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Find a pair of orthogonal vectors in $R^4$ that are also orthogonal to the vector (1,1,-2,3).

What i have tried so far: enter image description here

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If you're talking about the usual, euclidean inner product in $\;\Bbb R^4\;$ , just solve a simple system of equations: you want two vectors $\;(x_1,...,x_4)\,,\,\,(y_1,...,y_4)\;$ s.t.

$$\begin{cases}x_1y_1+x_2y_2+x_3y_3+x_4y_4=0\\{}\\x_1+x_2-2x_3+3x_4=0\\{}\\y_1+y_2-2y_3+3y_4=0\end{cases}$$

Further hint/advice: first solve parametrically the last two linear equations, and then substitute in the first one to get one particular solution.

Added on request: Take equations 1-2 and equal them:

$$x_1y_1+\ldots+x_4y_4=x_1+x_2-2x_3+3x_4\implies$$

$$ x_1(y_1-1)+x_2(y_2-1)+x_3(y_3+2)+x_4(y_4+3)=0$$

Choose, for example, $\;y_3=-2\,,\,\,y_4=-3\;,\;\;x_1=x_2=0\;$

Third equation now gives

$$y_1+y_2+4-9=0\implies y_1+y_2=5$$

so we can choose, say $\;y_1=0,y_2=5\;$ .

Likewise, from second equation we get

$$-2x_3+3x_4=0\implies \;\text{we can choose}\;\;x_3=3\,,\,\,x_4=2$$

and we're left with

$$\vec x=(0,0,3,2)\;,\;\;\;\vec y=(0,5,-2,-3)$$

Note that we made several "choices": why is this possible...?

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  • $\begingroup$ I'm still confused but i guess i'll just take another look at it tomorrow, maybe my tired brain is over thinking things. $\endgroup$ – Linear Algebra Noob Apr 29 '15 at 8:16
  • $\begingroup$ @LinearAlgebraNoob Give it a shot tomorrow, when fresh. If you have some problem write back here. $\endgroup$ – Timbuc Apr 29 '15 at 8:19
  • $\begingroup$ I am still very much confused.. The algebra that came out of trying to solve that system just became horrendous. Do you possibly have another hint or advice? $\endgroup$ – Linear Algebra Noob Apr 30 '15 at 7:05
  • $\begingroup$ @LinearAlgebraNoob Read the addition I made to my answer. $\endgroup$ – Timbuc May 10 '15 at 7:28

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