# Perpendicular unit vector

$$x = 4i + 2j − 3k$$ 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.

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

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.

• 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)?
– 000
Apr 15, 2012 at 5:06
• @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. Apr 15, 2012 at 5:16
• 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?
– 000
Apr 15, 2012 at 5:17
• @RagibZaman: That's just awesome then, if I got it correct. Thanks! :)
– hey
Apr 15, 2012 at 5:41
• @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. Apr 15, 2012 at 5:45