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March 1st, 2008, 08:26 PM  #1 
Newbie Joined: Mar 2008 From: Canada Posts: 2 Thanks: 0  orthogonal unit vectors
i am having a lot of trouble with the following question: Find two vectors of norm 1 that are orthogonal to the three vectors u=(2,1,4,0) v=(1,1,2,2) w=(3,2,5,4) i have figured out (but could be wrong ) that the orthogonal vectors must be unit vectors if they have a norm of 1. also, the dot products of the unit vector and u,v, and w must all equal zero. however, i can't figure out a unit vector that makes the dot products equal to zero with u,v, and w! if someone could help me out i would really appreciate it 
March 2nd, 2008, 07:11 AM  #2 
Member Joined: Aug 2007 Posts: 93 Thanks: 0 
have you tried converting u,v & w to unit vectors (i.e. dividing by their length)?

March 2nd, 2008, 08:20 AM  #3  
Senior Member Joined: Oct 2007 From: Chicago Posts: 1,701 Thanks: 3  Quote:
Intuitively, I've gotten so far, which may be of use: If these three vectors are neither colinear (which they obviously aren't), nor coplanar, (thus, co 3planar, by definition), then there is only one vector in 4space that is perpendicular. This means the 3 points need to be coplanar, right? Unless, the two unit vectors are z and z, which would be a ridiculous requirement. So, since the three vectors must be coplanar (I hope), each one intersects each of the others at some point. so u=v, v=z, and u=z at 3 distinct points, or all 3 cross at a single point (but this point wouldn't be enough to define a plane, so hopefully this isn't the case). I hope this will help, or I may be misunderstanding some subtle difference between 3space and 4space, but there's only one degree of freedom for a 3plane in 4space, so there should be only one line unless they create a plane. Anyway, DMY Sommerville has a lot of work on higher dimensions; you may want to look into his work.  
March 2nd, 2008, 02:19 PM  #4  
Global Moderator Joined: May 2007 Posts: 6,416 Thanks: 558  Re: orthogonal unit vectors Quote:
z.u=z.v=z.w=0 gives you three equations in four unknowns. Solving this linear system will give you three of the unknowns in terms of the fourth. Finally using z.z=1 can give you the solution you want. There will be a sign ambiguity, which gives two possibilites.  
March 2nd, 2008, 06:40 PM  #5 
Member Joined: Aug 2007 Posts: 93 Thanks: 0 
My thought was that the cross product formula in 4space would work (I do not know for sure) then you would have to calculate determinants on a 4 x 4 matrix. My idea was to that as a unit vector is a vector of cosines of direction angles,calculating the unit vectors for u,v & w one could set up an easier system of equations that would just add pi/2 to each vector. If I find the time I will see if this works.

March 2nd, 2008, 07:51 PM  #6 
Senior Member Joined: Oct 2007 From: Chicago Posts: 1,701 Thanks: 3 
I was thinking that, and it may work, but I don't know enough linear algebra to know if that's valid.

March 3rd, 2008, 02:59 PM  #7 
Newbie Joined: Mar 2008 From: Canada Posts: 2 Thanks: 0 
thanks for your help!


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