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How to find orthogonal vectors?

by meteorologist1
Tags: orthogonal, vectors
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meteorologist1
#1
Jan26-05, 10:23 PM
P: 101
Hi, this might be very easy, but I forgot how to do the following: I have a vector in R^6: (x1, x2, x3, x4, x5, x6). How do I find a vector such that their dot product vanishes? I know how to do it for the two dimensional case: (x1, x2), so the vector that is perpendicular to it is c(-x2, x1) where c is a scalar. Could someone show me how to do it for the vector in R^6 case?
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BobG
#2
Jan26-05, 10:28 PM
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Take the cross product of your vector and any other vector that has the same origin (any unit vector usually fits the bill and is easiest since only one axis is non-zero).
meteorologist1
#3
Jan26-05, 10:39 PM
P: 101
Ok I see. I'm not sure how to do a cross product for a six-component vector though. Could you please show me?

Justin Lazear
#4
Jan26-05, 10:55 PM
P: 290
How to find orthogonal vectors?

The only condition on the vector is that its dot product with x vanishes?

Just solve

[tex]\vec{x}\cdot\vec{y} = 0[/tex]

--J
robphy
#5
Jan26-05, 11:05 PM
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Given a vector in n-dimensions, there is a whole (n-1)-dimensional hyperplane of vectors that are orthogonal to that vector.

Let me do the 4D-case. The 6D-case is easily obtained by generalizing.
Given V=(x1,x2,x3,x4), choose W=(w1,w2,w3,w4) such that
0=(V dot W)=x1w1+x2w2+x3w3+x4w4.

One choice is w1=-x2, w2=x1, w3=0, w4=0. (This is a generalization of your 2D example).

Basically, you can choose values for three components of W. The fourth is determined by the requirement that 0=(V dot W).
vincentchan
#6
Jan26-05, 11:58 PM
P: 610
since someone has already posted the answer... I will show you another alternative method for this problem

let's say you have vector x in R^n, and wanna find a vector y orthogonal to it
we know the dot product of these two vector must be zero

[tex] \vec{x} \cdot \vec{y} = 0 [/tex]

[tex] \Rightarrow x_{1} y_{1} + x_{2} y_{2} + x_{3} y_{3} + .....+ x_{n-1} y_{n-1} + x_{n} y_{n} = 0 [/tex]

[tex]\Rightarrow x_{1} y_{1}+ x_{2} y_{2} + x_{3} y_{3} + .....+ x_{n-1} y_{n-1} = - x_{n} y_{n} [/tex]

[tex]\Rightarrow y_{n} = -(x_{1} y_{1} + x_{2} y_{2} + x_{3} y_{3} +.....+ x_{n-1} y_{n-1}) / x_{n} [/tex]

now, you have more freedom to vary [itex] y_{1}.y_{2}.....y_{n-1} [/itex] , as long as you the [itex] y_{n} [/itex] follows the formulas above
HallsofIvy
#7
Jan27-05, 06:54 AM
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Thanks
PF Gold
P: 39,339
If you know how to do it with 2 dim vectors, just take a vector with all except two components 0 and fix the other 2 to give a 0 dot product!

For example, to find a vector orthogonal to <1, -3, 4, 1 , -1, 2>, take all except the first two components 0, take the first two 3 and 1 respectively: <3, 1, 0, 0, 0, 0>.

Of course, there are many independent vectors orthogonal to a given six dimensional vector (not true in 2 dimensions). You could also choose to make all except the first and last components 0: <2, 0, 0, 0, 0, -1>.

OR <0, 4, 3, 0, 0, 0>

OR ....
meteorologist1
#8
Jan27-05, 08:28 AM
P: 101
All right thanks all.


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