1
$\begingroup$

$$x = 4i + 2j − 3k$$ I need to find a unit vector perpendicular to $x$.

I calculated and got the answer $\sqrt{1/5}*i - 2\sqrt{1/5}*j$

Is that correct? I assumed $k = 0$, so I could solve equation.

$\endgroup$
2
  • 1
    $\begingroup$ $\mathrm{k}$ is the vector $(0,0,1)$. What do you mean you "assumed $\mathrm{k}=0$"? $\endgroup$ Apr 15, 2012 at 4:50
  • $\begingroup$ @AntonioVargas: Yeah, you are absolutely right, my mistake. $\endgroup$
    – hey
    Apr 15, 2012 at 5:40

1 Answer 1

1
$\begingroup$

You want to find a unit vector $ Y= ai + bj + ck $ such that $X\cdot Y = 0.$ That is; $ 4a + 2b -3c = 0 . $ Clearly we can make some arbitrary choices here since many combinations of $a,b,c$ satisfy that requirement. It appears you chose $c=0$ (not the same as $k=0$, $k$ is a predefined vector remember), so then $a=1,b=-2.$

Thus you've found that $ i - 2j $ is a vector perpendicular to $X.$ Now you want to make it a unit vector, so divide by the length of the vector, which is $ \sqrt{ 1^2 + (-2)^2} = \sqrt{5}.$ Thus the required vector is $ Y = \frac{i}{\sqrt{5}} - \frac{2}{\sqrt{5}} j $ as you found.

$\endgroup$
6
  • $\begingroup$ Pardon the stupidity in this question, $Y=i-2j$ still qualifies as a vector in $3$-space? (I presume its ordered pair would be $(1,-2,0)$?) Also, can we find a unit vector $(a,b,c)$ where $a$, $b$, and $c$ are nonzero (and is perpendicular to our original vector)? $\endgroup$
    – 000
    Apr 15, 2012 at 5:06
  • $\begingroup$ @Limitless No stupidity involved! Yes, $i-2j$ is a vector is 3-space, with the ordered pair you mentioned. Indeed we can find non-zero a,b,c, it just makes the work slightly harder. For example, $a=b=1, c=2 $ works as well. So $i+j -2c $ is perpendicular to $X$, and we divide by its length to make it a unit vector. $\endgroup$ Apr 15, 2012 at 5:16
  • $\begingroup$ That's very interesting. Are there any unit vectors in this case which have special importance? I presume that the vast amount of them implies there are certain special ones. Also, are there an infinite amount? $\endgroup$
    – 000
    Apr 15, 2012 at 5:17
  • 1
    $\begingroup$ @RagibZaman: That's just awesome then, if I got it correct. Thanks! :) $\endgroup$
    – hey
    Apr 15, 2012 at 5:41
  • $\begingroup$ @Limitless There are indeed an infinite number of them, as there are an infinite number of solutions to $4a+2b-3c=0$ and each solution gives you a perpendicular vector. Or more geometrically, if you imagine the vector $X$ is 3-space, and then extend it infinitely in both directions to form a line, you can see why there are infinitely many vectors perpendicular to that line. As for which ones are special, I guess it depends on your particular problem, maybe there is some other condition, but as it is I don't think any are more special than the others. $\endgroup$ Apr 15, 2012 at 5:45

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.