Proof 3d vector cross products

1. Apr 1, 2007

chocbizkt

2 questions i have;

1. proove that; p x ( q + r ) = p x q + p x r

2. and p x ( q x r ) = ( p x q ) x r

where;

p = p1i + p2j + p3k
q = q1i + q2j + q3k
r = r1i + r2j + r3k

2. Apr 1, 2007

cristo

Staff Emeritus
Well, start with telling us what you know. What is the cross product between two vectors?

3. Apr 1, 2007

robphy

Sounds like homework.

Do you know the result of these expressions: i x i , i x j, etc... , ?

4. Apr 1, 2007

chocbizkt

yes well so far for Qn1.
i have left handside p x (q + r) = (p1i + p2j + p3k)( (q1+ r1)i + (q2+ r2)j + (q3+ r3)k)

am i on the right track?

5. Apr 1, 2007

robphy

p x (q + r) = (p1i + p2j + p3k) x ( (q1+ r1)i + (q2+ r2)j + (q3+ r3)k)
Don't forget these important symbols.

6. Apr 1, 2007

chocbizkt

thanks i dont know where to go from there

7. Apr 1, 2007

robphy

Can you now explicitly calculate the cross-product of two vectors?

8. Apr 2, 2007

SeReNiTy

Cross product is not associative, so i don't see how

px(qxr) = (pxq)xr

9. Apr 2, 2007

D H

Staff Emeritus
SeReNiTy is correct. $\vec p\times(\vec q \times\vec r) = \vec (p\times\vec q) \times\vect$ only under some special circumstances. Tthe two forms are not equal in general. Google "vector triple product".

10. Apr 3, 2007

chocbizkt

true, i found out; px(qxr) = (pxq)xr
if you let q=p

then px(pxr)=(pxp)xr

such that (pxp)=0

therefore its not true statement

11. Apr 3, 2007

Corneo

Just grind through that. By that I mean set up the determinant and perform the algebra.

12. Apr 4, 2007

Vinodh

Then wat is distributive law?

Vinodh

13. Apr 4, 2007

cristo

Staff Emeritus
Number 1. in the original post is the distributivity (under addition) of the cross product: i.e. ax(b+c)=axb +axc

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