Let's assume that I have previously found an orthonormal basis for the plane (dim 2) whose cartesian equation I want to find. Is it as simple as using Gram-Schmidt a third time to find a vector that's orthogonal to both basis vectors, and then plugging in the coordinates in the equation $ax+by+cz+dw=0$ (substituting $a$, $b$, $c$ and $d$)?
Finding cartesian equation for a plane in $\mathbb R^4$
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linear-algebra
analytic-geometry
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3A plane in $\mathbf R^4$ requires *two* linear equations. – 2017-01-27
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0... which are the equations of 2 distinct hyperplanes (dim. 3) in 4D containing that plane. This the same in $\mathbb{R^3}$: a line necessitates 2 equations. – 2017-01-27
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0@AstlyDichrar do you really want subspaces passing through the origin ? Otherwise, you need equations of the form $ax+by+cz+dw=e$. – 2017-01-27
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0I think I understand it now. I need to plug in the coordinates of $b_1$ (first orthogonal vector) in $ax+by+cz+dw=0$ ($0$ because it's an orthogonal complement of a subspace, and thus being a subspace needs to contain the null vector) and $b_2$ (second orthogonal vector) in another equation of the same type, correct? – 2017-01-27
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There are a couple of complementary ways to specify a subspace of $\mathbb R^n$ (or any vector space $V$, for that matter).
One is by listing a set of vectors that generate the space. If these vectors are linearly independent, then they are a basis for that space, of course.
The complementary method is via a system of homogeneous linear equations. These equations can be written in the form $\mathbf n\cdot\mathbf x=0$, so what you’re really doing is listing a set of vectors orthogonal to the space. More generally, you can specify a subspace $W$ of $V$ by listing a set of vectors $\mathbf\alpha_i$ in the dual space $V^*$ that annihilate the subspace, i.e., such that for all $\mathbf w\in W$, $\mathbf\alpha[\mathbf v]=0$. For a finite-dimensional vector space, you will need $\dim V-\dim W$ such dual vectors/equations.
So, you can find a set of linear equations that elements of your subspace satisfy by finding a basis $\mathbf n_i$ for the orthogonal complement of that space. You have a two-dimensional subspace of $\mathbb R^4$, so you’ll need two equations. Since $W^{\perp\perp}=W$ this amounts to finding the kernel (null space) of the matrix that has your two basis vectors as rows.