Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 $v_2=(-1,1,1)$

share|cite|improve this question
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

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


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

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 $v=(2,0,-1)$

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.