How can three vectors be orthogonal to each other?Three planar vectors $x,y,z$ such that $x$ is orthogonal to $y + z$ and $z$I see some contradiction in the definition of orthogonal vectorsHow would one prove that the row space and null space are orthogonal compliments of each other?Vector perpendicular to three other vectorsProving the two given vectors are orthogonalVector orthogonal to the set of vectors3 vectors making equal angles with each other in 3d space?why these two vectors are orthogonal
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How can three vectors be orthogonal to each other?
Three planar vectors $x,y,z$ such that $x$ is orthogonal to $y + z$ and $z$I see some contradiction in the definition of orthogonal vectorsHow would one prove that the row space and null space are orthogonal compliments of each other?Vector perpendicular to three other vectorsProving the two given vectors are orthogonalVector orthogonal to the set of vectors3 vectors making equal angles with each other in 3d space?why these two vectors are orthogonal
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In a scenario, say that:
Vectors $mathbfU$, $mathbfV$ and $mathbfW$ are all orthogonal such that the dot product between each of these $(mathbfUV;mathbfVW;mathbfWU)$ is equal to zero.
I imagine that for any potential vector space $mathbfR$ this would only be possible in two situations.
1) $mathbfU$, $mathbfW$ and/or $mathbfV$ is the zero vector.
2) $mathbfU=(1, 0, 0)$, $mathbfV = (0, 1, 0)$ and $mathbfW = (0, 0, 1)$.
Is there any other situation where three vectors are all orthogonal to each other?
linear-algebra vectors orthogonality
edited Sep 29 at 16:45
Ernie060
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asked Sep 29 at 0:20
HeavenlyPandaHeavenlyPanda
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show 1 more comment
$begingroup$
In a scenario, say that:
Vectors $mathbfU$, $mathbfV$ and $mathbfW$ are all orthogonal such that the dot product between each of these $(mathbfUV;mathbfVW;mathbfWU)$ is equal to zero.
I imagine that for any potential vector space $mathbfR$ this would only be possible in two situations.
1) $mathbfU$, $mathbfW$ and/or $mathbfV$ is the zero vector.
2) $mathbfU=(1, 0, 0)$, $mathbfV = (0, 1, 0)$ and $mathbfW = (0, 0, 1)$.
Is there any other situation where three vectors are all orthogonal to each other?
linear-algebra vectors orthogonality
edited Sep 29 at 16:45
Ernie060
4,1071 gold badge9 silver badges22 bronze badges
asked Sep 29 at 0:20
HeavenlyPandaHeavenlyPanda
495 bronze badges
$endgroup$
$begingroup$
What about $u=big(tfrac1sqrt 2,tfrac1sqrt 2,0big)$, $v=big(tfrac1sqrt 2,-tfrac1sqrt 2,0big)$ and $w=(0,0,1)$?
$endgroup$
– Azif00
Sep 29 at 0:24
1
$begingroup$
Yes, infinitely many. For example, rotate your basis in 2.
$endgroup$
– AnyAD
Sep 29 at 0:25
4
$begingroup$
Pick two non-zero and non-parallel vectors $x$ and $u$ in $mathbb R^3$. Then set $v=xtimes u$ and $w=utimes v$. Now $u,v,w$ are mutually orthogonal.
$endgroup$
– user1551
Sep 29 at 3:53
1
$begingroup$
@Azif00 More generally, $(a,b,0)$, $(b,-a,0)$, $(0,0,c)$ for any values of $a$, $b$, and $c$. Orthogonal vectors don't have to be unit vectors. The OP also seems to have made that mistake in the second example.
$endgroup$
– alephzero
Sep 29 at 10:34
12
$begingroup$
Hold out your hand. Arrange your fingers so that your index finger, middle finger, and thumb are all mutually perpendicular. Now rotate your hand around. All of these configurations correspond to a distinct set of 3 mutually orthogonal vectors.
$endgroup$
– Steven Gubkin
Sep 29 at 12:53
|
show 1 more comment
$begingroup$
In a scenario, say that:
Vectors $mathbfU$, $mathbfV$ and $mathbfW$ are all orthogonal such that the dot product between each of these $(mathbfUV;mathbfVW;mathbfWU)$ is equal to zero.
I imagine that for any potential vector space $mathbfR$ this would only be possible in two situations.
1) $mathbfU$, $mathbfW$ and/or $mathbfV$ is the zero vector.
2) $mathbfU=(1, 0, 0)$, $mathbfV = (0, 1, 0)$ and $mathbfW = (0, 0, 1)$.
Is there any other situation where three vectors are all orthogonal to each other?
linear-algebra vectors orthogonality
edited Sep 29 at 16:45
Ernie060
4,1071 gold badge9 silver badges22 bronze badges
asked Sep 29 at 0:20
HeavenlyPandaHeavenlyPanda
495 bronze badges
$endgroup$
In a scenario, say that:
Vectors $mathbfU$, $mathbfV$ and $mathbfW$ are all orthogonal such that the dot product between each of these $(mathbfUV;mathbfVW;mathbfWU)$ is equal to zero.
I imagine that for any potential vector space $mathbfR$ this would only be possible in two situations.
1) $mathbfU$, $mathbfW$ and/or $mathbfV$ is the zero vector.
2) $mathbfU=(1, 0, 0)$, $mathbfV = (0, 1, 0)$ and $mathbfW = (0, 0, 1)$.
Is there any other situation where three vectors are all orthogonal to each other?
linear-algebra vectors orthogonality
linear-algebra vectors orthogonality
edited Sep 29 at 16:45
Ernie060
4,1071 gold badge9 silver badges22 bronze badges
asked Sep 29 at 0:20
HeavenlyPandaHeavenlyPanda
495 bronze badges
edited Sep 29 at 16:45
Ernie060
4,1071 gold badge9 silver badges22 bronze badges
asked Sep 29 at 0:20
HeavenlyPandaHeavenlyPanda
495 bronze badges
edited Sep 29 at 16:45
Ernie060
4,1071 gold badge9 silver badges22 bronze badges
edited Sep 29 at 16:45
Ernie060
4,1071 gold badge9 silver badges22 bronze badges
edited Sep 29 at 16:45
Ernie060
4,1071 gold badge9 silver badges22 bronze badges
4,1071 gold badge9 silver badges22 bronze badges
asked Sep 29 at 0:20
HeavenlyPandaHeavenlyPanda
495 bronze badges
asked Sep 29 at 0:20
HeavenlyPandaHeavenlyPanda
495 bronze badges
asked Sep 29 at 0:20
HeavenlyPandaHeavenlyPanda
495 bronze badges
495 bronze badges
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What about $u=big(tfrac1sqrt 2,tfrac1sqrt 2,0big)$, $v=big(tfrac1sqrt 2,-tfrac1sqrt 2,0big)$ and $w=(0,0,1)$?
$endgroup$
– Azif00
Sep 29 at 0:24
1
$begingroup$
Yes, infinitely many. For example, rotate your basis in 2.
$endgroup$
– AnyAD
Sep 29 at 0:25
4
$begingroup$
Pick two non-zero and non-parallel vectors $x$ and $u$ in $mathbb R^3$. Then set $v=xtimes u$ and $w=utimes v$. Now $u,v,w$ are mutually orthogonal.
$endgroup$
– user1551
Sep 29 at 3:53
1
$begingroup$
@Azif00 More generally, $(a,b,0)$, $(b,-a,0)$, $(0,0,c)$ for any values of $a$, $b$, and $c$. Orthogonal vectors don't have to be unit vectors. The OP also seems to have made that mistake in the second example.
$endgroup$
– alephzero
Sep 29 at 10:34
12
$begingroup$
Hold out your hand. Arrange your fingers so that your index finger, middle finger, and thumb are all mutually perpendicular. Now rotate your hand around. All of these configurations correspond to a distinct set of 3 mutually orthogonal vectors.
$endgroup$
– Steven Gubkin
Sep 29 at 12:53
|
show 1 more comment
$begingroup$
What about $u=big(tfrac1sqrt 2,tfrac1sqrt 2,0big)$, $v=big(tfrac1sqrt 2,-tfrac1sqrt 2,0big)$ and $w=(0,0,1)$?
$endgroup$
– Azif00
Sep 29 at 0:24
1
$begingroup$
Yes, infinitely many. For example, rotate your basis in 2.
$endgroup$
– AnyAD
Sep 29 at 0:25
4
$begingroup$
Pick two non-zero and non-parallel vectors $x$ and $u$ in $mathbb R^3$. Then set $v=xtimes u$ and $w=utimes v$. Now $u,v,w$ are mutually orthogonal.
$endgroup$
– user1551
Sep 29 at 3:53
1
$begingroup$
@Azif00 More generally, $(a,b,0)$, $(b,-a,0)$, $(0,0,c)$ for any values of $a$, $b$, and $c$. Orthogonal vectors don't have to be unit vectors. The OP also seems to have made that mistake in the second example.
$endgroup$
– alephzero
Sep 29 at 10:34
12
$begingroup$
Hold out your hand. Arrange your fingers so that your index finger, middle finger, and thumb are all mutually perpendicular. Now rotate your hand around. All of these configurations correspond to a distinct set of 3 mutually orthogonal vectors.
$endgroup$
– Steven Gubkin
Sep 29 at 12:53
$begingroup$
What about $u=big(tfrac1sqrt 2,tfrac1sqrt 2,0big)$, $v=big(tfrac1sqrt 2,-tfrac1sqrt 2,0big)$ and $w=(0,0,1)$?
$endgroup$
– Azif00
Sep 29 at 0:24
$begingroup$
What about $u=big(tfrac1sqrt 2,tfrac1sqrt 2,0big)$, $v=big(tfrac1sqrt 2,-tfrac1sqrt 2,0big)$ and $w=(0,0,1)$?
$endgroup$
– Azif00
Sep 29 at 0:24
1
1
$begingroup$
Yes, infinitely many. For example, rotate your basis in 2.
$endgroup$
– AnyAD
Sep 29 at 0:25
$begingroup$
Yes, infinitely many. For example, rotate your basis in 2.
$endgroup$
– AnyAD
Sep 29 at 0:25
4
4
$begingroup$
Pick two non-zero and non-parallel vectors $x$ and $u$ in $mathbb R^3$. Then set $v=xtimes u$ and $w=utimes v$. Now $u,v,w$ are mutually orthogonal.
$endgroup$
– user1551
Sep 29 at 3:53
$begingroup$
Pick two non-zero and non-parallel vectors $x$ and $u$ in $mathbb R^3$. Then set $v=xtimes u$ and $w=utimes v$. Now $u,v,w$ are mutually orthogonal.
$endgroup$
– user1551
Sep 29 at 3:53
1
1
$begingroup$
@Azif00 More generally, $(a,b,0)$, $(b,-a,0)$, $(0,0,c)$ for any values of $a$, $b$, and $c$. Orthogonal vectors don't have to be unit vectors. The OP also seems to have made that mistake in the second example.
$endgroup$
– alephzero
Sep 29 at 10:34
$begingroup$
@Azif00 More generally, $(a,b,0)$, $(b,-a,0)$, $(0,0,c)$ for any values of $a$, $b$, and $c$. Orthogonal vectors don't have to be unit vectors. The OP also seems to have made that mistake in the second example.
$endgroup$
– alephzero
Sep 29 at 10:34
12
12
$begingroup$
Hold out your hand. Arrange your fingers so that your index finger, middle finger, and thumb are all mutually perpendicular. Now rotate your hand around. All of these configurations correspond to a distinct set of 3 mutually orthogonal vectors.
$endgroup$
– Steven Gubkin
Sep 29 at 12:53
$begingroup$
Hold out your hand. Arrange your fingers so that your index finger, middle finger, and thumb are all mutually perpendicular. Now rotate your hand around. All of these configurations correspond to a distinct set of 3 mutually orthogonal vectors.
$endgroup$
– Steven Gubkin
Sep 29 at 12:53
|
show 1 more comment
6 Answers
6
active
oldest
votes
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Think it like this: for a given vector $uneq 0$ in $mathbbR^3$, what is the space of all vectors perpendicular to it? It is a plane $P$ containing the origin, whose perpendicular direction is obviously $u$.
Now take a $vneq 0$ in that plane. The space of all vectors perpendicular to $v$ is a new plane $P'$. In order to get a vector $w$ perpendicular to both $u,v$, we need $win Pcap P'$. What does $Pcap P'$ look like?
By the way, notice that $v$ is ANY non-zero vector, not only $(1,0,0), (0,1,0)$ or $(0,0,1)$
answered Sep 29 at 0:37
rmdmc89rmdmc89
5,6611 gold badge9 silver badges29 bronze badges
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There are infinitely many triple of non zero orthogonal vectors obtained by the three you have indicated by scaling of each one and rotations of the triple all togheter.
To construct any othogonal triple we can proceed as follows:
- choose a first vector $v_1=(a,b,c)$
- find a second vector orthogonal to $v_1$ that is e.g. $v_2=(-b,a,0)$
- determine the third by cross product $v_3=v_1times v_2$
answered Sep 29 at 0:25
useruser
112k10 gold badges49 silver badges103 bronze badges
$endgroup$
$begingroup$
You can also have $v_3 = v_2 times v_1 = - v_1 times v_2$, but of course you can just as easily interchange $v_1$ and $v_2$ to do that, so it's not really an extra degree of freedom. Still, chirality is sometimes useful to talk about.
$endgroup$
– Kevin
Sep 30 at 3:33
$begingroup$
@Kevin We ca see that as a simple scaling by a factor $-1$ for $v_3$. There are of course many ways! It was just to give a simple procedure. To be more genral we can assume $v_1=(a,b,c)$, $v_2=(d,e,-ad-ce)$ and $v_3$by $v_1times v_2$.
$endgroup$
– user
Sep 30 at 5:54
add a comment
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How about $v_1= (frac1sqrt2, frac1sqrt2,0),v_2=(-frac1sqrt2,frac1sqrt2,0),v_3=(0,0,1)$ ?
Geometrically take any frame and rotate in any direction.
answered Sep 29 at 0:26
Monkey.D.LuffyMonkey.D.Luffy
1255 bronze badges
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There are infinitely many possibilities. $(1,0,0), (0,1,-1)$ and $(0,1,1)$ is one example.
answered Sep 29 at 0:25
Kabo MurphyKabo Murphy
143k8 gold badges48 silver badges106 bronze badges
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The word "orthogonal" really just corresponds to the intuitive notion of vectors being perpendicular to each other. Draw out the unit vectors in the $x$, $y$ and $z$ directions respectively--those are one set of three mutually orthogonal (i.e. perpendicular) vectors, just like you observed. But if you rotate those three vectors together in any way that you like in 3D space, without changing the angles between them, then of course they will still remain orthogonal.
Furthermore, if we scale these vectors by changing their lengths independently by any arbitrary nonzero factor, we will still end up with a set of orthogonal vectors, since the only thing that matters is the directions and not the lengths of the vectors. In this manner we end up with a description for an infinite family of orthogonal vectors, which hopefully makes it easy for you to convince yourself intuitively.
In a more general vector space, of course, this sort of pictorial intuition might no longer hold, but the idea of orthogonality can be easily generalised. That's the reason we define orthogonality abstractly and independent of the usual geometric notion of perpendicularity.
answered Sep 29 at 8:44
YiFanYiFan
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Here is an example of just what you seek:
edited Sep 29 at 15:09
Tanner Swett
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answered Sep 29 at 1:05
David G. StorkDavid G. Stork
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17
$begingroup$
I don't know what this GIF looks like to you, but on my setup, it flickers between some sort of geometry animation and a mostly-white frame with a red rectangle in the upper left, before eventually settling down to just the white frame with red rectangle.
$endgroup$
– user2357112 supports Monica
Sep 29 at 9:25
$begingroup$
I don't know how you created that gif, but it had a lot of blank frames in it seemingly at random, causing it to appear to flicker. I've taken the liberty of editing the gif to remove the blank frames.
$endgroup$
– Tanner Swett
Sep 29 at 15:10
1
$begingroup$
Gif works fine for me
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– idle mathematician
Sep 29 at 17:11
1
$begingroup$
@B.Swan Yes, it's fixed now. You can see the edit history to compare with the old version.
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– Joonas Ilmavirta
Sep 29 at 17:45
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6 Answers
6
active
oldest
votes
6 Answers
6
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Think it like this: for a given vector $uneq 0$ in $mathbbR^3$, what is the space of all vectors perpendicular to it? It is a plane $P$ containing the origin, whose perpendicular direction is obviously $u$.
Now take a $vneq 0$ in that plane. The space of all vectors perpendicular to $v$ is a new plane $P'$. In order to get a vector $w$ perpendicular to both $u,v$, we need $win Pcap P'$. What does $Pcap P'$ look like?
By the way, notice that $v$ is ANY non-zero vector, not only $(1,0,0), (0,1,0)$ or $(0,0,1)$
answered Sep 29 at 0:37
rmdmc89rmdmc89
5,6611 gold badge9 silver badges29 bronze badges
$endgroup$
add a comment
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$begingroup$
Think it like this: for a given vector $uneq 0$ in $mathbbR^3$, what is the space of all vectors perpendicular to it? It is a plane $P$ containing the origin, whose perpendicular direction is obviously $u$.
Now take a $vneq 0$ in that plane. The space of all vectors perpendicular to $v$ is a new plane $P'$. In order to get a vector $w$ perpendicular to both $u,v$, we need $win Pcap P'$. What does $Pcap P'$ look like?
By the way, notice that $v$ is ANY non-zero vector, not only $(1,0,0), (0,1,0)$ or $(0,0,1)$
answered Sep 29 at 0:37
rmdmc89rmdmc89
5,6611 gold badge9 silver badges29 bronze badges
$endgroup$
add a comment
|
$begingroup$
Think it like this: for a given vector $uneq 0$ in $mathbbR^3$, what is the space of all vectors perpendicular to it? It is a plane $P$ containing the origin, whose perpendicular direction is obviously $u$.
Now take a $vneq 0$ in that plane. The space of all vectors perpendicular to $v$ is a new plane $P'$. In order to get a vector $w$ perpendicular to both $u,v$, we need $win Pcap P'$. What does $Pcap P'$ look like?
By the way, notice that $v$ is ANY non-zero vector, not only $(1,0,0), (0,1,0)$ or $(0,0,1)$
answered Sep 29 at 0:37
rmdmc89rmdmc89
5,6611 gold badge9 silver badges29 bronze badges
$endgroup$
Think it like this: for a given vector $uneq 0$ in $mathbbR^3$, what is the space of all vectors perpendicular to it? It is a plane $P$ containing the origin, whose perpendicular direction is obviously $u$.
Now take a $vneq 0$ in that plane. The space of all vectors perpendicular to $v$ is a new plane $P'$. In order to get a vector $w$ perpendicular to both $u,v$, we need $win Pcap P'$. What does $Pcap P'$ look like?
By the way, notice that $v$ is ANY non-zero vector, not only $(1,0,0), (0,1,0)$ or $(0,0,1)$
answered Sep 29 at 0:37
rmdmc89rmdmc89
5,6611 gold badge9 silver badges29 bronze badges
edited Sep 29 at 2:37
answered Sep 29 at 0:37
rmdmc89rmdmc89
5,6611 gold badge9 silver badges29 bronze badges
answered Sep 29 at 0:37
rmdmc89rmdmc89
5,6611 gold badge9 silver badges29 bronze badges
answered Sep 29 at 0:37
rmdmc89rmdmc89
5,6611 gold badge9 silver badges29 bronze badges
5,6611 gold badge9 silver badges29 bronze badges
add a comment
|
add a comment
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$begingroup$
There are infinitely many triple of non zero orthogonal vectors obtained by the three you have indicated by scaling of each one and rotations of the triple all togheter.
To construct any othogonal triple we can proceed as follows:
- choose a first vector $v_1=(a,b,c)$
- find a second vector orthogonal to $v_1$ that is e.g. $v_2=(-b,a,0)$
- determine the third by cross product $v_3=v_1times v_2$
answered Sep 29 at 0:25
useruser
112k10 gold badges49 silver badges103 bronze badges
$endgroup$
$begingroup$
You can also have $v_3 = v_2 times v_1 = - v_1 times v_2$, but of course you can just as easily interchange $v_1$ and $v_2$ to do that, so it's not really an extra degree of freedom. Still, chirality is sometimes useful to talk about.
$endgroup$
– Kevin
Sep 30 at 3:33
$begingroup$
@Kevin We ca see that as a simple scaling by a factor $-1$ for $v_3$. There are of course many ways! It was just to give a simple procedure. To be more genral we can assume $v_1=(a,b,c)$, $v_2=(d,e,-ad-ce)$ and $v_3$by $v_1times v_2$.
$endgroup$
– user
Sep 30 at 5:54
add a comment
|
$begingroup$
There are infinitely many triple of non zero orthogonal vectors obtained by the three you have indicated by scaling of each one and rotations of the triple all togheter.
To construct any othogonal triple we can proceed as follows:
- choose a first vector $v_1=(a,b,c)$
- find a second vector orthogonal to $v_1$ that is e.g. $v_2=(-b,a,0)$
- determine the third by cross product $v_3=v_1times v_2$
answered Sep 29 at 0:25
useruser
112k10 gold badges49 silver badges103 bronze badges
$endgroup$
$begingroup$
You can also have $v_3 = v_2 times v_1 = - v_1 times v_2$, but of course you can just as easily interchange $v_1$ and $v_2$ to do that, so it's not really an extra degree of freedom. Still, chirality is sometimes useful to talk about.
$endgroup$
– Kevin
Sep 30 at 3:33
$begingroup$
@Kevin We ca see that as a simple scaling by a factor $-1$ for $v_3$. There are of course many ways! It was just to give a simple procedure. To be more genral we can assume $v_1=(a,b,c)$, $v_2=(d,e,-ad-ce)$ and $v_3$by $v_1times v_2$.
$endgroup$
– user
Sep 30 at 5:54
add a comment
|
$begingroup$
There are infinitely many triple of non zero orthogonal vectors obtained by the three you have indicated by scaling of each one and rotations of the triple all togheter.
To construct any othogonal triple we can proceed as follows:
- choose a first vector $v_1=(a,b,c)$
- find a second vector orthogonal to $v_1$ that is e.g. $v_2=(-b,a,0)$
- determine the third by cross product $v_3=v_1times v_2$
answered Sep 29 at 0:25
useruser
112k10 gold badges49 silver badges103 bronze badges
$endgroup$
There are infinitely many triple of non zero orthogonal vectors obtained by the three you have indicated by scaling of each one and rotations of the triple all togheter.
To construct any othogonal triple we can proceed as follows:
- choose a first vector $v_1=(a,b,c)$
- find a second vector orthogonal to $v_1$ that is e.g. $v_2=(-b,a,0)$
- determine the third by cross product $v_3=v_1times v_2$
answered Sep 29 at 0:25
useruser
112k10 gold badges49 silver badges103 bronze badges
edited Sep 29 at 0:31
answered Sep 29 at 0:25
useruser
112k10 gold badges49 silver badges103 bronze badges
answered Sep 29 at 0:25
useruser
112k10 gold badges49 silver badges103 bronze badges
answered Sep 29 at 0:25
useruser
112k10 gold badges49 silver badges103 bronze badges
112k10 gold badges49 silver badges103 bronze badges
$begingroup$
You can also have $v_3 = v_2 times v_1 = - v_1 times v_2$, but of course you can just as easily interchange $v_1$ and $v_2$ to do that, so it's not really an extra degree of freedom. Still, chirality is sometimes useful to talk about.
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– Kevin
Sep 30 at 3:33
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@Kevin We ca see that as a simple scaling by a factor $-1$ for $v_3$. There are of course many ways! It was just to give a simple procedure. To be more genral we can assume $v_1=(a,b,c)$, $v_2=(d,e,-ad-ce)$ and $v_3$by $v_1times v_2$.
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– user
Sep 30 at 5:54
add a comment
|
$begingroup$
You can also have $v_3 = v_2 times v_1 = - v_1 times v_2$, but of course you can just as easily interchange $v_1$ and $v_2$ to do that, so it's not really an extra degree of freedom. Still, chirality is sometimes useful to talk about.
$endgroup$
– Kevin
Sep 30 at 3:33
$begingroup$
@Kevin We ca see that as a simple scaling by a factor $-1$ for $v_3$. There are of course many ways! It was just to give a simple procedure. To be more genral we can assume $v_1=(a,b,c)$, $v_2=(d,e,-ad-ce)$ and $v_3$by $v_1times v_2$.
$endgroup$
– user
Sep 30 at 5:54
$begingroup$
You can also have $v_3 = v_2 times v_1 = - v_1 times v_2$, but of course you can just as easily interchange $v_1$ and $v_2$ to do that, so it's not really an extra degree of freedom. Still, chirality is sometimes useful to talk about.
$endgroup$
– Kevin
Sep 30 at 3:33
$begingroup$
You can also have $v_3 = v_2 times v_1 = - v_1 times v_2$, but of course you can just as easily interchange $v_1$ and $v_2$ to do that, so it's not really an extra degree of freedom. Still, chirality is sometimes useful to talk about.
$endgroup$
– Kevin
Sep 30 at 3:33
$begingroup$
@Kevin We ca see that as a simple scaling by a factor $-1$ for $v_3$. There are of course many ways! It was just to give a simple procedure. To be more genral we can assume $v_1=(a,b,c)$, $v_2=(d,e,-ad-ce)$ and $v_3$by $v_1times v_2$.
$endgroup$
– user
Sep 30 at 5:54
$begingroup$
@Kevin We ca see that as a simple scaling by a factor $-1$ for $v_3$. There are of course many ways! It was just to give a simple procedure. To be more genral we can assume $v_1=(a,b,c)$, $v_2=(d,e,-ad-ce)$ and $v_3$by $v_1times v_2$.
$endgroup$
– user
Sep 30 at 5:54
add a comment
|
$begingroup$
How about $v_1= (frac1sqrt2, frac1sqrt2,0),v_2=(-frac1sqrt2,frac1sqrt2,0),v_3=(0,0,1)$ ?
Geometrically take any frame and rotate in any direction.
answered Sep 29 at 0:26
Monkey.D.LuffyMonkey.D.Luffy
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How about $v_1= (frac1sqrt2, frac1sqrt2,0),v_2=(-frac1sqrt2,frac1sqrt2,0),v_3=(0,0,1)$ ?
Geometrically take any frame and rotate in any direction.
answered Sep 29 at 0:26
Monkey.D.LuffyMonkey.D.Luffy
1255 bronze badges
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How about $v_1= (frac1sqrt2, frac1sqrt2,0),v_2=(-frac1sqrt2,frac1sqrt2,0),v_3=(0,0,1)$ ?
Geometrically take any frame and rotate in any direction.
answered Sep 29 at 0:26
Monkey.D.LuffyMonkey.D.Luffy
1255 bronze badges
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How about $v_1= (frac1sqrt2, frac1sqrt2,0),v_2=(-frac1sqrt2,frac1sqrt2,0),v_3=(0,0,1)$ ?
Geometrically take any frame and rotate in any direction.
answered Sep 29 at 0:26
Monkey.D.LuffyMonkey.D.Luffy
1255 bronze badges
answered Sep 29 at 0:26
Monkey.D.LuffyMonkey.D.Luffy
1255 bronze badges
answered Sep 29 at 0:26
Monkey.D.LuffyMonkey.D.Luffy
1255 bronze badges
answered Sep 29 at 0:26
Monkey.D.LuffyMonkey.D.Luffy
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There are infinitely many possibilities. $(1,0,0), (0,1,-1)$ and $(0,1,1)$ is one example.
answered Sep 29 at 0:25
Kabo MurphyKabo Murphy
143k8 gold badges48 silver badges106 bronze badges
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There are infinitely many possibilities. $(1,0,0), (0,1,-1)$ and $(0,1,1)$ is one example.
answered Sep 29 at 0:25
Kabo MurphyKabo Murphy
143k8 gold badges48 silver badges106 bronze badges
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add a comment
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There are infinitely many possibilities. $(1,0,0), (0,1,-1)$ and $(0,1,1)$ is one example.
answered Sep 29 at 0:25
Kabo MurphyKabo Murphy
143k8 gold badges48 silver badges106 bronze badges
$endgroup$
There are infinitely many possibilities. $(1,0,0), (0,1,-1)$ and $(0,1,1)$ is one example.
answered Sep 29 at 0:25
Kabo MurphyKabo Murphy
143k8 gold badges48 silver badges106 bronze badges
answered Sep 29 at 0:25
Kabo MurphyKabo Murphy
143k8 gold badges48 silver badges106 bronze badges
answered Sep 29 at 0:25
Kabo MurphyKabo Murphy
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answered Sep 29 at 0:25
Kabo MurphyKabo Murphy
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143k8 gold badges48 silver badges106 bronze badges
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$begingroup$
The word "orthogonal" really just corresponds to the intuitive notion of vectors being perpendicular to each other. Draw out the unit vectors in the $x$, $y$ and $z$ directions respectively--those are one set of three mutually orthogonal (i.e. perpendicular) vectors, just like you observed. But if you rotate those three vectors together in any way that you like in 3D space, without changing the angles between them, then of course they will still remain orthogonal.
Furthermore, if we scale these vectors by changing their lengths independently by any arbitrary nonzero factor, we will still end up with a set of orthogonal vectors, since the only thing that matters is the directions and not the lengths of the vectors. In this manner we end up with a description for an infinite family of orthogonal vectors, which hopefully makes it easy for you to convince yourself intuitively.
In a more general vector space, of course, this sort of pictorial intuition might no longer hold, but the idea of orthogonality can be easily generalised. That's the reason we define orthogonality abstractly and independent of the usual geometric notion of perpendicularity.
answered Sep 29 at 8:44
YiFanYiFan
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$begingroup$
The word "orthogonal" really just corresponds to the intuitive notion of vectors being perpendicular to each other. Draw out the unit vectors in the $x$, $y$ and $z$ directions respectively--those are one set of three mutually orthogonal (i.e. perpendicular) vectors, just like you observed. But if you rotate those three vectors together in any way that you like in 3D space, without changing the angles between them, then of course they will still remain orthogonal.
Furthermore, if we scale these vectors by changing their lengths independently by any arbitrary nonzero factor, we will still end up with a set of orthogonal vectors, since the only thing that matters is the directions and not the lengths of the vectors. In this manner we end up with a description for an infinite family of orthogonal vectors, which hopefully makes it easy for you to convince yourself intuitively.
In a more general vector space, of course, this sort of pictorial intuition might no longer hold, but the idea of orthogonality can be easily generalised. That's the reason we define orthogonality abstractly and independent of the usual geometric notion of perpendicularity.
answered Sep 29 at 8:44
YiFanYiFan
11.3k4 gold badges16 silver badges44 bronze badges
$endgroup$
add a comment
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$begingroup$
The word "orthogonal" really just corresponds to the intuitive notion of vectors being perpendicular to each other. Draw out the unit vectors in the $x$, $y$ and $z$ directions respectively--those are one set of three mutually orthogonal (i.e. perpendicular) vectors, just like you observed. But if you rotate those three vectors together in any way that you like in 3D space, without changing the angles between them, then of course they will still remain orthogonal.
Furthermore, if we scale these vectors by changing their lengths independently by any arbitrary nonzero factor, we will still end up with a set of orthogonal vectors, since the only thing that matters is the directions and not the lengths of the vectors. In this manner we end up with a description for an infinite family of orthogonal vectors, which hopefully makes it easy for you to convince yourself intuitively.
In a more general vector space, of course, this sort of pictorial intuition might no longer hold, but the idea of orthogonality can be easily generalised. That's the reason we define orthogonality abstractly and independent of the usual geometric notion of perpendicularity.
answered Sep 29 at 8:44
YiFanYiFan
11.3k4 gold badges16 silver badges44 bronze badges
$endgroup$
The word "orthogonal" really just corresponds to the intuitive notion of vectors being perpendicular to each other. Draw out the unit vectors in the $x$, $y$ and $z$ directions respectively--those are one set of three mutually orthogonal (i.e. perpendicular) vectors, just like you observed. But if you rotate those three vectors together in any way that you like in 3D space, without changing the angles between them, then of course they will still remain orthogonal.
Furthermore, if we scale these vectors by changing their lengths independently by any arbitrary nonzero factor, we will still end up with a set of orthogonal vectors, since the only thing that matters is the directions and not the lengths of the vectors. In this manner we end up with a description for an infinite family of orthogonal vectors, which hopefully makes it easy for you to convince yourself intuitively.
In a more general vector space, of course, this sort of pictorial intuition might no longer hold, but the idea of orthogonality can be easily generalised. That's the reason we define orthogonality abstractly and independent of the usual geometric notion of perpendicularity.
answered Sep 29 at 8:44
YiFanYiFan
11.3k4 gold badges16 silver badges44 bronze badges
answered Sep 29 at 8:44
YiFanYiFan
11.3k4 gold badges16 silver badges44 bronze badges
answered Sep 29 at 8:44
YiFanYiFan
11.3k4 gold badges16 silver badges44 bronze badges
answered Sep 29 at 8:44
YiFanYiFan
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11.3k4 gold badges16 silver badges44 bronze badges
add a comment
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$begingroup$
Here is an example of just what you seek:
edited Sep 29 at 15:09
Tanner Swett
6,93122 silver badges44 bronze badges
answered Sep 29 at 1:05
David G. StorkDavid G. Stork
16.7k4 gold badges20 silver badges41 bronze badges
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17
$begingroup$
I don't know what this GIF looks like to you, but on my setup, it flickers between some sort of geometry animation and a mostly-white frame with a red rectangle in the upper left, before eventually settling down to just the white frame with red rectangle.
$endgroup$
– user2357112 supports Monica
Sep 29 at 9:25
$begingroup$
I don't know how you created that gif, but it had a lot of blank frames in it seemingly at random, causing it to appear to flicker. I've taken the liberty of editing the gif to remove the blank frames.
$endgroup$
– Tanner Swett
Sep 29 at 15:10
1
$begingroup$
Gif works fine for me
$endgroup$
– idle mathematician
Sep 29 at 17:11
1
$begingroup$
@B.Swan Yes, it's fixed now. You can see the edit history to compare with the old version.
$endgroup$
– Joonas Ilmavirta
Sep 29 at 17:45
add a comment
|
$begingroup$
Here is an example of just what you seek:
edited Sep 29 at 15:09
Tanner Swett
6,93122 silver badges44 bronze badges
answered Sep 29 at 1:05
David G. StorkDavid G. Stork
16.7k4 gold badges20 silver badges41 bronze badges
$endgroup$
17
$begingroup$
I don't know what this GIF looks like to you, but on my setup, it flickers between some sort of geometry animation and a mostly-white frame with a red rectangle in the upper left, before eventually settling down to just the white frame with red rectangle.
$endgroup$
– user2357112 supports Monica
Sep 29 at 9:25
$begingroup$
I don't know how you created that gif, but it had a lot of blank frames in it seemingly at random, causing it to appear to flicker. I've taken the liberty of editing the gif to remove the blank frames.
$endgroup$
– Tanner Swett
Sep 29 at 15:10
1
$begingroup$
Gif works fine for me
$endgroup$
– idle mathematician
Sep 29 at 17:11
1
$begingroup$
@B.Swan Yes, it's fixed now. You can see the edit history to compare with the old version.
$endgroup$
– Joonas Ilmavirta
Sep 29 at 17:45
add a comment
|
$begingroup$
Here is an example of just what you seek:
edited Sep 29 at 15:09
Tanner Swett
6,93122 silver badges44 bronze badges
answered Sep 29 at 1:05
David G. StorkDavid G. Stork
16.7k4 gold badges20 silver badges41 bronze badges
$endgroup$
Here is an example of just what you seek:
edited Sep 29 at 15:09
Tanner Swett
6,93122 silver badges44 bronze badges
answered Sep 29 at 1:05
David G. StorkDavid G. Stork
16.7k4 gold badges20 silver badges41 bronze badges
edited Sep 29 at 15:09
Tanner Swett
6,93122 silver badges44 bronze badges
edited Sep 29 at 15:09
Tanner Swett
6,93122 silver badges44 bronze badges
edited Sep 29 at 15:09
Tanner Swett
6,93122 silver badges44 bronze badges
6,93122 silver badges44 bronze badges
answered Sep 29 at 1:05
David G. StorkDavid G. Stork
16.7k4 gold badges20 silver badges41 bronze badges
answered Sep 29 at 1:05
David G. StorkDavid G. Stork
16.7k4 gold badges20 silver badges41 bronze badges
answered Sep 29 at 1:05
David G. StorkDavid G. Stork
16.7k4 gold badges20 silver badges41 bronze badges
16.7k4 gold badges20 silver badges41 bronze badges
17
$begingroup$
I don't know what this GIF looks like to you, but on my setup, it flickers between some sort of geometry animation and a mostly-white frame with a red rectangle in the upper left, before eventually settling down to just the white frame with red rectangle.
$endgroup$
– user2357112 supports Monica
Sep 29 at 9:25
$begingroup$
I don't know how you created that gif, but it had a lot of blank frames in it seemingly at random, causing it to appear to flicker. I've taken the liberty of editing the gif to remove the blank frames.
$endgroup$
– Tanner Swett
Sep 29 at 15:10
1
$begingroup$
Gif works fine for me
$endgroup$
– idle mathematician
Sep 29 at 17:11
1
$begingroup$
@B.Swan Yes, it's fixed now. You can see the edit history to compare with the old version.
$endgroup$
– Joonas Ilmavirta
Sep 29 at 17:45
add a comment
|
17
$begingroup$
I don't know what this GIF looks like to you, but on my setup, it flickers between some sort of geometry animation and a mostly-white frame with a red rectangle in the upper left, before eventually settling down to just the white frame with red rectangle.
$endgroup$
– user2357112 supports Monica
Sep 29 at 9:25
$begingroup$
I don't know how you created that gif, but it had a lot of blank frames in it seemingly at random, causing it to appear to flicker. I've taken the liberty of editing the gif to remove the blank frames.
$endgroup$
– Tanner Swett
Sep 29 at 15:10
1
$begingroup$
Gif works fine for me
$endgroup$
– idle mathematician
Sep 29 at 17:11
1
$begingroup$
@B.Swan Yes, it's fixed now. You can see the edit history to compare with the old version.
$endgroup$
– Joonas Ilmavirta
Sep 29 at 17:45
17
17
$begingroup$
I don't know what this GIF looks like to you, but on my setup, it flickers between some sort of geometry animation and a mostly-white frame with a red rectangle in the upper left, before eventually settling down to just the white frame with red rectangle.
$endgroup$
– user2357112 supports Monica
Sep 29 at 9:25
$begingroup$
I don't know what this GIF looks like to you, but on my setup, it flickers between some sort of geometry animation and a mostly-white frame with a red rectangle in the upper left, before eventually settling down to just the white frame with red rectangle.
$endgroup$
– user2357112 supports Monica
Sep 29 at 9:25
$begingroup$
I don't know how you created that gif, but it had a lot of blank frames in it seemingly at random, causing it to appear to flicker. I've taken the liberty of editing the gif to remove the blank frames.
$endgroup$
– Tanner Swett
Sep 29 at 15:10
$begingroup$
I don't know how you created that gif, but it had a lot of blank frames in it seemingly at random, causing it to appear to flicker. I've taken the liberty of editing the gif to remove the blank frames.
$endgroup$
– Tanner Swett
Sep 29 at 15:10
1
1
$begingroup$
Gif works fine for me
$endgroup$
– idle mathematician
Sep 29 at 17:11
$begingroup$
Gif works fine for me
$endgroup$
– idle mathematician
Sep 29 at 17:11
1
1
$begingroup$
@B.Swan Yes, it's fixed now. You can see the edit history to compare with the old version.
$endgroup$
– Joonas Ilmavirta
Sep 29 at 17:45
$begingroup$
@B.Swan Yes, it's fixed now. You can see the edit history to compare with the old version.
$endgroup$
– Joonas Ilmavirta
Sep 29 at 17:45
add a comment
|
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$begingroup$
What about $u=big(tfrac1sqrt 2,tfrac1sqrt 2,0big)$, $v=big(tfrac1sqrt 2,-tfrac1sqrt 2,0big)$ and $w=(0,0,1)$?
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– Azif00
Sep 29 at 0:24
1
$begingroup$
Yes, infinitely many. For example, rotate your basis in 2.
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– AnyAD
Sep 29 at 0:25
4
$begingroup$
Pick two non-zero and non-parallel vectors $x$ and $u$ in $mathbb R^3$. Then set $v=xtimes u$ and $w=utimes v$. Now $u,v,w$ are mutually orthogonal.
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– user1551
Sep 29 at 3:53
1
$begingroup$
@Azif00 More generally, $(a,b,0)$, $(b,-a,0)$, $(0,0,c)$ for any values of $a$, $b$, and $c$. Orthogonal vectors don't have to be unit vectors. The OP also seems to have made that mistake in the second example.
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– alephzero
Sep 29 at 10:34
12
$begingroup$
Hold out your hand. Arrange your fingers so that your index finger, middle finger, and thumb are all mutually perpendicular. Now rotate your hand around. All of these configurations correspond to a distinct set of 3 mutually orthogonal vectors.
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– Steven Gubkin
Sep 29 at 12:53