1
$\begingroup$

given u=[0,4,3]T a unit vector that is prependicular to u is

How can I find the normal?

I am new to these stuff.

Thanks

$\endgroup$
  • $\begingroup$ In tridimensional space there are a plane of perpendicular vectors to $u$. $\endgroup$ – Sigur Dec 6 '13 at 17:33
  • $\begingroup$ Just find "by inspection" any vector which has a zero scalar product with your vector. Then divide by its length to get a unit vector. $\endgroup$ – Old John Dec 6 '13 at 17:39
  • $\begingroup$ what if they ask for a vector parallel to u $\endgroup$ – user107895 Dec 6 '13 at 17:46
  • $\begingroup$ math.stackexchange.com/questions/133177/… there is complete answer ,please see it $\endgroup$ – dato datuashvili Dec 6 '13 at 17:55
  • $\begingroup$ please see answer if it is enough $\endgroup$ – dato datuashvili Dec 6 '13 at 18:03
0
$\begingroup$

ok in your case vector is simple $u=[0,4,3]$,for some vector s,we say that $s$ is perpendicular to $u$,if $s*u=0$,so let us suppose that

$s=(s_1,s_2,s_3)$ ,so we have

$S_1*0+s_2*4+s_3*3=0$,

after that follow this rule

Finding a unit vector perpendicular to another vector

$\endgroup$
  • $\begingroup$ so if i want to find normal to <-10,-2,-8> -10x-2b-8c=0 what do I do after? $\endgroup$ – user107895 Dec 6 '13 at 19:58
  • $\begingroup$ choose any values of $x$,$b$ and $c$ which satisfy these solution,but to gain unit vector,divide this vector by its length $\endgroup$ – dato datuashvili Dec 6 '13 at 20:07

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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