In the plane vectors a, b, a =(4,-3),|b|=1, and the product of a and b =5, then the vector b =?

In the plane vectors a, b, a =(4,-3),|b|=1, and the product of a and b =5, then the vector b =?

Let b (x, y)
Since the number product of a and b=5,4x-3y=5
X^2+y^2=1 under the root because |b|=1
Solution x=0.8 y=-0.6
So b (0.8,-0.6)

If the vector a, b satisfies |a-b|=5, a=(7/2,1/2),|b|=√2/2, then the product of a, b is?

/A-b/=5
A^2+b^2-2ab=25
A =(7/2,1/2)
A^2=50/4, b^=1/2
A* b=(5 Nos.2-49)/4

Given that the module length of vector a is equal to 1, the module length of vector b is equal to 6, and the number product of vector a and vector (b-a) is 2, then is the angle between vector a and vector b?

The quantity product of vector a and vector (b-a) is a-module b-module cosa-a^2
Substitute data to obtain cosa=1/2, a=π/3

The quantity product of vector a and vector (b-a), i.eule b-module cosa-a^2
Substitute data to obtain cosa=1/2, a=π/3

If the number product of high and medium plane vectors is 3, vector a=(1,2), vector b=(x,-2), and a⊥(a-b), then the real number x is equal to High and medium plane vector quantity product 3. Given vector a=(1,2), vector b=(x,-2), and a⊥(a-b), then the real number x is equal to? 4, If vector a=(1,1), b=(-1,2), then a*b equals?

3.9
4.1

If the module of vector a is equal to zero, then vector a is equal to zero vector

Yeah.
If the vector modulus is equal to 0, then the vector is a zero vector.
Assume vector a=(x, y, z)3D vector
Then modulus of vector = under root (x^2+y^2+z^2)=0
Then only x = y = z =0, so the vector a =(0,0,0)
A is the zero vector.

Yes.
If the vector modulus is equal to 0, then the vector is a zero vector.
Assume vector a=(x, y, z)3D vector
Then modulus of vector = under root (x^2+y^2+z^2)=0
Then we only have x=y=z=0, so the vector a=(0,0,0)
A is the zero vector.

The modulus of a vector plus b vector is greater than or equal to the modulus of a vector minus b vector

No! No!
Because vectors are directional.
Even a scalar is a false proposition

No! No!
Because vectors have directions.
Even a scalar is a false proposition