Scalar product

[prashant singh]

Why A.A = ||A||^2 , I know that from product rule we can prove this where theta =0 , I am asking this because I have seen many proves for A.B = ||A||||B||cos(theta) and to prove this they have used A.A = ||A||^2, how can they use this , this is the result of dot product formula. I havee seen every where even on Wikipedia they have used this method only

 

[blue_leaf77]

Why A.A = ||A||^2

That's the definition of a norm.

 

[prashant singh]

Okk , can u give me the proof of vector product,or tell me the about the difficulty or knowledge requires to prove this

That's the definition of a norm.

 

[blue_leaf77]

vector product

Do you mean "scalar product"?

 

[prashant singh]

No ,vector product, A X B = ||A||||B||sin(theta)

Do you mean "scalar product"?

 

[blue_leaf77]

No ,vector product, A X B = ||A||||B||sin(theta)

Do you want to prove that the magnitude of $\vec{A}\times \vec{B}$ is equal to $|\vec{A}| |\vec{B}| \sin \theta$?

 

[prashant singh]

Yess

Do you want to prove that the magnitude of $\vec{A}\times \vec{B}$ is equal to $|\vec{A}| |\vec{B}| \sin \theta$?

 

[blue_leaf77]

The proof is tedious and might look unattractive if one were to outline it here, for this purpose I will simply refer you to this link.

 

[Mark44]

No ,vector product, A X B = ||A||||B||sin(theta)

This formula is not correct. On the left side, A X B is a vector. On the right side $|A||B|\sin(\theta)$ is a scalar. As already noted, $|A||B|\sin(\theta)$ is equal to the magnitude of A X B; that is, |A X B|.

 

[mathman]

Why A.A = ||A||^2 , I know that from product rule we can prove this where theta =0 , I am asking this because I have seen many proves for A.B = ||A||||B||cos(theta) and to prove this they have used A.A = ||A||^2, how can they use this , this is the result of dot product formula. I havee seen every where even on Wikipedia they have used this method only

http://clas.sa.ucsb.edu/staff/alex/DotProductDerivation.pdf

Above may help.

 

[SammyS]

Why A.A = ||A||^2 , I know that from product rule we can prove this where theta =0 , I am asking this because I have seen many proves for A.B = ||A||||B||cos(theta) and to prove this they have used A.A = ||A||^2, how can they use this , this is the result of dot product formula. I have seen every where even on Wikipedia they have used this method only

No ,vector product, (magnitude of) A X B = ||A||||B||sin(theta)

If you want to prove either:

A⋅B = |A| |B| cos(θ)​

or

|A×B| = |A| |B| sin(θ)​

, you need to give the definitions you are using for scalar product and vector product.

 

[prashant singh]

I forget to write it as |A X B|

This formula is not correct. On the left side, A X B is a vector. On the right side $|A||B|\sin(\theta)$ is a scalar. As already noted, $|A||B|\sin(\theta)$ is equal to the magnitude of A X B; that is, |A X B|.

 

[Mark44]

I forget to write it as |A X B|

If you forget to write the absolute value symbols, you will mistake a vector for a scalar (a number).

 

[prashant singh]

I know that

If you forget to write the absolute value symbols, you will mistake a vector for a scalar (a number).