# Proof 3d vector cross products

## Main Question or Discussion Point

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

robphy
Homework Helper
Gold Member
Sounds like homework.

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

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

thanks i dont know where to go from there

robphy
Homework Helper
Gold Member
thanks i dont know where to go from there
Can you now explicitly calculate the cross-product of two vectors?

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

px(qxr) = (pxq)xr

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".

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

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.

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

px(qxr) = (pxq)xr

Then wat is distributive law?

Vinodh

cristo
Staff Emeritus