How to Prove Distributive Laws for 3D Vector Cross Products?

[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

 

[cristo]

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

 

[robphy]

Sounds like homework.

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

 

[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?

 

[robphy]

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

 

[chocbizkt]

thanks i don't know where to go from there

 

[robphy]

thanks i don't know where to go from there

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

 

[SeReNiTy]

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

px(qxr) = (pxq)xr

 

[D H]

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".

 

[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

 

[Corneo]

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

am i on the right track?

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

 

[Vinodh]

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

px(qxr) = (pxq)xr

Then wat is distributive law?

Vinodh

 

[cristo]

Then wat is distributive law?

Vinodh

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