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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
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2 Answers 2

up vote 2 down vote accepted

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
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There are lots of ways to do this. Here are a few:

  • Calculate the cross product of CB and CD. This is automatically perpendicular to both. (If you don't know what the cross product is, this may seem a little like a magic formula, and you should check that it's true if you use it.)
  • Find a vector equation of the plane passing through B, C and D, and use it to find a Cartesian equation of the same plane. Read off a normal vector. This is perpendicular to everything in the plane, and CB and CD are in the plane.
  • Find CB, and read off two linearly independent vectors u and v perpendicular to CB; these span a plane of "directions perpendicular to CB". Some vector au + bv should then be perpendicular to CD. (In fact, lots probably will. Try choosing a=1 or b=1 to simplify your life a little.)
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