Proving $(bf xtimes ycdot N) z+(ytimes zcdot N) x+(ztimes x cdot N) y= 0$ when $bf x,y,z$ are coplanar and $bf N$ is a unit normal vector The Next CEO of Stack OverflowNorm and Determinant relationProduct of reflections is a rotation, by elementary vector methodsComputing the unit vector for a generalised helixHow to rewrite this trigonometric formula in terms of scalar and vector products between vectors?Second derivative of the position vector in a spherical coordinate systemWhat is the logic/rationale behind the vector cross product?Simplify vector equation $2mathbf c - (mathbf a + mathbf b)times(mathbf a - mathbf b)$Unit Vectors ProblemIs this true, that the angle tangent betweem two vectors is equal to their cross product norm divided by it's inner product?Vector Cross Product.
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Proving $(bf xtimes ycdot N) z+(ytimes zcdot N) x+(ztimes x cdot N) y= 0$ when $bf x,y,z$ are coplanar and $bf N$ is a unit normal vector
The Next CEO of Stack OverflowNorm and Determinant relationProduct of reflections is a rotation, by elementary vector methodsComputing the unit vector for a generalised helixHow to rewrite this trigonometric formula in terms of scalar and vector products between vectors?Second derivative of the position vector in a spherical coordinate systemWhat is the logic/rationale behind the vector cross product?Simplify vector equation $2mathbf c - (mathbf a + mathbf b)times(mathbf a - mathbf b)$Unit Vectors ProblemIs this true, that the angle tangent betweem two vectors is equal to their cross product norm divided by it's inner product?Vector Cross Product.
$begingroup$
Prove that if $mathbfx,mathbfy,mathbfz in mathbbR^3$ are coplanar vectors and $mathbfN$ is a unit normal vector to the plane then $$(mathbfxtimesmathbfy cdot mathbfN) mathbfz + (mathbfytimesmathbfz cdot mathbfN) mathbfx + (mathbfztimesmathbfx cdot mathbfN) mathbfy=mathbf0.$$
This is an elementary identity involving cross products which is used in the proof of the Gauss-Bonnet Theorem and whose proof was left as an exercise. I've tried it unsuccessfully. Initially I tried writing $mathbfN=fracmathbfxtimesmathbfy=fracmathbfytimesmathbfz=fracmathbfztimesmathbfx mathbfztimesmathbfx$ and substituting into the equation to get $| mathbfxtimesmathbfy|z +| mathbfytimesmathbfz|mathbfx+| mathbfztimesmathbfx|mathbfy=mathbf0$ but then I realised these terms are only correct up to $pm$ signs. You could write the norms in terms of sines of angles and divide by norms to get unit vectors with coefficients $sintheta,sinpsi,sin(theta+psi)$ (or $2pi -(theta+psi)$ I suppose) but I don't know what to do from there, especially when the terms are only correct up to sign. Any hints how to prove this identity? Perhaps there is a clever trick to it but I can't see it. Edit: Maybe writing $mathbfz=lambdamathbfx+mumathbfy$ will help.
linear-algebra vectors cross-product
edited Mar 21 at 9:56
Asaf Karagila♦
307k33440773
asked Mar 20 at 11:16
AlephNullAlephNull
557110
$endgroup$
add a comment |
$begingroup$
Prove that if $mathbfx,mathbfy,mathbfz in mathbbR^3$ are coplanar vectors and $mathbfN$ is a unit normal vector to the plane then $$(mathbfxtimesmathbfy cdot mathbfN) mathbfz + (mathbfytimesmathbfz cdot mathbfN) mathbfx + (mathbfztimesmathbfx cdot mathbfN) mathbfy=mathbf0.$$
This is an elementary identity involving cross products which is used in the proof of the Gauss-Bonnet Theorem and whose proof was left as an exercise. I've tried it unsuccessfully. Initially I tried writing $mathbfN=fracmathbfxtimesmathbfy=fracmathbfytimesmathbfz=fracmathbfztimesmathbfx mathbfztimesmathbfx$ and substituting into the equation to get $| mathbfxtimesmathbfy|z +| mathbfytimesmathbfz|mathbfx+| mathbfztimesmathbfx|mathbfy=mathbf0$ but then I realised these terms are only correct up to $pm$ signs. You could write the norms in terms of sines of angles and divide by norms to get unit vectors with coefficients $sintheta,sinpsi,sin(theta+psi)$ (or $2pi -(theta+psi)$ I suppose) but I don't know what to do from there, especially when the terms are only correct up to sign. Any hints how to prove this identity? Perhaps there is a clever trick to it but I can't see it. Edit: Maybe writing $mathbfz=lambdamathbfx+mumathbfy$ will help.
linear-algebra vectors cross-product
edited Mar 21 at 9:56
Asaf Karagila♦
307k33440773
asked Mar 20 at 11:16
AlephNullAlephNull
557110
$endgroup$
1
$begingroup$
What does $x times y cdot N$ mean? Dot product $(x times y) cdot N$ ?
$endgroup$
– Widawensen
Mar 20 at 12:51
4
$begingroup$
@Widawensen Yes, what else could it mean?
$endgroup$
– Marc van Leeuwen
Mar 20 at 12:53
1
$begingroup$
@MarcvanLeeuwen: that could mean a badly written problem. That happens here sometimes.
$endgroup$
– Taladris
Mar 21 at 2:54
add a comment |
$begingroup$
Prove that if $mathbfx,mathbfy,mathbfz in mathbbR^3$ are coplanar vectors and $mathbfN$ is a unit normal vector to the plane then $$(mathbfxtimesmathbfy cdot mathbfN) mathbfz + (mathbfytimesmathbfz cdot mathbfN) mathbfx + (mathbfztimesmathbfx cdot mathbfN) mathbfy=mathbf0.$$
This is an elementary identity involving cross products which is used in the proof of the Gauss-Bonnet Theorem and whose proof was left as an exercise. I've tried it unsuccessfully. Initially I tried writing $mathbfN=fracmathbfxtimesmathbfy=fracmathbfytimesmathbfz=fracmathbfztimesmathbfx mathbfztimesmathbfx$ and substituting into the equation to get $| mathbfxtimesmathbfy|z +| mathbfytimesmathbfz|mathbfx+| mathbfztimesmathbfx|mathbfy=mathbf0$ but then I realised these terms are only correct up to $pm$ signs. You could write the norms in terms of sines of angles and divide by norms to get unit vectors with coefficients $sintheta,sinpsi,sin(theta+psi)$ (or $2pi -(theta+psi)$ I suppose) but I don't know what to do from there, especially when the terms are only correct up to sign. Any hints how to prove this identity? Perhaps there is a clever trick to it but I can't see it. Edit: Maybe writing $mathbfz=lambdamathbfx+mumathbfy$ will help.
linear-algebra vectors cross-product
edited Mar 21 at 9:56
Asaf Karagila♦
307k33440773
asked Mar 20 at 11:16
AlephNullAlephNull
557110
$endgroup$
Prove that if $mathbfx,mathbfy,mathbfz in mathbbR^3$ are coplanar vectors and $mathbfN$ is a unit normal vector to the plane then $$(mathbfxtimesmathbfy cdot mathbfN) mathbfz + (mathbfytimesmathbfz cdot mathbfN) mathbfx + (mathbfztimesmathbfx cdot mathbfN) mathbfy=mathbf0.$$
This is an elementary identity involving cross products which is used in the proof of the Gauss-Bonnet Theorem and whose proof was left as an exercise. I've tried it unsuccessfully. Initially I tried writing $mathbfN=fracmathbfxtimesmathbfy=fracmathbfytimesmathbfz=fracmathbfztimesmathbfx mathbfztimesmathbfx$ and substituting into the equation to get $| mathbfxtimesmathbfy|z +| mathbfytimesmathbfz|mathbfx+| mathbfztimesmathbfx|mathbfy=mathbf0$ but then I realised these terms are only correct up to $pm$ signs. You could write the norms in terms of sines of angles and divide by norms to get unit vectors with coefficients $sintheta,sinpsi,sin(theta+psi)$ (or $2pi -(theta+psi)$ I suppose) but I don't know what to do from there, especially when the terms are only correct up to sign. Any hints how to prove this identity? Perhaps there is a clever trick to it but I can't see it. Edit: Maybe writing $mathbfz=lambdamathbfx+mumathbfy$ will help.
linear-algebra vectors cross-product
linear-algebra vectors cross-product
edited Mar 21 at 9:56
Asaf Karagila♦
307k33440773
asked Mar 20 at 11:16
AlephNullAlephNull
557110
edited Mar 21 at 9:56
Asaf Karagila♦
307k33440773
asked Mar 20 at 11:16
AlephNullAlephNull
557110
edited Mar 21 at 9:56
Asaf Karagila♦
307k33440773
edited Mar 21 at 9:56
Asaf Karagila♦
307k33440773
edited Mar 21 at 9:56
Asaf Karagila♦
307k33440773
307k33440773
asked Mar 20 at 11:16
AlephNullAlephNull
557110
asked Mar 20 at 11:16
AlephNullAlephNull
557110
asked Mar 20 at 11:16
AlephNullAlephNull
557110
557110
1
$begingroup$
What does $x times y cdot N$ mean? Dot product $(x times y) cdot N$ ?
$endgroup$
– Widawensen
Mar 20 at 12:51
4
$begingroup$
@Widawensen Yes, what else could it mean?
$endgroup$
– Marc van Leeuwen
Mar 20 at 12:53
1
$begingroup$
@MarcvanLeeuwen: that could mean a badly written problem. That happens here sometimes.
$endgroup$
– Taladris
Mar 21 at 2:54
add a comment |
1
$begingroup$
What does $x times y cdot N$ mean? Dot product $(x times y) cdot N$ ?
$endgroup$
– Widawensen
Mar 20 at 12:51
4
$begingroup$
@Widawensen Yes, what else could it mean?
$endgroup$
– Marc van Leeuwen
Mar 20 at 12:53
1
$begingroup$
@MarcvanLeeuwen: that could mean a badly written problem. That happens here sometimes.
$endgroup$
– Taladris
Mar 21 at 2:54
1
1
$begingroup$
What does $x times y cdot N$ mean? Dot product $(x times y) cdot N$ ?
$endgroup$
– Widawensen
Mar 20 at 12:51
$begingroup$
What does $x times y cdot N$ mean? Dot product $(x times y) cdot N$ ?
$endgroup$
– Widawensen
Mar 20 at 12:51
4
4
$begingroup$
@Widawensen Yes, what else could it mean?
$endgroup$
– Marc van Leeuwen
Mar 20 at 12:53
$begingroup$
@Widawensen Yes, what else could it mean?
$endgroup$
– Marc van Leeuwen
Mar 20 at 12:53
1
1
$begingroup$
@MarcvanLeeuwen: that could mean a badly written problem. That happens here sometimes.
$endgroup$
– Taladris
Mar 21 at 2:54
$begingroup$
@MarcvanLeeuwen: that could mean a badly written problem. That happens here sometimes.
$endgroup$
– Taladris
Mar 21 at 2:54
add a comment |
7 Answers
7
active
oldest
votes
$begingroup$
Here's an observation: If $Q$ is a rotation matrix, then
$$
(Qx) times (Qy) = Q(x times y)
$$
You have to prove that, of course, but it's not too tough. Similarly,
$$
(Qx) cdot (Qy) = x cdot y
$$
and, for a scalar $alpha$, we have
$$
Q (alpha x) = alpha (Q x)
$$
Now suppose that for some vector $v$, we have
$$
(mathbfxtimesmathbfy cdot mathbfN) mathbfz + (mathbfytimesmathbfz cdot mathbfN) mathbfx + (mathbfztimesmathbfx cdot mathbfN) mathbfy=mathbfv.
$$
Key idea 1: You can apply the rules above to show that for any rotation matrix $Q$, you can apply $Q$ to all the elements on the left to get $Qv$.
Key idea 2: You can choose $Q$ so that it takes $N$ to the vector $(0,0,1)$, and puts $x, y,$ and $z$ into the plane consisting of vectors of the form $(a, b, 0)$. And in that plane, it's easy to see that you get $0$, so $Qv = 0$. Hence $v = 0$, and you're done.
In short: by a change of basis, you can assume that $N$ is the vector $(0,0,1)$ and that the other vectors all lie in the $(a, b, 0)$ plane, and things get easy.
answered Mar 20 at 11:45
John HughesJohn Hughes
65.1k24293
$endgroup$
$begingroup$
Ah so it is a clever approach with orthogonal matrices. I'm familiar with those identities. But I don't see how $Qv=0$. I can't see how we apply anything other than the third identity $Q(alpha x)=alpha (Qx)$.
$endgroup$
– AlephNull
Mar 20 at 11:55
$begingroup$
Oh I see, you're talking about the elements, not the terms. I understand the solution now.
$endgroup$
– AlephNull
Mar 20 at 13:42
2
$begingroup$
By the way, there's a general principle at work here, much loved by physicists, but worth remembering even if that's not your domain: "find the right coordinate system for your problem." (The math version is mostly "choose the right basis/generating set/...")
$endgroup$
– John Hughes
Mar 20 at 16:51
add a comment |
$begingroup$
If what is required is only to prove the validity of the given identity, there is another approach. Observe that if $x$ and $y$ are linearly dependent, i.e. for some $c$, $x=c y$ or $y=c x$, then the identity holds trivially because $wtimes v =-(vtimes w)$ and $v times v=0$ for all $v,w$. Thus, we may assume $x$ and $y$ are linearly independent and hence $z$ is a linear combination of $x$ and $y$, that is, $z=ax+by$ for some $a,b$. Now, since the given identity is linear in each variable and it holds for both $z=x$ and $z=y$, it is also true for $z=ax+by$. This proves the identity. It can be also noted that $N$ being perpendicular to the plane containing $x,y,z$ plays no role in this proof.
answered Mar 20 at 16:15
SongSong
18.5k21651
$endgroup$
2
$begingroup$
very nice solution!
$endgroup$
– John Hughes
Mar 20 at 16:54
$begingroup$
Indeed, this is very elegant. So my last remark had some significance!
$endgroup$
– AlephNull
Mar 20 at 17:06
$begingroup$
Thank you both :-)
$endgroup$
– Song
Mar 21 at 7:07
add a comment |
$begingroup$
Writing $x=ahati+bhatj,,y=chati+dhatj,,z=ehati+fhatj,,N=Nhatk$ reduces the sum to $$N((ad-bc)(ehati+fhatj)+(cf-de)(ahati+bhatj)+(be-af)(chati+dhatj)).$$The $hati$ coefficient is $N(ade-bce+acf-ade+bce-acf)=0$. The $hatj$ coefficient can be handled similarly.
answered Mar 20 at 12:45
J.G.J.G.
32.5k23250
$endgroup$
$begingroup$
I accepted a different answer but +1 because I appreciate that this finishes off the solution in that answer.
$endgroup$
– AlephNull
Mar 20 at 13:45
add a comment |
$begingroup$
Since $bf x, bf y, bf z$ are coplanar, they are linearly dependent. Since the result to be proved is symmetric in $bf x, bf y, bf z$, withouht loss of generality we can write $bf z = lambda bf x + mu bf y$ for some scalars $lambda, mu$.
Now, $$beginalign
& (bf y times bf z cdot bf N); bf x \
= & (bf y times (lambda bf x + mu bf y) cdot bf N); bf x \
= & (bf y times lambda bf x cdot bf N); bf x \
= & (bf y times bf x cdot bf N), (lambda bf x)
endalign$$
and similarly $$beginalign
& (bf z times bf x cdot bf N); bf y \
= & (bf y times bf x cdot bf N), (mu bf y)
endalign$$
So $$beginalign
& (bf y times bf z cdot bf N); bf x +
(bf z times bf x cdot bf N); bf y \
= & (bf y times bf x cdot bf N), (lambda bf x + mu bf y)\
= & -(bf x times bf y cdot bf N);z
endalign $$
and the result follows.
answered Mar 20 at 19:43
alephzeroalephzero
72037
$endgroup$
add a comment |
$begingroup$
If you know a little about the exterior algebra we can see this almost immediately, and in a way that generalizes substantially.
Pick any plane $Pi$ containing $bf x, bf y, bf z$. The map on $Pi$ defined by $$(bf a, bf b, bf c) mapsto [(bf a times bf b) cdot bf N] bf c + [(bf b times bf c) cdot bf N] bf a + [(bf c times bf a) cdot bf N] bf b$$
is visibly trilinear and totally skew in its arguments, so it is a (vector-valued) $3$-form on a $2$-dimensional vector space and hence is the zero map.
NB this argument doesn't use any properties of $bf N$.
answered Mar 21 at 5:02
TravisTravis
63.8k769151
$endgroup$
add a comment |
$begingroup$
By the properties of the triple product ( circluar shift) we can rearrange formula:
$ (mathbfxtimesmathbfy) cdot mathbfN) mathbfz + (mathbfytimesmathbfz) cdot mathbfN) mathbfx + (mathbfztimesmathbfx) cdot mathbfN) mathbfy \ =(mathbfNtimesmathbfx) cdot mathbfy) mathbfz + (mathbfNtimesmathbfy) cdot mathbfz) mathbfx + (mathbfNtimesmathbfz) cdot mathbfx) mathbfy $
All cross product vectors $$v_1=(mathbfNtimesmathbfx),v_2=(mathbfNtimesmathbfy), v_3=(mathbfNtimesmathbfz)$$
lie in the plane of coplanar vectors $mathbfx,mathbfy,mathbfz$ and they are vectors $mathbfx,mathbfy,mathbfz$ rotated by $pi/2$ in this plane.
So we can limit themselves to this plane and take any vectors with components $mathbfx=[ x_1 x_2]^T,mathbfy=[ y_1 y_2]^T,mathbfz =[ z_1 z_2]^T$.
Transform them with the rotation matrix $R=beginbmatrix 0 & -1 \ 1 & 0 endbmatrix$ , calculate appropriate dot products and finally check the formula with these assumed general components.
Namely we need to calculate:
$$(y^TRx)z+(z^TRy)x+(x^TRz)y$$
answered Mar 20 at 13:42
WidawensenWidawensen
4,75831446
$endgroup$
add a comment |
$begingroup$
Another approach to the problem uses a formula for triple product.
$ mathbfacdot(mathbfbtimes mathbfc) = det beginbmatrix
a_1 & b_1 & c_1 \
a_2 & b_2 & c_2 \
a_3 & b_3 & c_3 \
endbmatrix $
Then consider determinant
$beginvmatrix n_1 & x_1 & y_1 & z_1 \ n_2 & x_2 & y_2 & z_2 \ n_3 & x_3 & y_3 & z_3 \ n_1 & x_1 & y_1 & z_1 endvmatrix $
where columns consist of vectors $ mathbfN ,mathbfx,mathbfy,mathbfz$ components (the fourth row repeats the first one).
Of course such determinant equals to $0$.
Developing the determinant along the fourth row we obtain:
$-n_1beginvmatrix x_1 & y_1 & z_1 \ x_2 & y_2 & z_2 \ x_3 & y_3 & z_3 \ endvmatrix +x_1beginvmatrix n_1 & y_1 & z_1 \ n_2 & y_2 & z_2 \ n_3 & y_3 & z_3 \ endvmatrix -y_1beginvmatrix n_1 & x_1 & z_1 \ n_2 & x_2 & z_2 \ n_3 & x_3 & z_3 \ endvmatrix +z_1beginvmatrix n_1 & x_1 & y_1 \ n_2 & x_2 & y_2 \ n_3 & x_3 & y_3 \ endvmatrix=0$
from which the formula for the first component of the vector given in the question follows
(the first summand is equal to $0$ as the vectors $mathbfx,mathbfy,mathbfz$ are collinear, the columns can be permuted (required for the third summand) if needed to give appropriate sign in expression)
Similarly the determinants
$beginvmatrix n_1 & x_1 & y_1 & z_1 \ n_2 & x_2 & y_2 & z_2 \ n_3 & x_3 & y_3 & z_3 \ n_2 & x_2 & y_2 & z_2 endvmatrix $ and $beginvmatrix n_1 & x_1 & y_1 & z_1 \ n_2 & x_2 & y_2 & z_2 \ n_3 & x_3 & y_3 & z_3 \ n_3 & x_3 & y_3 & z_3 endvmatrix $
give the second and the third component of the question vector, equal to $0$.
answered Mar 29 at 13:57
WidawensenWidawensen
4,75831446
$endgroup$
add a comment |
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7 Answers
7
active
oldest
votes
7 Answers
7
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Here's an observation: If $Q$ is a rotation matrix, then
$$
(Qx) times (Qy) = Q(x times y)
$$
You have to prove that, of course, but it's not too tough. Similarly,
$$
(Qx) cdot (Qy) = x cdot y
$$
and, for a scalar $alpha$, we have
$$
Q (alpha x) = alpha (Q x)
$$
Now suppose that for some vector $v$, we have
$$
(mathbfxtimesmathbfy cdot mathbfN) mathbfz + (mathbfytimesmathbfz cdot mathbfN) mathbfx + (mathbfztimesmathbfx cdot mathbfN) mathbfy=mathbfv.
$$
Key idea 1: You can apply the rules above to show that for any rotation matrix $Q$, you can apply $Q$ to all the elements on the left to get $Qv$.
Key idea 2: You can choose $Q$ so that it takes $N$ to the vector $(0,0,1)$, and puts $x, y,$ and $z$ into the plane consisting of vectors of the form $(a, b, 0)$. And in that plane, it's easy to see that you get $0$, so $Qv = 0$. Hence $v = 0$, and you're done.
In short: by a change of basis, you can assume that $N$ is the vector $(0,0,1)$ and that the other vectors all lie in the $(a, b, 0)$ plane, and things get easy.
answered Mar 20 at 11:45
John HughesJohn Hughes
65.1k24293
$endgroup$
$begingroup$
Ah so it is a clever approach with orthogonal matrices. I'm familiar with those identities. But I don't see how $Qv=0$. I can't see how we apply anything other than the third identity $Q(alpha x)=alpha (Qx)$.
$endgroup$
– AlephNull
Mar 20 at 11:55
$begingroup$
Oh I see, you're talking about the elements, not the terms. I understand the solution now.
$endgroup$
– AlephNull
Mar 20 at 13:42
2
$begingroup$
By the way, there's a general principle at work here, much loved by physicists, but worth remembering even if that's not your domain: "find the right coordinate system for your problem." (The math version is mostly "choose the right basis/generating set/...")
$endgroup$
– John Hughes
Mar 20 at 16:51
add a comment |
$begingroup$
Here's an observation: If $Q$ is a rotation matrix, then
$$
(Qx) times (Qy) = Q(x times y)
$$
You have to prove that, of course, but it's not too tough. Similarly,
$$
(Qx) cdot (Qy) = x cdot y
$$
and, for a scalar $alpha$, we have
$$
Q (alpha x) = alpha (Q x)
$$
Now suppose that for some vector $v$, we have
$$
(mathbfxtimesmathbfy cdot mathbfN) mathbfz + (mathbfytimesmathbfz cdot mathbfN) mathbfx + (mathbfztimesmathbfx cdot mathbfN) mathbfy=mathbfv.
$$
Key idea 1: You can apply the rules above to show that for any rotation matrix $Q$, you can apply $Q$ to all the elements on the left to get $Qv$.
Key idea 2: You can choose $Q$ so that it takes $N$ to the vector $(0,0,1)$, and puts $x, y,$ and $z$ into the plane consisting of vectors of the form $(a, b, 0)$. And in that plane, it's easy to see that you get $0$, so $Qv = 0$. Hence $v = 0$, and you're done.
In short: by a change of basis, you can assume that $N$ is the vector $(0,0,1)$ and that the other vectors all lie in the $(a, b, 0)$ plane, and things get easy.
answered Mar 20 at 11:45
John HughesJohn Hughes
65.1k24293
$endgroup$
$begingroup$
Ah so it is a clever approach with orthogonal matrices. I'm familiar with those identities. But I don't see how $Qv=0$. I can't see how we apply anything other than the third identity $Q(alpha x)=alpha (Qx)$.
$endgroup$
– AlephNull
Mar 20 at 11:55
$begingroup$
Oh I see, you're talking about the elements, not the terms. I understand the solution now.
$endgroup$
– AlephNull
Mar 20 at 13:42
2
$begingroup$
By the way, there's a general principle at work here, much loved by physicists, but worth remembering even if that's not your domain: "find the right coordinate system for your problem." (The math version is mostly "choose the right basis/generating set/...")
$endgroup$
– John Hughes
Mar 20 at 16:51
add a comment |
$begingroup$
Here's an observation: If $Q$ is a rotation matrix, then
$$
(Qx) times (Qy) = Q(x times y)
$$
You have to prove that, of course, but it's not too tough. Similarly,
$$
(Qx) cdot (Qy) = x cdot y
$$
and, for a scalar $alpha$, we have
$$
Q (alpha x) = alpha (Q x)
$$
Now suppose that for some vector $v$, we have
$$
(mathbfxtimesmathbfy cdot mathbfN) mathbfz + (mathbfytimesmathbfz cdot mathbfN) mathbfx + (mathbfztimesmathbfx cdot mathbfN) mathbfy=mathbfv.
$$
Key idea 1: You can apply the rules above to show that for any rotation matrix $Q$, you can apply $Q$ to all the elements on the left to get $Qv$.
Key idea 2: You can choose $Q$ so that it takes $N$ to the vector $(0,0,1)$, and puts $x, y,$ and $z$ into the plane consisting of vectors of the form $(a, b, 0)$. And in that plane, it's easy to see that you get $0$, so $Qv = 0$. Hence $v = 0$, and you're done.
In short: by a change of basis, you can assume that $N$ is the vector $(0,0,1)$ and that the other vectors all lie in the $(a, b, 0)$ plane, and things get easy.
answered Mar 20 at 11:45
John HughesJohn Hughes
65.1k24293
$endgroup$
Here's an observation: If $Q$ is a rotation matrix, then
$$
(Qx) times (Qy) = Q(x times y)
$$
You have to prove that, of course, but it's not too tough. Similarly,
$$
(Qx) cdot (Qy) = x cdot y
$$
and, for a scalar $alpha$, we have
$$
Q (alpha x) = alpha (Q x)
$$
Now suppose that for some vector $v$, we have
$$
(mathbfxtimesmathbfy cdot mathbfN) mathbfz + (mathbfytimesmathbfz cdot mathbfN) mathbfx + (mathbfztimesmathbfx cdot mathbfN) mathbfy=mathbfv.
$$
Key idea 1: You can apply the rules above to show that for any rotation matrix $Q$, you can apply $Q$ to all the elements on the left to get $Qv$.
Key idea 2: You can choose $Q$ so that it takes $N$ to the vector $(0,0,1)$, and puts $x, y,$ and $z$ into the plane consisting of vectors of the form $(a, b, 0)$. And in that plane, it's easy to see that you get $0$, so $Qv = 0$. Hence $v = 0$, and you're done.
In short: by a change of basis, you can assume that $N$ is the vector $(0,0,1)$ and that the other vectors all lie in the $(a, b, 0)$ plane, and things get easy.
answered Mar 20 at 11:45
John HughesJohn Hughes
65.1k24293
answered Mar 20 at 11:45
John HughesJohn Hughes
65.1k24293
answered Mar 20 at 11:45
John HughesJohn Hughes
65.1k24293
answered Mar 20 at 11:45
John HughesJohn Hughes
65.1k24293
65.1k24293
$begingroup$
Ah so it is a clever approach with orthogonal matrices. I'm familiar with those identities. But I don't see how $Qv=0$. I can't see how we apply anything other than the third identity $Q(alpha x)=alpha (Qx)$.
$endgroup$
– AlephNull
Mar 20 at 11:55
$begingroup$
Oh I see, you're talking about the elements, not the terms. I understand the solution now.
$endgroup$
– AlephNull
Mar 20 at 13:42
2
$begingroup$
By the way, there's a general principle at work here, much loved by physicists, but worth remembering even if that's not your domain: "find the right coordinate system for your problem." (The math version is mostly "choose the right basis/generating set/...")
$endgroup$
– John Hughes
Mar 20 at 16:51
add a comment |
$begingroup$
Ah so it is a clever approach with orthogonal matrices. I'm familiar with those identities. But I don't see how $Qv=0$. I can't see how we apply anything other than the third identity $Q(alpha x)=alpha (Qx)$.
$endgroup$
– AlephNull
Mar 20 at 11:55
$begingroup$
Oh I see, you're talking about the elements, not the terms. I understand the solution now.
$endgroup$
– AlephNull
Mar 20 at 13:42
2
$begingroup$
By the way, there's a general principle at work here, much loved by physicists, but worth remembering even if that's not your domain: "find the right coordinate system for your problem." (The math version is mostly "choose the right basis/generating set/...")
$endgroup$
– John Hughes
Mar 20 at 16:51
$begingroup$
Ah so it is a clever approach with orthogonal matrices. I'm familiar with those identities. But I don't see how $Qv=0$. I can't see how we apply anything other than the third identity $Q(alpha x)=alpha (Qx)$.
$endgroup$
– AlephNull
Mar 20 at 11:55
$begingroup$
Ah so it is a clever approach with orthogonal matrices. I'm familiar with those identities. But I don't see how $Qv=0$. I can't see how we apply anything other than the third identity $Q(alpha x)=alpha (Qx)$.
$endgroup$
– AlephNull
Mar 20 at 11:55
$begingroup$
Oh I see, you're talking about the elements, not the terms. I understand the solution now.
$endgroup$
– AlephNull
Mar 20 at 13:42
$begingroup$
Oh I see, you're talking about the elements, not the terms. I understand the solution now.
$endgroup$
– AlephNull
Mar 20 at 13:42
2
2
$begingroup$
By the way, there's a general principle at work here, much loved by physicists, but worth remembering even if that's not your domain: "find the right coordinate system for your problem." (The math version is mostly "choose the right basis/generating set/...")
$endgroup$
– John Hughes
Mar 20 at 16:51
$begingroup$
By the way, there's a general principle at work here, much loved by physicists, but worth remembering even if that's not your domain: "find the right coordinate system for your problem." (The math version is mostly "choose the right basis/generating set/...")
$endgroup$
– John Hughes
Mar 20 at 16:51
add a comment |
$begingroup$
If what is required is only to prove the validity of the given identity, there is another approach. Observe that if $x$ and $y$ are linearly dependent, i.e. for some $c$, $x=c y$ or $y=c x$, then the identity holds trivially because $wtimes v =-(vtimes w)$ and $v times v=0$ for all $v,w$. Thus, we may assume $x$ and $y$ are linearly independent and hence $z$ is a linear combination of $x$ and $y$, that is, $z=ax+by$ for some $a,b$. Now, since the given identity is linear in each variable and it holds for both $z=x$ and $z=y$, it is also true for $z=ax+by$. This proves the identity. It can be also noted that $N$ being perpendicular to the plane containing $x,y,z$ plays no role in this proof.
answered Mar 20 at 16:15
SongSong
18.5k21651
$endgroup$
2
$begingroup$
very nice solution!
$endgroup$
– John Hughes
Mar 20 at 16:54
$begingroup$
Indeed, this is very elegant. So my last remark had some significance!
$endgroup$
– AlephNull
Mar 20 at 17:06
$begingroup$
Thank you both :-)
$endgroup$
– Song
Mar 21 at 7:07
add a comment |
$begingroup$
If what is required is only to prove the validity of the given identity, there is another approach. Observe that if $x$ and $y$ are linearly dependent, i.e. for some $c$, $x=c y$ or $y=c x$, then the identity holds trivially because $wtimes v =-(vtimes w)$ and $v times v=0$ for all $v,w$. Thus, we may assume $x$ and $y$ are linearly independent and hence $z$ is a linear combination of $x$ and $y$, that is, $z=ax+by$ for some $a,b$. Now, since the given identity is linear in each variable and it holds for both $z=x$ and $z=y$, it is also true for $z=ax+by$. This proves the identity. It can be also noted that $N$ being perpendicular to the plane containing $x,y,z$ plays no role in this proof.
answered Mar 20 at 16:15
SongSong
18.5k21651
$endgroup$
2
$begingroup$
very nice solution!
$endgroup$
– John Hughes
Mar 20 at 16:54
$begingroup$
Indeed, this is very elegant. So my last remark had some significance!
$endgroup$
– AlephNull
Mar 20 at 17:06
$begingroup$
Thank you both :-)
$endgroup$
– Song
Mar 21 at 7:07
add a comment |
$begingroup$
If what is required is only to prove the validity of the given identity, there is another approach. Observe that if $x$ and $y$ are linearly dependent, i.e. for some $c$, $x=c y$ or $y=c x$, then the identity holds trivially because $wtimes v =-(vtimes w)$ and $v times v=0$ for all $v,w$. Thus, we may assume $x$ and $y$ are linearly independent and hence $z$ is a linear combination of $x$ and $y$, that is, $z=ax+by$ for some $a,b$. Now, since the given identity is linear in each variable and it holds for both $z=x$ and $z=y$, it is also true for $z=ax+by$. This proves the identity. It can be also noted that $N$ being perpendicular to the plane containing $x,y,z$ plays no role in this proof.
answered Mar 20 at 16:15
SongSong
18.5k21651
$endgroup$
If what is required is only to prove the validity of the given identity, there is another approach. Observe that if $x$ and $y$ are linearly dependent, i.e. for some $c$, $x=c y$ or $y=c x$, then the identity holds trivially because $wtimes v =-(vtimes w)$ and $v times v=0$ for all $v,w$. Thus, we may assume $x$ and $y$ are linearly independent and hence $z$ is a linear combination of $x$ and $y$, that is, $z=ax+by$ for some $a,b$. Now, since the given identity is linear in each variable and it holds for both $z=x$ and $z=y$, it is also true for $z=ax+by$. This proves the identity. It can be also noted that $N$ being perpendicular to the plane containing $x,y,z$ plays no role in this proof.
answered Mar 20 at 16:15
SongSong
18.5k21651
edited Mar 20 at 16:22
answered Mar 20 at 16:15
SongSong
18.5k21651
answered Mar 20 at 16:15
SongSong
18.5k21651
answered Mar 20 at 16:15
SongSong
18.5k21651
18.5k21651
2
$begingroup$
very nice solution!
$endgroup$
– John Hughes
Mar 20 at 16:54
$begingroup$
Indeed, this is very elegant. So my last remark had some significance!
$endgroup$
– AlephNull
Mar 20 at 17:06
$begingroup$
Thank you both :-)
$endgroup$
– Song
Mar 21 at 7:07
add a comment |
2
$begingroup$
very nice solution!
$endgroup$
– John Hughes
Mar 20 at 16:54
$begingroup$
Indeed, this is very elegant. So my last remark had some significance!
$endgroup$
– AlephNull
Mar 20 at 17:06
$begingroup$
Thank you both :-)
$endgroup$
– Song
Mar 21 at 7:07
2
2
$begingroup$
very nice solution!
$endgroup$
– John Hughes
Mar 20 at 16:54
$begingroup$
very nice solution!
$endgroup$
– John Hughes
Mar 20 at 16:54
$begingroup$
Indeed, this is very elegant. So my last remark had some significance!
$endgroup$
– AlephNull
Mar 20 at 17:06
$begingroup$
Indeed, this is very elegant. So my last remark had some significance!
$endgroup$
– AlephNull
Mar 20 at 17:06
$begingroup$
Thank you both :-)
$endgroup$
– Song
Mar 21 at 7:07
$begingroup$
Thank you both :-)
$endgroup$
– Song
Mar 21 at 7:07
add a comment |
$begingroup$
Writing $x=ahati+bhatj,,y=chati+dhatj,,z=ehati+fhatj,,N=Nhatk$ reduces the sum to $$N((ad-bc)(ehati+fhatj)+(cf-de)(ahati+bhatj)+(be-af)(chati+dhatj)).$$The $hati$ coefficient is $N(ade-bce+acf-ade+bce-acf)=0$. The $hatj$ coefficient can be handled similarly.
answered Mar 20 at 12:45
J.G.J.G.
32.5k23250
$endgroup$
$begingroup$
I accepted a different answer but +1 because I appreciate that this finishes off the solution in that answer.
$endgroup$
– AlephNull
Mar 20 at 13:45
add a comment |
$begingroup$
Writing $x=ahati+bhatj,,y=chati+dhatj,,z=ehati+fhatj,,N=Nhatk$ reduces the sum to $$N((ad-bc)(ehati+fhatj)+(cf-de)(ahati+bhatj)+(be-af)(chati+dhatj)).$$The $hati$ coefficient is $N(ade-bce+acf-ade+bce-acf)=0$. The $hatj$ coefficient can be handled similarly.
answered Mar 20 at 12:45
J.G.J.G.
32.5k23250
$endgroup$
$begingroup$
I accepted a different answer but +1 because I appreciate that this finishes off the solution in that answer.
$endgroup$
– AlephNull
Mar 20 at 13:45
add a comment |
$begingroup$
Writing $x=ahati+bhatj,,y=chati+dhatj,,z=ehati+fhatj,,N=Nhatk$ reduces the sum to $$N((ad-bc)(ehati+fhatj)+(cf-de)(ahati+bhatj)+(be-af)(chati+dhatj)).$$The $hati$ coefficient is $N(ade-bce+acf-ade+bce-acf)=0$. The $hatj$ coefficient can be handled similarly.
answered Mar 20 at 12:45
J.G.J.G.
32.5k23250
$endgroup$
Writing $x=ahati+bhatj,,y=chati+dhatj,,z=ehati+fhatj,,N=Nhatk$ reduces the sum to $$N((ad-bc)(ehati+fhatj)+(cf-de)(ahati+bhatj)+(be-af)(chati+dhatj)).$$The $hati$ coefficient is $N(ade-bce+acf-ade+bce-acf)=0$. The $hatj$ coefficient can be handled similarly.
answered Mar 20 at 12:45
J.G.J.G.
32.5k23250
answered Mar 20 at 12:45
J.G.J.G.
32.5k23250
answered Mar 20 at 12:45
J.G.J.G.
32.5k23250
answered Mar 20 at 12:45
J.G.J.G.
32.5k23250
32.5k23250
$begingroup$
I accepted a different answer but +1 because I appreciate that this finishes off the solution in that answer.
$endgroup$
– AlephNull
Mar 20 at 13:45
add a comment |
$begingroup$
I accepted a different answer but +1 because I appreciate that this finishes off the solution in that answer.
$endgroup$
– AlephNull
Mar 20 at 13:45
$begingroup$
I accepted a different answer but +1 because I appreciate that this finishes off the solution in that answer.
$endgroup$
– AlephNull
Mar 20 at 13:45
$begingroup$
I accepted a different answer but +1 because I appreciate that this finishes off the solution in that answer.
$endgroup$
– AlephNull
Mar 20 at 13:45
add a comment |
$begingroup$
Since $bf x, bf y, bf z$ are coplanar, they are linearly dependent. Since the result to be proved is symmetric in $bf x, bf y, bf z$, withouht loss of generality we can write $bf z = lambda bf x + mu bf y$ for some scalars $lambda, mu$.
Now, $$beginalign
& (bf y times bf z cdot bf N); bf x \
= & (bf y times (lambda bf x + mu bf y) cdot bf N); bf x \
= & (bf y times lambda bf x cdot bf N); bf x \
= & (bf y times bf x cdot bf N), (lambda bf x)
endalign$$
and similarly $$beginalign
& (bf z times bf x cdot bf N); bf y \
= & (bf y times bf x cdot bf N), (mu bf y)
endalign$$
So $$beginalign
& (bf y times bf z cdot bf N); bf x +
(bf z times bf x cdot bf N); bf y \
= & (bf y times bf x cdot bf N), (lambda bf x + mu bf y)\
= & -(bf x times bf y cdot bf N);z
endalign $$
and the result follows.
answered Mar 20 at 19:43
alephzeroalephzero
72037
$endgroup$
add a comment |
$begingroup$
Since $bf x, bf y, bf z$ are coplanar, they are linearly dependent. Since the result to be proved is symmetric in $bf x, bf y, bf z$, withouht loss of generality we can write $bf z = lambda bf x + mu bf y$ for some scalars $lambda, mu$.
Now, $$beginalign
& (bf y times bf z cdot bf N); bf x \
= & (bf y times (lambda bf x + mu bf y) cdot bf N); bf x \
= & (bf y times lambda bf x cdot bf N); bf x \
= & (bf y times bf x cdot bf N), (lambda bf x)
endalign$$
and similarly $$beginalign
& (bf z times bf x cdot bf N); bf y \
= & (bf y times bf x cdot bf N), (mu bf y)
endalign$$
So $$beginalign
& (bf y times bf z cdot bf N); bf x +
(bf z times bf x cdot bf N); bf y \
= & (bf y times bf x cdot bf N), (lambda bf x + mu bf y)\
= & -(bf x times bf y cdot bf N);z
endalign $$
and the result follows.
answered Mar 20 at 19:43
alephzeroalephzero
72037
$endgroup$
add a comment |
$begingroup$
Since $bf x, bf y, bf z$ are coplanar, they are linearly dependent. Since the result to be proved is symmetric in $bf x, bf y, bf z$, withouht loss of generality we can write $bf z = lambda bf x + mu bf y$ for some scalars $lambda, mu$.
Now, $$beginalign
& (bf y times bf z cdot bf N); bf x \
= & (bf y times (lambda bf x + mu bf y) cdot bf N); bf x \
= & (bf y times lambda bf x cdot bf N); bf x \
= & (bf y times bf x cdot bf N), (lambda bf x)
endalign$$
and similarly $$beginalign
& (bf z times bf x cdot bf N); bf y \
= & (bf y times bf x cdot bf N), (mu bf y)
endalign$$
So $$beginalign
& (bf y times bf z cdot bf N); bf x +
(bf z times bf x cdot bf N); bf y \
= & (bf y times bf x cdot bf N), (lambda bf x + mu bf y)\
= & -(bf x times bf y cdot bf N);z
endalign $$
and the result follows.
answered Mar 20 at 19:43
alephzeroalephzero
72037
$endgroup$
Since $bf x, bf y, bf z$ are coplanar, they are linearly dependent. Since the result to be proved is symmetric in $bf x, bf y, bf z$, withouht loss of generality we can write $bf z = lambda bf x + mu bf y$ for some scalars $lambda, mu$.
Now, $$beginalign
& (bf y times bf z cdot bf N); bf x \
= & (bf y times (lambda bf x + mu bf y) cdot bf N); bf x \
= & (bf y times lambda bf x cdot bf N); bf x \
= & (bf y times bf x cdot bf N), (lambda bf x)
endalign$$
and similarly $$beginalign
& (bf z times bf x cdot bf N); bf y \
= & (bf y times bf x cdot bf N), (mu bf y)
endalign$$
So $$beginalign
& (bf y times bf z cdot bf N); bf x +
(bf z times bf x cdot bf N); bf y \
= & (bf y times bf x cdot bf N), (lambda bf x + mu bf y)\
= & -(bf x times bf y cdot bf N);z
endalign $$
and the result follows.
answered Mar 20 at 19:43
alephzeroalephzero
72037
answered Mar 20 at 19:43
alephzeroalephzero
72037
answered Mar 20 at 19:43
alephzeroalephzero
72037
answered Mar 20 at 19:43
alephzeroalephzero
72037
72037
add a comment |
add a comment |
$begingroup$
If you know a little about the exterior algebra we can see this almost immediately, and in a way that generalizes substantially.
Pick any plane $Pi$ containing $bf x, bf y, bf z$. The map on $Pi$ defined by $$(bf a, bf b, bf c) mapsto [(bf a times bf b) cdot bf N] bf c + [(bf b times bf c) cdot bf N] bf a + [(bf c times bf a) cdot bf N] bf b$$
is visibly trilinear and totally skew in its arguments, so it is a (vector-valued) $3$-form on a $2$-dimensional vector space and hence is the zero map.
NB this argument doesn't use any properties of $bf N$.
answered Mar 21 at 5:02
TravisTravis
63.8k769151
$endgroup$
add a comment |
$begingroup$
If you know a little about the exterior algebra we can see this almost immediately, and in a way that generalizes substantially.
Pick any plane $Pi$ containing $bf x, bf y, bf z$. The map on $Pi$ defined by $$(bf a, bf b, bf c) mapsto [(bf a times bf b) cdot bf N] bf c + [(bf b times bf c) cdot bf N] bf a + [(bf c times bf a) cdot bf N] bf b$$
is visibly trilinear and totally skew in its arguments, so it is a (vector-valued) $3$-form on a $2$-dimensional vector space and hence is the zero map.
NB this argument doesn't use any properties of $bf N$.
answered Mar 21 at 5:02
TravisTravis
63.8k769151
$endgroup$
add a comment |
$begingroup$
If you know a little about the exterior algebra we can see this almost immediately, and in a way that generalizes substantially.
Pick any plane $Pi$ containing $bf x, bf y, bf z$. The map on $Pi$ defined by $$(bf a, bf b, bf c) mapsto [(bf a times bf b) cdot bf N] bf c + [(bf b times bf c) cdot bf N] bf a + [(bf c times bf a) cdot bf N] bf b$$
is visibly trilinear and totally skew in its arguments, so it is a (vector-valued) $3$-form on a $2$-dimensional vector space and hence is the zero map.
NB this argument doesn't use any properties of $bf N$.
answered Mar 21 at 5:02
TravisTravis
63.8k769151
$endgroup$
If you know a little about the exterior algebra we can see this almost immediately, and in a way that generalizes substantially.
Pick any plane $Pi$ containing $bf x, bf y, bf z$. The map on $Pi$ defined by $$(bf a, bf b, bf c) mapsto [(bf a times bf b) cdot bf N] bf c + [(bf b times bf c) cdot bf N] bf a + [(bf c times bf a) cdot bf N] bf b$$
is visibly trilinear and totally skew in its arguments, so it is a (vector-valued) $3$-form on a $2$-dimensional vector space and hence is the zero map.
NB this argument doesn't use any properties of $bf N$.
answered Mar 21 at 5:02
TravisTravis
63.8k769151
answered Mar 21 at 5:02
TravisTravis
63.8k769151
answered Mar 21 at 5:02
TravisTravis
63.8k769151
answered Mar 21 at 5:02
TravisTravis
63.8k769151
63.8k769151
add a comment |
add a comment |
$begingroup$
By the properties of the triple product ( circluar shift) we can rearrange formula:
$ (mathbfxtimesmathbfy) cdot mathbfN) mathbfz + (mathbfytimesmathbfz) cdot mathbfN) mathbfx + (mathbfztimesmathbfx) cdot mathbfN) mathbfy \ =(mathbfNtimesmathbfx) cdot mathbfy) mathbfz + (mathbfNtimesmathbfy) cdot mathbfz) mathbfx + (mathbfNtimesmathbfz) cdot mathbfx) mathbfy $
All cross product vectors $$v_1=(mathbfNtimesmathbfx),v_2=(mathbfNtimesmathbfy), v_3=(mathbfNtimesmathbfz)$$
lie in the plane of coplanar vectors $mathbfx,mathbfy,mathbfz$ and they are vectors $mathbfx,mathbfy,mathbfz$ rotated by $pi/2$ in this plane.
So we can limit themselves to this plane and take any vectors with components $mathbfx=[ x_1 x_2]^T,mathbfy=[ y_1 y_2]^T,mathbfz =[ z_1 z_2]^T$.
Transform them with the rotation matrix $R=beginbmatrix 0 & -1 \ 1 & 0 endbmatrix$ , calculate appropriate dot products and finally check the formula with these assumed general components.
Namely we need to calculate:
$$(y^TRx)z+(z^TRy)x+(x^TRz)y$$
answered Mar 20 at 13:42
WidawensenWidawensen
4,75831446
$endgroup$
add a comment |
$begingroup$
By the properties of the triple product ( circluar shift) we can rearrange formula:
$ (mathbfxtimesmathbfy) cdot mathbfN) mathbfz + (mathbfytimesmathbfz) cdot mathbfN) mathbfx + (mathbfztimesmathbfx) cdot mathbfN) mathbfy \ =(mathbfNtimesmathbfx) cdot mathbfy) mathbfz + (mathbfNtimesmathbfy) cdot mathbfz) mathbfx + (mathbfNtimesmathbfz) cdot mathbfx) mathbfy $
All cross product vectors $$v_1=(mathbfNtimesmathbfx),v_2=(mathbfNtimesmathbfy), v_3=(mathbfNtimesmathbfz)$$
lie in the plane of coplanar vectors $mathbfx,mathbfy,mathbfz$ and they are vectors $mathbfx,mathbfy,mathbfz$ rotated by $pi/2$ in this plane.
So we can limit themselves to this plane and take any vectors with components $mathbfx=[ x_1 x_2]^T,mathbfy=[ y_1 y_2]^T,mathbfz =[ z_1 z_2]^T$.
Transform them with the rotation matrix $R=beginbmatrix 0 & -1 \ 1 & 0 endbmatrix$ , calculate appropriate dot products and finally check the formula with these assumed general components.
Namely we need to calculate:
$$(y^TRx)z+(z^TRy)x+(x^TRz)y$$
answered Mar 20 at 13:42
WidawensenWidawensen
4,75831446
$endgroup$
add a comment |
$begingroup$
By the properties of the triple product ( circluar shift) we can rearrange formula:
$ (mathbfxtimesmathbfy) cdot mathbfN) mathbfz + (mathbfytimesmathbfz) cdot mathbfN) mathbfx + (mathbfztimesmathbfx) cdot mathbfN) mathbfy \ =(mathbfNtimesmathbfx) cdot mathbfy) mathbfz + (mathbfNtimesmathbfy) cdot mathbfz) mathbfx + (mathbfNtimesmathbfz) cdot mathbfx) mathbfy $
All cross product vectors $$v_1=(mathbfNtimesmathbfx),v_2=(mathbfNtimesmathbfy), v_3=(mathbfNtimesmathbfz)$$
lie in the plane of coplanar vectors $mathbfx,mathbfy,mathbfz$ and they are vectors $mathbfx,mathbfy,mathbfz$ rotated by $pi/2$ in this plane.
So we can limit themselves to this plane and take any vectors with components $mathbfx=[ x_1 x_2]^T,mathbfy=[ y_1 y_2]^T,mathbfz =[ z_1 z_2]^T$.
Transform them with the rotation matrix $R=beginbmatrix 0 & -1 \ 1 & 0 endbmatrix$ , calculate appropriate dot products and finally check the formula with these assumed general components.
Namely we need to calculate:
$$(y^TRx)z+(z^TRy)x+(x^TRz)y$$
answered Mar 20 at 13:42
WidawensenWidawensen
4,75831446
$endgroup$
By the properties of the triple product ( circluar shift) we can rearrange formula:
$ (mathbfxtimesmathbfy) cdot mathbfN) mathbfz + (mathbfytimesmathbfz) cdot mathbfN) mathbfx + (mathbfztimesmathbfx) cdot mathbfN) mathbfy \ =(mathbfNtimesmathbfx) cdot mathbfy) mathbfz + (mathbfNtimesmathbfy) cdot mathbfz) mathbfx + (mathbfNtimesmathbfz) cdot mathbfx) mathbfy $
All cross product vectors $$v_1=(mathbfNtimesmathbfx),v_2=(mathbfNtimesmathbfy), v_3=(mathbfNtimesmathbfz)$$
lie in the plane of coplanar vectors $mathbfx,mathbfy,mathbfz$ and they are vectors $mathbfx,mathbfy,mathbfz$ rotated by $pi/2$ in this plane.
So we can limit themselves to this plane and take any vectors with components $mathbfx=[ x_1 x_2]^T,mathbfy=[ y_1 y_2]^T,mathbfz =[ z_1 z_2]^T$.
Transform them with the rotation matrix $R=beginbmatrix 0 & -1 \ 1 & 0 endbmatrix$ , calculate appropriate dot products and finally check the formula with these assumed general components.
Namely we need to calculate:
$$(y^TRx)z+(z^TRy)x+(x^TRz)y$$
answered Mar 20 at 13:42
WidawensenWidawensen
4,75831446
edited Mar 21 at 9:01
answered Mar 20 at 13:42
WidawensenWidawensen
4,75831446
answered Mar 20 at 13:42
WidawensenWidawensen
4,75831446
answered Mar 20 at 13:42
WidawensenWidawensen
4,75831446
4,75831446
add a comment |
add a comment |
$begingroup$
Another approach to the problem uses a formula for triple product.
$ mathbfacdot(mathbfbtimes mathbfc) = det beginbmatrix
a_1 & b_1 & c_1 \
a_2 & b_2 & c_2 \
a_3 & b_3 & c_3 \
endbmatrix $
Then consider determinant
$beginvmatrix n_1 & x_1 & y_1 & z_1 \ n_2 & x_2 & y_2 & z_2 \ n_3 & x_3 & y_3 & z_3 \ n_1 & x_1 & y_1 & z_1 endvmatrix $
where columns consist of vectors $ mathbfN ,mathbfx,mathbfy,mathbfz$ components (the fourth row repeats the first one).
Of course such determinant equals to $0$.
Developing the determinant along the fourth row we obtain:
$-n_1beginvmatrix x_1 & y_1 & z_1 \ x_2 & y_2 & z_2 \ x_3 & y_3 & z_3 \ endvmatrix +x_1beginvmatrix n_1 & y_1 & z_1 \ n_2 & y_2 & z_2 \ n_3 & y_3 & z_3 \ endvmatrix -y_1beginvmatrix n_1 & x_1 & z_1 \ n_2 & x_2 & z_2 \ n_3 & x_3 & z_3 \ endvmatrix +z_1beginvmatrix n_1 & x_1 & y_1 \ n_2 & x_2 & y_2 \ n_3 & x_3 & y_3 \ endvmatrix=0$
from which the formula for the first component of the vector given in the question follows
(the first summand is equal to $0$ as the vectors $mathbfx,mathbfy,mathbfz$ are collinear, the columns can be permuted (required for the third summand) if needed to give appropriate sign in expression)
Similarly the determinants
$beginvmatrix n_1 & x_1 & y_1 & z_1 \ n_2 & x_2 & y_2 & z_2 \ n_3 & x_3 & y_3 & z_3 \ n_2 & x_2 & y_2 & z_2 endvmatrix $ and $beginvmatrix n_1 & x_1 & y_1 & z_1 \ n_2 & x_2 & y_2 & z_2 \ n_3 & x_3 & y_3 & z_3 \ n_3 & x_3 & y_3 & z_3 endvmatrix $
give the second and the third component of the question vector, equal to $0$.
answered Mar 29 at 13:57
WidawensenWidawensen
4,75831446
$endgroup$
add a comment |
$begingroup$
Another approach to the problem uses a formula for triple product.
$ mathbfacdot(mathbfbtimes mathbfc) = det beginbmatrix
a_1 & b_1 & c_1 \
a_2 & b_2 & c_2 \
a_3 & b_3 & c_3 \
endbmatrix $
Then consider determinant
$beginvmatrix n_1 & x_1 & y_1 & z_1 \ n_2 & x_2 & y_2 & z_2 \ n_3 & x_3 & y_3 & z_3 \ n_1 & x_1 & y_1 & z_1 endvmatrix $
where columns consist of vectors $ mathbfN ,mathbfx,mathbfy,mathbfz$ components (the fourth row repeats the first one).
Of course such determinant equals to $0$.
Developing the determinant along the fourth row we obtain:
$-n_1beginvmatrix x_1 & y_1 & z_1 \ x_2 & y_2 & z_2 \ x_3 & y_3 & z_3 \ endvmatrix +x_1beginvmatrix n_1 & y_1 & z_1 \ n_2 & y_2 & z_2 \ n_3 & y_3 & z_3 \ endvmatrix -y_1beginvmatrix n_1 & x_1 & z_1 \ n_2 & x_2 & z_2 \ n_3 & x_3 & z_3 \ endvmatrix +z_1beginvmatrix n_1 & x_1 & y_1 \ n_2 & x_2 & y_2 \ n_3 & x_3 & y_3 \ endvmatrix=0$
from which the formula for the first component of the vector given in the question follows
(the first summand is equal to $0$ as the vectors $mathbfx,mathbfy,mathbfz$ are collinear, the columns can be permuted (required for the third summand) if needed to give appropriate sign in expression)
Similarly the determinants
$beginvmatrix n_1 & x_1 & y_1 & z_1 \ n_2 & x_2 & y_2 & z_2 \ n_3 & x_3 & y_3 & z_3 \ n_2 & x_2 & y_2 & z_2 endvmatrix $ and $beginvmatrix n_1 & x_1 & y_1 & z_1 \ n_2 & x_2 & y_2 & z_2 \ n_3 & x_3 & y_3 & z_3 \ n_3 & x_3 & y_3 & z_3 endvmatrix $
give the second and the third component of the question vector, equal to $0$.
answered Mar 29 at 13:57
WidawensenWidawensen
4,75831446
$endgroup$
add a comment |
$begingroup$
Another approach to the problem uses a formula for triple product.
$ mathbfacdot(mathbfbtimes mathbfc) = det beginbmatrix
a_1 & b_1 & c_1 \
a_2 & b_2 & c_2 \
a_3 & b_3 & c_3 \
endbmatrix $
Then consider determinant
$beginvmatrix n_1 & x_1 & y_1 & z_1 \ n_2 & x_2 & y_2 & z_2 \ n_3 & x_3 & y_3 & z_3 \ n_1 & x_1 & y_1 & z_1 endvmatrix $
where columns consist of vectors $ mathbfN ,mathbfx,mathbfy,mathbfz$ components (the fourth row repeats the first one).
Of course such determinant equals to $0$.
Developing the determinant along the fourth row we obtain:
$-n_1beginvmatrix x_1 & y_1 & z_1 \ x_2 & y_2 & z_2 \ x_3 & y_3 & z_3 \ endvmatrix +x_1beginvmatrix n_1 & y_1 & z_1 \ n_2 & y_2 & z_2 \ n_3 & y_3 & z_3 \ endvmatrix -y_1beginvmatrix n_1 & x_1 & z_1 \ n_2 & x_2 & z_2 \ n_3 & x_3 & z_3 \ endvmatrix +z_1beginvmatrix n_1 & x_1 & y_1 \ n_2 & x_2 & y_2 \ n_3 & x_3 & y_3 \ endvmatrix=0$
from which the formula for the first component of the vector given in the question follows
(the first summand is equal to $0$ as the vectors $mathbfx,mathbfy,mathbfz$ are collinear, the columns can be permuted (required for the third summand) if needed to give appropriate sign in expression)
Similarly the determinants
$beginvmatrix n_1 & x_1 & y_1 & z_1 \ n_2 & x_2 & y_2 & z_2 \ n_3 & x_3 & y_3 & z_3 \ n_2 & x_2 & y_2 & z_2 endvmatrix $ and $beginvmatrix n_1 & x_1 & y_1 & z_1 \ n_2 & x_2 & y_2 & z_2 \ n_3 & x_3 & y_3 & z_3 \ n_3 & x_3 & y_3 & z_3 endvmatrix $
give the second and the third component of the question vector, equal to $0$.
answered Mar 29 at 13:57
WidawensenWidawensen
4,75831446
$endgroup$
Another approach to the problem uses a formula for triple product.
$ mathbfacdot(mathbfbtimes mathbfc) = det beginbmatrix
a_1 & b_1 & c_1 \
a_2 & b_2 & c_2 \
a_3 & b_3 & c_3 \
endbmatrix $
Then consider determinant
$beginvmatrix n_1 & x_1 & y_1 & z_1 \ n_2 & x_2 & y_2 & z_2 \ n_3 & x_3 & y_3 & z_3 \ n_1 & x_1 & y_1 & z_1 endvmatrix $
where columns consist of vectors $ mathbfN ,mathbfx,mathbfy,mathbfz$ components (the fourth row repeats the first one).
Of course such determinant equals to $0$.
Developing the determinant along the fourth row we obtain:
$-n_1beginvmatrix x_1 & y_1 & z_1 \ x_2 & y_2 & z_2 \ x_3 & y_3 & z_3 \ endvmatrix +x_1beginvmatrix n_1 & y_1 & z_1 \ n_2 & y_2 & z_2 \ n_3 & y_3 & z_3 \ endvmatrix -y_1beginvmatrix n_1 & x_1 & z_1 \ n_2 & x_2 & z_2 \ n_3 & x_3 & z_3 \ endvmatrix +z_1beginvmatrix n_1 & x_1 & y_1 \ n_2 & x_2 & y_2 \ n_3 & x_3 & y_3 \ endvmatrix=0$
from which the formula for the first component of the vector given in the question follows
(the first summand is equal to $0$ as the vectors $mathbfx,mathbfy,mathbfz$ are collinear, the columns can be permuted (required for the third summand) if needed to give appropriate sign in expression)
Similarly the determinants
$beginvmatrix n_1 & x_1 & y_1 & z_1 \ n_2 & x_2 & y_2 & z_2 \ n_3 & x_3 & y_3 & z_3 \ n_2 & x_2 & y_2 & z_2 endvmatrix $ and $beginvmatrix n_1 & x_1 & y_1 & z_1 \ n_2 & x_2 & y_2 & z_2 \ n_3 & x_3 & y_3 & z_3 \ n_3 & x_3 & y_3 & z_3 endvmatrix $
give the second and the third component of the question vector, equal to $0$.
answered Mar 29 at 13:57
WidawensenWidawensen
4,75831446
edited Mar 29 at 14:04
answered Mar 29 at 13:57
WidawensenWidawensen
4,75831446
answered Mar 29 at 13:57
WidawensenWidawensen
4,75831446
answered Mar 29 at 13:57
WidawensenWidawensen
4,75831446
4,75831446
add a comment |
add a comment |
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$begingroup$
What does $x times y cdot N$ mean? Dot product $(x times y) cdot N$ ?
$endgroup$
– Widawensen
Mar 20 at 12:51
4
$begingroup$
@Widawensen Yes, what else could it mean?
$endgroup$
– Marc van Leeuwen
Mar 20 at 12:53
1
$begingroup$
@MarcvanLeeuwen: that could mean a badly written problem. That happens here sometimes.
$endgroup$
– Taladris
Mar 21 at 2:54