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Find a vector perpendicular to $Oy$ axis. Knowing that $v\cdot v_1=8$ and $v\cdot v_2=-3$, where $v_1=(3,1,-2)$ and $v_2=(-1,1,1)$

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Are you aware of the practice of accepting answers to your question? (See this meta question.) –  joriki Jul 27 '12 at 20:43
    
Yes, but I don´t understand your commentaire. Was I rude with someone here. If I was, I´m sorry about it. –  Vinicius L. Beserra Jul 28 '12 at 22:29

1 Answer 1

ley $v=(x,y,z)$ perpendicular to $OY$ axis means that $v*(0,1,0)=0$ $v*v_1=3*x+1*y-2*z=8$

$v*v_2=-1*x+1*y+1*z=-3$

$v*(0,1,0)=0 -->y=0$

so $3*x-2*z=8$

$-x+z=-3$ from second $z=-3+x$ put into first one $3*x-2*(-3+x)=8$
$x=2$ and $z=-1$ so we have $v=(2,0,-1)$

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Thanks. It was just what i wask thinking about. –  Vinicius L. Beserra Jul 27 '12 at 20:53
    
my friend i am happy that could help you,just please consider advices from other people and accept their answers,it is rule of this website ok?it is friendly advice –  dato datuashvili Jul 27 '12 at 20:54
    
Thanks dato. I will consider this in the future. –  Vinicius L. Beserra Jul 28 '12 at 22:36

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