I recognize that I have to use a sport of the dot product but I am not sure how to do that because I only have actually one vector?

Just take any kind of three live independence vectors $\vec v_1,\vec v_2,\vec v_3$, likewise independent of $\vec v$. You can use the conventional basis vectors (<1,0,0,0> etc.). Then task them far from $\vec v$.

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$$\vec b_1 = \vec v_1 - \left(\frac\vec v_1\cdot\vec v\vec v\cdot\vec v\right)\vec v$$$$\vec b_2 = \vec v_2 - \left(\frac\vec v_2\cdot\vec v\vec v\cdot\vec v\right)\vec v$$$$\vec b_3 = \vec v_3 - \left(\frac\vec v_3\cdot\vec v\vec v\cdot\vec v\right)\vec v$$

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

I think I have figured that out. When you take it v and also one the it"s orthogonal vectors the period product will provide you zero, thus let the orthogonal vector = (a,b,c,d) and also let v = (e,f,g,h), climate ae+bf+cg+dh = 0 climate if you isolate a friend can get a = ((-f)b+(-g)c+(-h)d)/e, you have the right to then change a in her orthogonal vector giving

((-f)b+(-g)c+(-h)d)/e , b , c , d)/ which have the right to the it is in decomposed into

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

P.S. Sorry because that the absence of notation and structure I"m not quite supplied to proofs yet.

Let $v=$

You want to uncover three linearly live independence vectors, which room perpendicular come $v$

For example you may take into consideration

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

in instance of $v_i \ne 0$ for $i=1,2,3,4$

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What walk "the orthogonal communication vectors extending the subspace perpendicular come vector $\vece_1$" mean?