Proof 3d vector cross products

  • Thread starter chocbizkt
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  • #1
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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
 

Answers and Replies

  • #2
cristo
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Well, start with telling us what you know. What is the cross product between two vectors?
 
  • #3
robphy
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Sounds like homework.

Do you know the result of these expressions: i x i , i x j, etc... , ?
 
  • #4
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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
robphy
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p x (q + r) = (p1i + p2j + p3k) x ( (q1+ r1)i + (q2+ r2)j + (q3+ r3)k)
Don't forget these important symbols.
 
  • #6
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thanks i dont know where to go from there
 
  • #7
robphy
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thanks i dont know where to go from there
Can you now explicitly calculate the cross-product of two vectors?
 
  • #8
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Cross product is not associative, so i don't see how

px(qxr) = (pxq)xr
 
  • #9
D H
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SeReNiTy is correct. [itex]\vec p\times(\vec q \times\vec r) = \vec (p\times\vec q) \times\vect [/itex] only under some special circumstances. Tthe two forms are not equal in general. Google "vector triple product".
 
  • #10
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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
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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.
 
  • #12
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Cross product is not associative, so i don't see how

px(qxr) = (pxq)xr

Then wat is distributive law?:confused: :confused:

Vinodh
 
  • #13
cristo
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Then wat is distributive law?:confused: :confused:

Vinodh
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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