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In three dimensions two planes are orthogonal when their normal vectors are orthogonal (their inner product is zero). For example, planes $xy$ and $xz$ are orthogonal because the vectors $\hat{z}$ and $\hat{y}$ which are normal to the planes, respectively, are orthogonal, i.e $\hat{z}\cdot \hat{y}=0$.

How we define orthogonality of planes in $n$ dimensions? I am talking about 2d planes through the origin, in n-dimensional Euclidean space, that are specified by orthonormal vectors $\hat{x}_1, \hat{x}_2,.., \hat{x}_n$.

In 4-D case we have four orthogonal axes x,y,z,w defined by orthonormal vectors $\hat{x}, \hat{y}, \hat{z}, \hat{w}$. These axes make six planes: $xy, xz, xw, yz, yw, zw$. Are these planes orthogonal? For example, the vectors $\hat{z}$ and $\hat{w}$ are perpendicular to the plane $xy$, but they are orthogonal, i.e $\hat{z}\cdot \hat{w}=0$, not parallel. How it is possible that they are not parallel when they are perpendicular to the same plane and how we check if the plane $xy$ is orthogonal to the plane $wz$?

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    Two planes can't be orthogonal in $3$ dimensions. What do you mean by "orthogonal"?2011-07-21
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    For a source of geometric intuition about the $4$-dimensional case, step down everything by $1$ dimension. So $4$-D becomes $3$-D, and planes become lines. It is a familiar fact that the $y$-axis and the $z$-axis are orthogonal to the $x$-axis, and to each other. In $d$ dimensions, the objects that behave the most like ordinary planes are the subspaces of dimension $d-1$. These are called *hyperplanes*.2011-07-21
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    @Eric Naslund Why can't? What I mean is said in first line. The inner product of their normal vectors is zero. But in four dimensions each plane has two normal vectors.2011-07-21
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    @Eric, yes they can. For example in $\mathbb{R}^3$, the y-z plane and the x-z plane are orthogonal because their normal axes, the x-axis and the y-axis, are orthogonal subspaces.2011-07-21
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    In n-dimensional space (n>3), the "plane" will be called as hyperplane, which actually is a n-1 dimensional subspace. For example, in 4-D space, given four orthogonal axes, the hyperplanes determined by the axes are yzw, xzw, xyw, xyz. see http://en.wikipedia.org/wiki/Hyperplane. Hope I'm not wrong:)2011-07-21
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    @Anku: No, in 3 dimensions each plane has two normal vectors; in 4 dimensions each plane has continuum many normal vectors.2011-07-21
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    Ok, but I'm not talking about hyperplanes but 2-d planes. We can still define them in more then three dimensions.2011-07-21
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    I'm talking about 2d planes too. In 3d, the normal vectors of the x-y plane are $\langle 0,0,-1\rangle$ and $\langle 0,0,+1\rangle$. In 4d, the normal vectors of the x-y plane are `$\{\langle 0,0,\operatorname{cos}(t),\operatorname{sin}(t)\rangle : -\pi < t \leq +\pi\}$`.2011-07-21
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    The last equation in the above comment is {<0,0,cos(t),sin(t)> : -π < t < +π}, I don't know why it's not rendering.2011-07-21
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    Actually they are infinite in 3d too. Take for example (0,0,a) where a is real number. But what I actually mean is not any normal vector and not any plane, but the orthonormal basis vectors of planes xy, xz etc. I need to reformulate my question. Sorry.2011-07-21
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    Ok, I made some changes to the question. Hope it makes more sense now.2011-07-21

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