# I Scalar product

1. May 1, 2016

### 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

2. May 2, 2016

### blue_leaf77

That's the definition of a norm.

3. May 2, 2016

### 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

4. May 2, 2016

### blue_leaf77

Do you mean "scalar product"?

5. May 2, 2016

### prashant singh

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

6. May 2, 2016

### blue_leaf77

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

7. May 2, 2016

Yess

8. May 2, 2016

### 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.

9. May 2, 2016

### Staff: Mentor

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|.

10. May 2, 2016

### mathman

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

Above may help.

11. May 3, 2016

### SammyS

Staff Emeritus
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.

12. May 3, 2016

### prashant singh

I forget to write it as |A X B|

13. May 4, 2016

### Staff: Mentor

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

14. May 4, 2016

I know that