Just take any 3 independent vectors $vec v_1,vec v_2,vec v_3$, additionally independent of $vec v$. You might usage the conventional basis vectors (<1,0,0,0> etc.). Then project them away from $vec v$.

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$$vec b_1 = vec v_1 - left(fracvec v_1cdotvec vvec vcdotvec v ight)vec v$$$$vec b_2 = vec v_2 - left(fracvec v_2cdotvec vvec vcdotvec v ight)vec v$$$$vec b_3 = vec v_3 - left(fracvec v_3cdotvec vvec vcdotvec v ight)vec v$$

You can use the Gram-Schmidt procedure rather, if you desire an orthonormal basis.

I think I have figured it out. When you take v and among it"s orthogonal vectors the dot product will certainly provide you zero, therefore let the orthogonal vector = (a,b,c,d) and let v = (e,f,g,h), then ae+bf+cg+dh = 0 then if you isolate a you have the right to get a = ((-f)b+(-g)c+(-h)d)/e, you can then relocation a in your orthogonal vector giving

((-f)b+(-g)c+(-h)d)/e , b , c , d)/ which can the be decomposed into

b ((-f)/e , 1 , 0 , 0) + c ((-g)/e , 0 , 1 , 0) + d ((-h)/e , 0 , 0 , 1)

P.S. Sorry for the lack of notation and structure I"m not quite provided to proofs yet.

Let $v=$

You want to find three livirtually independent vectors, which are perpendicular to $v$

For instance you might take into consideration

$$a=\b=\c= $$

in situation of $v_i e 0$ for $i=1,2,3,4$

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