# Find an orthogonal vector to 2 vector

I have the following problem:

A B C D are the 4 consecutive summit of a parallelogram, and have the following coordinates

A(1,-1,1);B(3,0,2);C(2,3,4);D(0,2,3)

I must find a vector that is orthogonal to both CB and CD.

How? Is there some kind of formula?

Thanks,

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Do you know the cross-product? en.wikipedia.org/wiki/Cross_product –  Abel Apr 11 '13 at 0:23
No, all the information I have is written up there. So I assume I must calculate the cross-product of CB and CD first, then what? –  Machinegon Apr 11 '13 at 0:24
Then you are done. The cross product of two vectors is always orthogonal to both. –  Abel Apr 11 '13 at 0:26

The cross product of two vectors is orthogonal to both, and has magnitude equal to the area of the parallelogram bounded on two sides by those vectors. Thus, if you have: $$\vec{CB} = \langle3-2, 0-3, 2-4\rangle = \langle1, -3, -2\rangle$$ $$\vec{CD} = \langle0-2, 2-3, 3-4\rangle = \langle-2, -1, -1\rangle$$

Compute the following, which is an answer to your question: $$\langle1, -3, -2\rangle\times\langle-2, -1, -1\rangle = \langle1, 5, -7\rangle$$

Note, though, that there are infinitely many vectors that are orthogonal to $\vec{CB}$ and $\vec{CD}$. However, these are all non-zero scalar multiples of the cross product. So, you can multiply your cross product by any (non-zero) scalar.

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Alright, if I compute the cross product correctly using en.wikipedia.org/wiki/Cofactor_expansion it gives me i + 5j - 7k, so (1, 5, -7) is this correct? –  Machinegon Apr 11 '13 at 0:42
@Machinegon That is correct. –  anorton Apr 11 '13 at 0:45