1/2), what is the coordinate of vertex C? 4) Given the two points 0(0,0) and A (6,3), if point P is on the straight line OA, vector PA=2 vector OP, and P is the midpoint of line OB, what is the coordinate of point B? 5) Given vector a=(1,2) vector b=0(1,1) and the angle between vector a and vector a +β vector b is acute, what is the value range of real number β?

1) Vector a, vector b satisfies |vector a|,|vector b|=10, then what is the value range of |vector a-vector b| 2) Given |vector a|=6,|vector b|=8,|vector a-vector b|=10, then |vector a+vector b| equals what 3) Given △ABC vertex A (-1,-1/2), B (2,3) and center of gravity coordinate G (1,1/2), what is the coordinate of vertex C? 4) Given the two points 0(0,0) and A (6,3), if point P is on the straight line OA, vector PA=2 vector OP, and P is the midpoint of line OB, what is the coordinate of point B? 5) Given vector a=(1,2) vector b=0(1,1) and the angle between vector a and vector a +β vector b is acute, what is the value range of real number β?

For ease of writing, the solution process does not write the word "vector ". Known:|a|=|b|=10,|a-b|^2=(a-b)2. =a^2-2ab+b^2.& nbs...

High school mathematics compulsory four, the second chapter plane vector involves all formulas

1. Addition of vectors
The addition of vectors satisfies parallelogram rule and triangle rule.
AB+BC=AC.
A+b=(x+x', y+y').
A+0=0+a=a.
The operation law of vector addition:
Commutative law: a+b=b+a;
Binding law:(a+b)+c=a+(b+c).
2. Subtraction of vectors
If a, b are mutually opposite vectors, then the inverse of a=-b, b=-a, a+b=0.0 is 0
AB-AC=CB. i.e." Common starting point, pointing to subtracted "
A=(x, y) b=(x', y') then a-b=(x-x', y-y').
4. Number multiplication vector
The product of the real number λ and the vector a is a vector denoted by λa, and a = a.
When λ>0,λa is in the same direction as a;
When λ<0,λa is opposite to a;
When λ=0,λ a =0, the direction is arbitrary.
When a =0,λ a =0 for any real number λ.
Note: by definition, if λa=0, then λ=0 or a=0.
The real number λ is called the coefficient of vector a, and the geometric meaning of multiplier vector λa is to extend or compress the directed line segment representing vector a.
When >1, the directed line segment of vector a extends to the original times in the original direction (λ>0) or the opposite direction (λ<0);
When <1时,表示向量a的有向线段在原方向(λ>0) or the opposite direction (λ<0) is shortened to the original times.
The multiplication of numbers and vectors satisfies the following operation law
Binding law:(λ a)•b=λ(a •b)=(a •b).
The distribution law of vector to number (the first distribution law):() a= a a.
The distribution law of number to vector (the second distribution law):λ(a+b)=λa b.
The elimination law of number multiplication vector:1 If the real number λ=0 and λa=λb, then a=b.2 If a=0 and λa=μa, then λ=μ.
3. Quantity product of vector
Definition: If two non-zero vectors a, b are given as OA=a, OB=b, then the angle AOB is called the angle between vector a and vector b, and is denoted as < a, b > and 0≤< a, b 〉≤π
Definition: The quantity product (inner product, dot product) of two vectors is a quantity, which is recorded as a•b. If a and b are not collinear, then a•b=|a b cos〈a, b〉; if a and b are collinear, then a•b=+- a b.
The coordinate of the quantity product of the vector is represented as: a • b = x • x'+ y • y'.
ON THE OPERATIONAL LAW OF THE QUANTITATIVE PRODUCT OF VECTOR
A•b=b•a (exchange law);
(λA)•b=λ(a•b)(on the associativity of number multiplication);
(A+b)•c=a•c+b•c (distribution law);
Properties of the Quantity Product of a Vector
A • a =| a | squared.
A⊥b〈=〉 a•b=0.
|A•b a b|.
The Main Difference Between the Quantity Product of Vector and the Operation of Real Number
1. The quantity product of a vector does not satisfy the binding law, i.e.(a•b)•c=a•(b•c); for example:(a•b)^2=a^2•b^2.
2. The quantity product of the vector does not satisfy the elimination law, that is, from a • b = a • c (a =0), b = c.
3.|A•b|=|a b|
4. By |a|=|b|, we can not deduce a=b or a=-b.
4. Vector product of vector
Definition: The vector product (outer product, cross product) of two vectors a and b is a vector, denoted as a×b. If a and b are not collinear, then the module of a×b is: a×b =|a b sin〈a, b〉; the direction of a×b is perpendicular to a and b, and a, b and a×b form a right-hand system in this order. If a and b are collinear, then a×b=0.
Vector product property of vector:
A×b is the parallelogram area with a and b sides.
A×a=0.
A‖b〈=〉 a×b=0.
Vector Product Operation Law of Vector
A×b=-b×a;
(λA)×b=λ(a×b)=a×(λb);
(A+b)×c=a×c+b×c.
Note: Vector has no division," vector AB/vector CD "is meaningless.
Triangle Inequality of Vector
1. A b a+b a b;
1 If and only if a and b are reversed, the left is equal;
2 If and only if a and b are in the same direction, the right is equal.
2. A b a b a b.
1 If and only if a and b are in the same direction, take the equal sign on the left;
2 If and only if a and b are reversed, the right is equal.
Scoring point
Fixed fraction formula (vector P1P= vector PP2)
Let P1 and P2 be two points on a straight line, and P be any point on l which is different from P1 and P2. Then there exists a real number λ, such that vector P1P= vector PP2,λ is called the ratio of the point P to the directed segment P1P2.
If P1(x1, y1), P2(x2, y2), P (x, y), then yes
OP=(OP1 OP2)(1);(vector formula of constant fraction)
X =(x1+λx2)/(1+λ),
Y =(y1+λy2)/(1+λ).(Formula for coordinate of dividing point)
We call the above formula the fractionation formula of the directed segment P1P2
Three-point collinear theorem
If OC =λOA +μOB, and =1, then A, B and C are collinear
Triangular gravity judgment formula
In △ABC, if GA+GB+GC=O, then G is the center of gravity of △ABC
Important conditions of collinearity of vector
If b=0, then the important condition of a//b is the existence of a unique real number λ such that a=λb.
The important condition of a//b is xy'-x' y=0.
The zero vector 0 is parallel to any vector.
Necessary and Sufficient Conditions for Vector Verticality
The sufficient and necessary condition for a⊥b is a•b=0.
The sufficient and necessary condition for a⊥b is xx'+yy'=0.
The zero vector 0 is perpendicular to any vector.

High School Mathematics Compulsory 4 Chapter 2 All Formulas of Plane Vector

Not in the book.

How to prove that three points are collinear

Given the coordinates of three points
Method 1: Take two points to establish a straight line
Calculate the analytic expression of the line
Refer to the third point coordinate to see whether the analytic formula is satisfied
Method 2: Set three points as A, B and C
Prove by vector: a times AB vector=AC vector (where a is a non-zero real number)
Method 3: Calculate AB slope and AC slope by point difference method
Equal means three points are collinear

A Proof of Plane Vector If (vector AD)=x×(vector AB),(vector AE)=y×(vector AC), it is proved that:1/x+1/y=3x+1/y=3 is that the line l is constant over the △ABC center of gravity G. (The original question has no figure)

If we can prove that 1/x+1/y=3, then we can explain the problem if and only if the linear L is a straight line of the overweight center G.
The following proves that the straight line L of G can derive 1/x+1/y=3
Extended AG to BC at M
From the linear vector form of the parameter equation:(" vector "is too troublesome, I do not write the word vector, write the beginning of the table before, write the end of the table after) AG=kAD+(1-k) AE
Because AD=xAB, AE=yAC
So AG=kxAB+(1-k) yAC1
G is the center of gravity of the triangle, so M is the center line of the triangle (i.e. M is the midpoint of BC)
So AM=1/2AB+1/2AC
And AG=2/3AM, get AG=1/3AB+1/3AC2
So 12:1/3AB+1/3AC=kxAB+(1-k) yAC
So 1/3=kx,1/3=(1-k) y
1/X+1/y=3 after eliminating k
I hope you are satisfied with my answer!

The Inverse Proving Process of Plane Vector Collinear Theorem When two vectors a (x1, y1), b (x2, y2) in the same plane are parallel to b, the formula x1y2—x2y1=0 can be obtained. So if we know that the two vectors in the plane satisfy x1y2-x2y1=0, how to prove that the two vectors are collinear?

There are three cases to discuss:(1) If one of a, b is a zero vector, then a, b are collinear (zero vector is collinear with any vector);(2) a, b are not zero vectors: if one of the vectors a (x1, y1) is 0, then let x2=0, then y2=0(x1, x2 are not 0, otherwise it is assumed to be non-zero vector.