![](data:image/svg+xml;utf8,<svg xmlns="http://www.w3.org/2000/svg" xmlns:xlink="http://www.w3.org/1999/xlink" viewBox="0 0 72 71" width="72"></svg>)
What is the difference between quantity product and vector product? Does it matter?
Symbol size direction
Product of quantity:. Product of module length * cos (included angle) None
Vector product:* product of module length * sin (included angle) right-hand rule
Right hand rule: a*b direction is:
The right thumb is pointing a, the index finger is pointing b, and the middle finger is perpendicular to the plane of the thumb and index finger
The middle finger direction is the vector product direction
Symbol size direction
Product of quantity:. Product of module length * cos (included angle) None
Vector product:* product of module length * sin (included angle) right-hand rule
Right hand rule: a*b direction is:
The right thumb is pointing a, the index finger is pointing b, and the middle finger is perpendicular to the plane of the thumb and index finger
The direction of the middle finger is the vector product direction
If vectors a and b are not collinear, a.b=0, and c=a-[(a.a)/(b.b)].b, then is the angle between vectors a and c?
The angle between A and C is 90 degrees. The detailed steps are as follows,)
The question was wrong. It should be c=a-[(a.a)/(a.b)].b
Therefore, a.c.=a.a-[(a.a)/(a.b)] a.b.=a.a-a.a=0
So A and C were orthogonal, so the angle between A and C was 90 degrees.
The angle between a and c is 90 degrees. The detailed steps are as follows,)
The title is wrong. It should be c=a-[(a.a)/(a.b)].b
Thus a.c=a.a-[(a.a)/(a.b)] a.b=a.a-a.a=0
So a and c are orthogonal, so the angle between a and c is 90 degrees.
If the vectors a and b are not collinear, a·b=0, and c=a-(a·a)/(a·b) b, what is the included angle between vectors a and c Why is a*a-[(a*a)/(a*b)]*(a*b)=0 because it is not collinear? If vectors a and b are not collinear, a·b=0, and c=a-(a·a)/(a·b) b, then what is the angle between vectors a and c Why is a*a-[(a*a)/(a*b)]*(a*b)=0 because it is not collinear?
Yeah, yeah.
0
That's not right upstairs:
C=a-(a·a/a·b) b=a-(|^2/a·b) b, a·c=a·(a-(|^2/a·b) b)=|^2-(|^2/a·b) a·b=0
Therefore: a and c are perpendicular, i.e. the included angle between a and c:=π/2
What do we do with a known vector for a triangle area? Given vector OA=(2,3), OB=(3,-3), then the area of triangle OAB is? I saw someone else's formula. Cosθ=vector AB*vector OB/|vector AB vector OB| I don't know what that means. We did n' t learn about the relationship between vectors and triangulations. How to Solve the Vector Operation
Two vector point multiplication (not multiplication), their relationship is: vector AB * vector OB=| vector AB *|vector OB |* cosθ,θ refers to ∠AOB, this is vector operation, since it is a vector problem, there should be learning vector operation, if not that does not matter, vector operation is high school learning, you will learn sooner or later, such as...
How is the triangle area represented by the corresponding vector on both sides? Corresponding vectors are a vector and b vector respectively, explain the reason
Let the angle between the two vectors be θ.
A vector · b vector
Cos θ=------------------
| A vector | b vector |
Find sin θ=√(1-cosθ)2
√[| A vector |2 b vector |2-(a vector ·b vector)2]
= ---------------------------------------------
| A vector | b vector |
Triangle area S=|a vector b vector sinθ/2
√[| A vector |2 b vector |2-(a vector ·b vector)2]
= --------------------------------------------------
2
RELATED INFORMATIONS
- 1. Combining vectors to solve triangles In △ABC, the opposite sides of angles A, B and C are a,b,c,tanC =3 times the root number 7, and cosC is obtained Combining vectors to solve triangles In △ABC, the opposite sides of angles A, B and C are a,b,c,tanC =3 times the root number 7, and cosC is obtained.
- 2. In the triangle, how to prove the "five-center" point with a vector? I'm talking about how each of them can prove it. For example, how to prove that the vertical center is the intersection of three high lines, only one. How do we prove the "five-center" intersection in a triangle with a vector? I'm talking about how each of them can prove it. For example, how to prove that the vertical center is the intersection of three high lines, only one. In the triangle, how to prove the "five hearts" with a vector? I'm talking about how each of them can prove it. For example, how to prove that the vertical center is the intersection of three high lines, only one.
- 3. Given the vector a=(sinx- cosx,2cosx), b=(sinx+cosx, sinx).(1) if a⊥b, find the value of tan2x if a*b=3/5, find the value of sin4x The vectors a=(sinx- cosx,2cosx), b=(sinx+cosx, sinx) are known. (1) If a⊥b, find the value of tan2x 2 If a*b=3/5, find the value of sin4x
- 4. Given vectors a=(1,2), b=(2,-1), then the angle between vector a and vector b
- 5. Given vector a=(2,3), vector b=(-1,-2), vector c=(2,1), then (b•C) a=?
- 6. Given vector a=(-1, y) vector b=(1,-3) and satisfy (2a+b)⊥b 1. Find the coordinates of vector a 2. Find the angle between vectors a and b Vector a=(-1, y) vector b=(1,-3) and satisfies (2a+b)⊥b 1. Find the coordinates of vector a 2. Find the angle between vectors a and b
- 7. If the vectors a, b, c satisfy a+b+c=0, and /a/=3,/b/=1,/c/=4, then how much is a*b+b*c+c*a? Because /a/+/b/=/c/, and because a+b+c=0, a and b must be in the same direction direction, and c in the opposite direction. So a*b+b*c+c*a=/a//b/-/b//c/-/a//c/=3-4-12=-13 Find the reason why a and b must be in the same direction and c must be in the opposite direction. ——
- 8. Three vertices A (0,2), B (-1,-2), C (3,1) of a quadrilateral ABCD are known, and BC =2 AD, the coordinates of vertex D are () A.(2,7 2) B.(2,−1 2) C.(3,2) D.(1,3) Three vertices A (0,2), B (-1,-2), C (3,1) of the quadrilateral ABCD are known, and BC =2 AD, the coordinates of vertex D are () A.(2,7 2) B.(2,−1 2) C.(3,2) D.(1,3)
- 9. Given the vector a=(1,1), b=(4, x), u=a+2b, v=2a+b, and u is parallel to v, then x=
- 10. Given that the module of a vector is equal to 1, the module of b vector is equal to 2, find (a vector+2b vector)(2a vector-b vector)
- 11. What is the geometric meaning of the product of the number of vectors? If there is any physical knowledge involved, please speak in plain English. What is the geometric meaning of the product of the number of vectors? If physical knowledge is involved, please speak in plain English.
- 12. If the vector a=(1,-2),|b|=4|a|, and a, b are collinear, then the possible coordinates of b are If the vector a=(1,-2),|b|=4|a|, and a, b are collinear, then the possible coordinates of b are (-4,8)
- 13. Given non-zero vectors a, b and a perpendicular to b, verify that (absolute value of vector a + absolute value of vector b)/(absolute value of vector a + b) is less than or equal to root 2
- 14. How to find the normal vector of a plane made of triangles?
- 15. Given vectors OA, OB, OC and trilateral a, b, c, I are the inner of a triangle, prove vector OI=(aOA+bOB+cOC)/(a+b+c) Given vectors OA, OB, OC and trilateral a, b, c, I are the inner of a triangle, prove that vector OI=(aOA+bOB+cOC)/(a+b+c)
- 16. Give a proof of the theorem of the relationship between the diagonal and the sides of a parallelogram (the sum of the squares of the two diagonals of a parallelogram is equal to the sum of the squares of its four sides)
- 17. It is proved that if vector a*b+b*c+c*a=0, then a, b, c are coplanar. What's the theorem, what's the definition, what's the idea
- 18. In this paper, the vector method is used to prove that the three center lines of the triangle ABC intersect at a point P and have a Vector OP=1/3(vector OA+vector OB+OC vector) Note: Vector method is required instead of coordinates In this paper, the vector method is used to prove that the three midlines of the triangle ABC intersect at point P Vector OP=1/3(vector OA+vector OB+OC vector) Note: Vector method is required instead of coordinates
- 19. Hyperbola x2/a2-y2/b2=1(a >0, b >0) is known to be directed by vector k =(6,6) If the midpoint of the chord cut by the straight line of is (4,1), then the value of the hyperbolic eccentricity is A.√5/2 B.√6/2 C.√10/3 D.2
- 20. λ, R, a and b vector are not collinear, nonzero vector c=λa b, if c‖a, then λ and μ satisfy the following conditions:? λ, R, a and b vector are not collinear, non-zero vector c=λa b, if c‖a, then λ and μ satisfy the condition?