# Scalar product

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

## Answers and Replies

blue_leaf77
Homework Helper
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 knowlege requires to prove this

That's the definition of a norm.

blue_leaf77
Homework Helper
vector product
Do you mean "scalar product"?

No ,vector product, A X B = ||A||||B||sin(theta)
Do you mean "scalar product"?

blue_leaf77
Homework Helper
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##?

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
Homework Helper
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
Mentor
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
Staff Emeritus
Homework Helper
Gold Member
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:
AB = |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 and ProfuselyQuarky
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
Mentor
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).

I know that

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