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Consider a vector space $V$ of dimension $d$, and suppose you are given $d-1$ linearly independent vectors $v_1, \dots, v_{d-1}$. Is there a simple explicit expression for a vector $v_d$ which completes the basis?

In dimensions two and three this is easy, for example in three dimensions one can just take the cross product. But I don't know what to do in, say, four or five dimensions.

To make the answer less arbitrary let's say the vectors $v_1, \dots, v_{d-1}$ are already orthonormal and you are trying to find the last orthonormal vector $v_d$ (there should only be two possibilities).

What I'm looking for is an explicit formula that can be easily computed, like the three-dimensional case.

i like xkcd
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    http://math.stackexchange.com/questions/904172/how-to-find-a-4d-vector-perpendicular-to-3-other-4d-vectors – Rocket Man Sep 14 '15 at 01:11
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    Well, you could take some vector not in the span of the ${v_i}_{i=1}^{d-1}$ and apply the Gram–Schmidt process... you have to get a new vector from somewhere to start with, though. – Chappers Sep 14 '15 at 01:13
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    Yes, but I was looking for an explicit formula. The solution linked to by @AJStas gives the answer. Thanks! – i like xkcd Sep 14 '15 at 01:14

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