In a scenario, say that:

Vectors $\mathbf{U}$, $\mathbf{V}$ and $\mathbf{W}$ are all orthogonal such that the dot product between each of these $(\mathbf{UV}\;\mathbf{VW}\;\mathbf{WU})$ is equal to zero.

I imagine that for any potential vector space $\mathbf{R}$ this would only be possible in two situations.

1) $\mathbf{U}$, $\mathbf{W}$ and/or $\mathbf{V}$ is the zero vector.

2) $\mathbf{U}=(1, 0, 0)$, $\mathbf{V} = (0, 1, 0)$ and $\mathbf{W} = (0, 0, 1)$.

Is there any other situation where three vectors are all orthogonal to each other?

  • $\begingroup$ What about $u=\big(\tfrac{1}{\sqrt 2},\tfrac{1}{\sqrt 2},0\big)$, $v=\big(\tfrac{1}{\sqrt 2},-\tfrac{1}{\sqrt 2},0\big)$ and $w=(0,0,1)$? $\endgroup$
    – azif00
    Sep 29 '19 at 0:24
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    $\begingroup$ Yes, infinitely many. For example, rotate your basis in 2. $\endgroup$
    – AnyAD
    Sep 29 '19 at 0:25
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    $\begingroup$ Pick two non-zero and non-parallel vectors $x$ and $u$ in $\mathbb R^3$. Then set $v=x\times u$ and $w=u\times v$. Now $u,v,w$ are mutually orthogonal. $\endgroup$
    – user1551
    Sep 29 '19 at 3:53
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    $\begingroup$ @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 '19 at 10:34
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    $\begingroup$ 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. $\endgroup$ Sep 29 '19 at 12:53

Think it like this: for a given vector $u\neq 0$ in $\mathbb{R}^3$, 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 $v\neq 0$ 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 $w\in P\cap P'$. What does $P\cap P'$ look like?

By the way, notice that $v$ is ANY non-zero vector, not only $(1,0,0), (0,1,0)$ or $(0,0,1)$


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:

  1. choose a first vector $v_1=(a,b,c)$
  2. find a second vector orthogonal to $v_1$ that is e.g. $v_2=(-b,a,0)$
  3. determine the third by cross product $v_3=v_1\times v_2$
  • $\begingroup$ 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 '19 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 $v_1\times v_2$. $\endgroup$
    – user
    Sep 30 '19 at 5:54

How about $v_1= (\frac{1}{\sqrt2}, \frac{1}{\sqrt2},0),v_2=(-\frac{1}{\sqrt2},\frac{1}{\sqrt2},0),v_3=(0,0,1)$ ? Geometrically take any frame and rotate in any direction.


Here is an example of just what you seek:

An animation showing three orthogonal vectors in 3D space. A user is moving sliders, causing the vectors to rotate as a whole.

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    $\begingroup$ 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. $\endgroup$ Sep 29 '19 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. $\endgroup$ Sep 29 '19 at 15:10
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    $\begingroup$ Gif works fine for me $\endgroup$
    – B.Swan
    Sep 29 '19 at 17:11
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    $\begingroup$ @B.Swan Yes, it's fixed now. You can see the edit history to compare with the old version. $\endgroup$ Sep 29 '19 at 17:45

There are infinitely many possibilities. $(1,0,0), (0,1,-1)$ and $(0,1,1)$ is one example.


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.


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