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If $A(3, 4, 5)$ and $B(7, 10, 12)$ are two end points of a line segment. I know that vector $V_1=B-A$ i.e $V_1=(4, 6, 7)$ then how to find the other two orthogonal vectors of this?

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This smells like Vector Geometry homework. –  Zéychin Feb 9 '12 at 22:20
no not a vector geometry homework ... i just provided some example for better understanding.........and just now understood that there will be infinite vectors orthogonal to the given vector –  user1198477 Feb 9 '12 at 22:57

3 Answers 3

Three non-colinear points are needed to define a plane, but you only really have two here. If you just want to find a couple of arbitrary vectors orthogonal vectors, just pick any third point, C, not colinear with A and B, then W = (A-C)×(B-C) (where × is the cross product) is orthogonal to the line segment connecting A and B since any plane containing A and B also contains the line segment connecting them.

If you need a third vector U orthogonal to V = A-B and to W, then just set U = V×W

The first step is probably easiest if you pick C as the origin (again if it's not colinear with A and B); this reduces to setting W = (A×B).

More information is available both at Wolfram and Wikipedia.

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could you explain this approach for the above mentioned example in question.....Thanks. –  user1198477 Feb 9 '12 at 19:27
@user1198477: So that I know how detailed to get, do you know what a cross product is? –  andand Feb 10 '12 at 0:53

Use vector product: V1 ^ V2, if they aren't collateral, it will give you an orthogonal vector.

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There are not two unique orthogonal vectors, there are an infinity of them.

Consider V1 as the vector normal to a plane (in Euclidean space). Any two orthogonal vectors lying in that surface will be orthogonal to V1.

Imagine a clock face. Put the hands at 12 and 3. You now have a set of 3 orthogonal vectors, V1, 12'o'clock, and 3'o'clock. Rotate the clock face, and you get a different set. Mirror-image the clock face, and you get another set. And so on...

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