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?
how to find orthogonal vectors of 3 Dimensional point
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0no 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 – 2012-02-09
3 Answers
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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0@user1198477: So that I know how detailed to get, do you know what a cross product is? – 2012-02-10
Use vector product: V1 ^ V2
, if they aren't collateral, it will give you an orthogonal vector.
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...