If we know the normal vector of the plane, can we take the cross product of this vector and any other vector ( e.g, 0,0,1 ), and then do the same thing ( n x 0,1,0 ) for example. Both of the resultant vectors are perpendicular to the normal, and hence lie in the plane. If they turn out to be linearly independent, can we use them to write out the equation of our plane in parametric form?
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No, my problem is different. I'm asking whether we can just cross the normal vector and another vector twice to get 2 linearly independent vectors in the plane, and the answer only says that we can find 2 points in the plane, and subtract it from the original fixed point to get 2 linearly independent vectors. – May 27 '17 at 00:13
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Yes, this works (mostly; see below). To see this, let $\mathbf{n}$ be a normal vector and $\mathbf{v}$ be arbitrary. Then the vectors $\mathbf{n}$, $\mathbf{v}\times\mathbf{n}$, and $\mathbf{n}\times(\mathbf{v}\times\mathbf{n})$ are mutually orthogonal. They're also linearly independent, so no need to worry about that. Your plane can then be parameterized as $$ \mathbf{P}(s,t) = (\mathbf{v}\times\mathbf{n})s + [\mathbf{n}\times(\mathbf{v}\times\mathbf{n})]t + \mathbf{x} $$ where $\mathbf{x}$ is any point on the plane.
However, this doesn't work if $\mathbf{v}\times\mathbf{n} = 0$. This can obviously be avoided by judicious choice of $\mathbf{v}$, but it's something to be careful of.
eyeballfrog
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What if one vector was v x n and the other was w x n ? Also, if v and n are perpendicular to begin with, could we conclude that n x (v x n ) = v? – May 27 '17 at 06:56
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In the first case, you'd have to check that $\mathbf{v}\times\mathbf{n}$ isn't parallel to $\mathbf{w}\times\mathbf{n}$. If they aren't parallel then it's fine. In the second, that is almost true: $\mathbf{n}\times(\mathbf{v}\times\mathbf{n}) = n^2\mathbf{v}$. If $\mathbf{n}$ is a unit vector then the statement is true. – eyeballfrog May 27 '17 at 15:44
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