Given that the vector a, b satisfies the absolute value a =3, the absolute value b =4, and the angle between a and b is 120 degrees, how much is a multiplied by b Given that the vector a, b satisfies the absolute value a =3, the absolute value b =4, and the angle between a and b is 120 degrees, then what is a multiplied by b equal to Given that the vector a, b satisfies the absolute value a=3, the absolute value b=4, and the angle between a and b is 120 degrees, how much is a multiplied by b

Given that the vector a, b satisfies the absolute value a =3, the absolute value b =4, and the angle between a and b is 120 degrees, how much is a multiplied by b Given that the vector a, b satisfies the absolute value a =3, the absolute value b =4, and the angle between a and b is 120 degrees, then what is a multiplied by b equal to Given that the vector a, b satisfies the absolute value a=3, the absolute value b=4, and the angle between a and b is 120 degrees, how much is a multiplied by b

A*b=|a b|*cos120=-6

If the angle between the zero vectors a, b is θ, then cosθ is equal to the quantity product of vectors a, b divided by the product of their moduli. But isn't the module of vector a divided by vector a unit vector of a? Is n' t the module of vector b divided by vector a unit vector of b? Isn't that the multiplication of two unit vectors? Is n' t that 1? If not the angle of zero vector a, b is θ, then cosθ is equal to the quantity product of vectors a, b divided by the product of their moduli. But isn't the module of vector a divided by vector a unit vector of a? Is n' t the module of vector b divided by vector a unit vector of b? Isn't that the multiplication of two unit vectors? Is n' t that one? If the angle between the zero vectors a, b is θ, then cosθ is equal to the quantity product of vectors a, b divided by the product of their moduli. But isn't the module of vector a divided by vector a unit vector of a? Is n' t the module of vector b divided by vector a unit vector of b? Isn't that the multiplication of two unit vectors? Is n' t that one?

No, it's not.
1*1*Cosθ
You draw an angle, take a unit on each side, and theta is their angle, and only if theta is 90 is 1

No, it's not.
1*1*Cosθ
You draw a corner, take a unit on each side,θ is their angle, and only if θ is 90 is 1

Not really.
1*1*Cosθ
You draw an angle, take a unit on each side,θ is their angle, and only if θ is 90 is 1

Why is the number product of vectors |a||b|cosθ On the contrary, the cosine theorem can be proved by the product of quantities Since it is derived from the cosine theorem, it can be deduced from the abstract.

This is based on the work that physics does on objects...
The work of a force on a body is equal to the product of the component force on the force displacement and the displacement.
Or the definition. There's no need for theory. Just remember
I think it's based on the cosine theorem. I do n' t know if it is
On the contrary, the cosine theorem can be proved by the product of quantities. See the proof process of the cosine theorem in high school textbooks

This is based on the work that physics does on objects...
The work of a force on a body is equal to the product of the force component on the force displacement and the displacement.
Or the definition. There's no need to think. Just remember
I think it's based on the cosine theorem. I do n' t know if it is
On the contrary, the cosine theorem can be proved by the product of quantities. See the proof process of the cosine theorem in high school textbooks

How to Obtain the Product of Number of Plane Vector |a.b a||b|

Let vector a be the angle of vector b A
Then |cosA 1
∴| A.b |
=|A||b||cosA|
A||b|
∴| A.b a||b|

Let vector a and vector b have the angle A
Then |cosA 1
∴| A.b |
=|A||b||cosA|
A||b|
∴| A.b a||b|

My friend, I've seen one of your answers:" the product of the number of vectors divided by the product of the vector modulus is the cosine of the angle between the vectors." When did you learn this knowledge?

High School Mathematics Required Course 4 Chapter 2

High School Mathematics Compulsory Four Chapter Two

Given 2 vector a+vector b=(2,-4), vector c=(1,-2), the number product of vector a and vector c is 6, the module of vector b is 2, and the angle between vector b and vector c is determined

The following.. represents the dot multiplication of the vector.
Because 2a+b =(2,-4), c =(1,-2),
So (2a+b)... c =10,
I.e.2a..c + b..c =10.
And because a..c=6,
So b..c =-2.
Because |b|=2,|c|=root 5,
So cos =(b..c)/(|b c|)=-(root 5)/5.
And because it belongs to (0, pi),
So = pi - arccos [(root 5)/5].
That is, the angle between vector b and vector c is pi-arccos [(root 5)/5].
= = = = = = = = =
The above calculation may be incorrect.