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Subspace / basis problem
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[QUOTE="BvU, post: 4943338, member: 499340"] What you attempt is to find a,b,c,d,e,f such that (a,b,c) x (d,e,f) = (1,2,3) Might seem reasonable, but it's a bit tedious: you want to find six variables from three equations. And you don't really need the lengths (that comes in part b). Orthogonal to a vector ##\vec v## is anything that has ##\vec a \cdot \vec v = 0 ## with ##|\vec a| > 0\;##. That's only one equation with three unknowns. Pick a vector v1 that satisfies that equation. Pick a different one and call that v2. I think then you have a basis already according to this [U][URL='http://en.wikipedia.org/wiki/Basis_(linear_algebra)']link[/URL][/U] (exercise doesn't ask for orthogonal or orthonormal basis !) But if you do want them orthonormal you can do v1 x (1,2,3) to get a v2 that is perpendicular to both. Then normalize v1 and v2. Re describing S: Why do you think there is the possibility that S is a line ? [/QUOTE]
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Subspace / basis problem
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