Prove that $sum_i vec r_i times vec uF_i = (sum_i vec r_iF_i) times vec u$

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Given an expression
$$sum_i vec r_i times vec uF_i $$
where $vec r_i$ are position vectors from the origin, $vec u$ is a constant unit vector and $F_i$ are constants, it is true that$$sum_i vec r_i times vec uF_i = (sum_i vec r_iF_i) times vec u.$$

How to prove that the above equation is true?

summation vectors vector-analysis physics

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asked Jul 26 at 4:57

Taenyfan

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Given an expression
$$sum_i vec r_i times vec uF_i $$
where $vec r_i$ are position vectors from the origin, $vec u$ is a constant unit vector and $F_i$ are constants, it is true that$$sum_i vec r_i times vec uF_i = (sum_i vec r_iF_i) times vec u.$$

How to prove that the above equation is true?

summation vectors vector-analysis physics

share|cite|improve this question

asked Jul 26 at 4:57

Taenyfan

216

add a comment | 

up vote
0
down vote

favorite

up vote
0
down vote

favorite

Given an expression
$$sum_i vec r_i times vec uF_i $$
where $vec r_i$ are position vectors from the origin, $vec u$ is a constant unit vector and $F_i$ are constants, it is true that$$sum_i vec r_i times vec uF_i = (sum_i vec r_iF_i) times vec u.$$

How to prove that the above equation is true?

summation vectors vector-analysis physics

share|cite|improve this question

asked Jul 26 at 4:57

Taenyfan

216

Given an expression
$$sum_i vec r_i times vec uF_i $$
where $vec r_i$ are position vectors from the origin, $vec u$ is a constant unit vector and $F_i$ are constants, it is true that$$sum_i vec r_i times vec uF_i = (sum_i vec r_iF_i) times vec u.$$

How to prove that the above equation is true?

summation vectors vector-analysis physics

share|cite|improve this question

asked Jul 26 at 4:57

Taenyfan

216

share|cite|improve this question

share|cite|improve this question

asked Jul 26 at 4:57

Taenyfan

216

asked Jul 26 at 4:57

Taenyfan

216

asked Jul 26 at 4:57

Taenyfan

216

216

add a comment | 

add a comment | 

2 Answers
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1
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accepted

Let's keep it simple. You need just the following two properties of the cross product:

$$vecatimes(kvecb)=(kveca)timesvecbtag1$$

$$(veca+vecb)timesvecc=vecatimesvecc+vecbtimesvecctag2$$

Both can be easily proved if you use the fact that the cross product of vectors $veca=a_xveci+a_yvecj+a_zveck$ and $vecb=b_xveci+b_yvecj+b_zveck$ is:

$$vecatimesvecb=(a_yb_z-a_zb_y)veci+(a_zb_x-a_xb_z)vecj+(a_xb_y-a_yb_x)veck$$

All this is explained in excellent detail on Wikipedia.

The proof is now simple:

$$sum_i vec r_i times (F_ivec u)=$$

$$sum_i (F_ivec r_i) times vec u=tag(1) applied$$

$$(sum_i F_ivec r_i) times vec utag(2) applied$$

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answered Jul 26 at 6:26

Oldboy

2,6101316

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I will use superscript for the index of the vector, so the subscript can be the element of a vector.
$$left(sum_i vecr^i times vecuF^iright)_j=$$
$$sum_ileft(vecr^i times vecuF^iright)_j=$$
$$sum_ileft(sum_k,lepsilon_jkl r^i_k u_lF^iright)=$$
$$sum_k,lleft(sum_iepsilon_jkl r^i_k u_lF^iright)=$$
$$sum_k,lepsilon_jklleft(sum_i r^i_k F^iright)u_l=$$
$$sum_k,lepsilon_jklleft(sum_i vecr^i F^iright)_ku_l=$$
$$left(left(sum_i vecr^i F^iright)timesvecuright)_j$$
So we have that
$$sum_i vecr^i times vecuF^i=left(sum_i vecr^i F^iright)timesvecu$$

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answered Jul 26 at 5:20

Botond

3,8432632

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2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
1
down vote



accepted

Let's keep it simple. You need just the following two properties of the cross product:

$$vecatimes(kvecb)=(kveca)timesvecbtag1$$

$$(veca+vecb)timesvecc=vecatimesvecc+vecbtimesvecctag2$$

Both can be easily proved if you use the fact that the cross product of vectors $veca=a_xveci+a_yvecj+a_zveck$ and $vecb=b_xveci+b_yvecj+b_zveck$ is:

$$vecatimesvecb=(a_yb_z-a_zb_y)veci+(a_zb_x-a_xb_z)vecj+(a_xb_y-a_yb_x)veck$$

All this is explained in excellent detail on Wikipedia.

The proof is now simple:

$$sum_i vec r_i times (F_ivec u)=$$

$$sum_i (F_ivec r_i) times vec u=tag(1) applied$$

$$(sum_i F_ivec r_i) times vec utag(2) applied$$

share|cite|improve this answer

answered Jul 26 at 6:26

Oldboy

2,6101316

add a comment | 

up vote
1
down vote



accepted

Let's keep it simple. You need just the following two properties of the cross product:

$$vecatimes(kvecb)=(kveca)timesvecbtag1$$

$$(veca+vecb)timesvecc=vecatimesvecc+vecbtimesvecctag2$$

Both can be easily proved if you use the fact that the cross product of vectors $veca=a_xveci+a_yvecj+a_zveck$ and $vecb=b_xveci+b_yvecj+b_zveck$ is:

$$vecatimesvecb=(a_yb_z-a_zb_y)veci+(a_zb_x-a_xb_z)vecj+(a_xb_y-a_yb_x)veck$$

All this is explained in excellent detail on Wikipedia.

The proof is now simple:

$$sum_i vec r_i times (F_ivec u)=$$

$$sum_i (F_ivec r_i) times vec u=tag(1) applied$$

$$(sum_i F_ivec r_i) times vec utag(2) applied$$

share|cite|improve this answer

answered Jul 26 at 6:26

Oldboy

2,6101316

add a comment | 

up vote
1
down vote



accepted

up vote
1
down vote



accepted

Let's keep it simple. You need just the following two properties of the cross product:

$$vecatimes(kvecb)=(kveca)timesvecbtag1$$

$$(veca+vecb)timesvecc=vecatimesvecc+vecbtimesvecctag2$$

Both can be easily proved if you use the fact that the cross product of vectors $veca=a_xveci+a_yvecj+a_zveck$ and $vecb=b_xveci+b_yvecj+b_zveck$ is:

$$vecatimesvecb=(a_yb_z-a_zb_y)veci+(a_zb_x-a_xb_z)vecj+(a_xb_y-a_yb_x)veck$$

All this is explained in excellent detail on Wikipedia.

The proof is now simple:

$$sum_i vec r_i times (F_ivec u)=$$

$$sum_i (F_ivec r_i) times vec u=tag(1) applied$$

$$(sum_i F_ivec r_i) times vec utag(2) applied$$

share|cite|improve this answer

answered Jul 26 at 6:26

Oldboy

2,6101316

Let's keep it simple. You need just the following two properties of the cross product:

$$vecatimes(kvecb)=(kveca)timesvecbtag1$$

$$(veca+vecb)timesvecc=vecatimesvecc+vecbtimesvecctag2$$

Both can be easily proved if you use the fact that the cross product of vectors $veca=a_xveci+a_yvecj+a_zveck$ and $vecb=b_xveci+b_yvecj+b_zveck$ is:

$$vecatimesvecb=(a_yb_z-a_zb_y)veci+(a_zb_x-a_xb_z)vecj+(a_xb_y-a_yb_x)veck$$

All this is explained in excellent detail on Wikipedia.

The proof is now simple:

$$sum_i vec r_i times (F_ivec u)=$$

$$sum_i (F_ivec r_i) times vec u=tag(1) applied$$

$$(sum_i F_ivec r_i) times vec utag(2) applied$$

share|cite|improve this answer

answered Jul 26 at 6:26

Oldboy

2,6101316

share|cite|improve this answer

share|cite|improve this answer

answered Jul 26 at 6:26

Oldboy

2,6101316

answered Jul 26 at 6:26

Oldboy

2,6101316

answered Jul 26 at 6:26

Oldboy

2,6101316

2,6101316

add a comment | 

add a comment | 

up vote
0
down vote

I will use superscript for the index of the vector, so the subscript can be the element of a vector.
$$left(sum_i vecr^i times vecuF^iright)_j=$$
$$sum_ileft(vecr^i times vecuF^iright)_j=$$
$$sum_ileft(sum_k,lepsilon_jkl r^i_k u_lF^iright)=$$
$$sum_k,lleft(sum_iepsilon_jkl r^i_k u_lF^iright)=$$
$$sum_k,lepsilon_jklleft(sum_i r^i_k F^iright)u_l=$$
$$sum_k,lepsilon_jklleft(sum_i vecr^i F^iright)_ku_l=$$
$$left(left(sum_i vecr^i F^iright)timesvecuright)_j$$
So we have that
$$sum_i vecr^i times vecuF^i=left(sum_i vecr^i F^iright)timesvecu$$

share|cite|improve this answer

answered Jul 26 at 5:20

Botond

3,8432632

add a comment | 

up vote
0
down vote

I will use superscript for the index of the vector, so the subscript can be the element of a vector.
$$left(sum_i vecr^i times vecuF^iright)_j=$$
$$sum_ileft(vecr^i times vecuF^iright)_j=$$
$$sum_ileft(sum_k,lepsilon_jkl r^i_k u_lF^iright)=$$
$$sum_k,lleft(sum_iepsilon_jkl r^i_k u_lF^iright)=$$
$$sum_k,lepsilon_jklleft(sum_i r^i_k F^iright)u_l=$$
$$sum_k,lepsilon_jklleft(sum_i vecr^i F^iright)_ku_l=$$
$$left(left(sum_i vecr^i F^iright)timesvecuright)_j$$
So we have that
$$sum_i vecr^i times vecuF^i=left(sum_i vecr^i F^iright)timesvecu$$

share|cite|improve this answer

answered Jul 26 at 5:20

Botond

3,8432632

add a comment | 

up vote
0
down vote

up vote
0
down vote

I will use superscript for the index of the vector, so the subscript can be the element of a vector.
$$left(sum_i vecr^i times vecuF^iright)_j=$$
$$sum_ileft(vecr^i times vecuF^iright)_j=$$
$$sum_ileft(sum_k,lepsilon_jkl r^i_k u_lF^iright)=$$
$$sum_k,lleft(sum_iepsilon_jkl r^i_k u_lF^iright)=$$
$$sum_k,lepsilon_jklleft(sum_i r^i_k F^iright)u_l=$$
$$sum_k,lepsilon_jklleft(sum_i vecr^i F^iright)_ku_l=$$
$$left(left(sum_i vecr^i F^iright)timesvecuright)_j$$
So we have that
$$sum_i vecr^i times vecuF^i=left(sum_i vecr^i F^iright)timesvecu$$

share|cite|improve this answer

answered Jul 26 at 5:20

Botond

3,8432632

I will use superscript for the index of the vector, so the subscript can be the element of a vector.
$$left(sum_i vecr^i times vecuF^iright)_j=$$
$$sum_ileft(vecr^i times vecuF^iright)_j=$$
$$sum_ileft(sum_k,lepsilon_jkl r^i_k u_lF^iright)=$$
$$sum_k,lleft(sum_iepsilon_jkl r^i_k u_lF^iright)=$$
$$sum_k,lepsilon_jklleft(sum_i r^i_k F^iright)u_l=$$
$$sum_k,lepsilon_jklleft(sum_i vecr^i F^iright)_ku_l=$$
$$left(left(sum_i vecr^i F^iright)timesvecuright)_j$$
So we have that
$$sum_i vecr^i times vecuF^i=left(sum_i vecr^i F^iright)timesvecu$$

share|cite|improve this answer

answered Jul 26 at 5:20

Botond

3,8432632

share|cite|improve this answer

share|cite|improve this answer

answered Jul 26 at 5:20

Botond

3,8432632

answered Jul 26 at 5:20

Botond

3,8432632

answered Jul 26 at 5:20

Botond

3,8432632

3,8432632

add a comment | 

add a comment | 

 

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