$$x = 4i + 2j − 3k$$ 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.
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1$\begingroup$ $\mathrm{k}$ is the vector $(0,0,1)$. What do you mean you "assumed $\mathrm{k}=0$"? $\endgroup$– Antonio VargasApr 15, 2012 at 4:50
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$\begingroup$ @AntonioVargas: Yeah, you are absolutely right, my mistake. $\endgroup$– heyApr 15, 2012 at 5:40
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1 Answer
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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.
answered Apr 15, 2012 at 5:01
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$\begingroup$ 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)? $\endgroup$– 000Apr 15, 2012 at 5:06
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$\begingroup$ @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
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$\begingroup$ 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? $\endgroup$– 000Apr 15, 2012 at 5:17
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1$\begingroup$ @RagibZaman: That's just awesome then, if I got it correct. Thanks! :) $\endgroup$– heyApr 15, 2012 at 5:41
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$\begingroup$ @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