- #1

- 5

- 0

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

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- Thread starter chocbizkt
- Start date

- #1

- 5

- 0

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

cristo

Staff Emeritus

Science Advisor

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

- #3

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Sounds like homework.

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

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

- #4

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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

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Don't forget these important symbols.

- #6

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thanks i dont know where to go from there

- #7

- 6,295

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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

px(qxr) = (pxq)xr

- #9

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- #10

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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

- 1

- 0

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

px(qxr) = (pxq)xr

Then wat is distributive law?

Vinodh

- #13

cristo

Staff Emeritus

Science Advisor

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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

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