# Proof 3d vector cross products

• chocbizkt
This is also true for multiplication.Number 2. In the second part of the question, the author is asking about the cross product between two vectors, px(qxr), and whether or not it is distributive. According to the distributivity of the cross product, if q=p, then px(pxr)=(pxp)xr. This means that (pxp)=0. Finally, this also means that the statement "px(qxr) = (pxq)xr" is not always true.

#### 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

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

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?

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 don't know where to go from there

chocbizkt said:
thanks i don't 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

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

chocbizkt said:
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.

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

px(qxr) = (pxq)xr

Then wat is distributive law?

Vinodh

Vinodh said:
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