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I am quite confused with this question. The question itself wasn't originally in English, so I'll try my best to translate it:

"Complete the following linearly independent families with vectors of $\mathbb R$$^3$ in a basis of $\mathbb R$$^3$."

So, is the question asking for a family of 3 vectors, since the basis is OF $\mathbb R$$^3$? I guess the source of my confusion is finding out whether they are referring to the dimensions of the vectors within the family rather than the dimension of the vector space itself.

Assuming that the family does indeed comprise 3 vectors, let's say that the unfinished family given is ((1,1,1)). What is the best way to go about this problem? Should I just look for some vectors that would make it linearly indepedent? I feel like that might not work with more complicated problems.

My best idea is to to find ($u_1$,$u_2$,$u_3$), ($v_1$,$v_2$,$v_3$) $\in$ $\mathbb R$$^3$, such that for any a,b,c $\in$ $\mathbb R$ we have a(1,1,1) + b($u_1$,$u_2$,$u_3$) + c($v_1$,$v_2$,$v_3$) = (0,0,0) $\implies$ a = 0, b = 0, c = 0.

We would have:

a + b$u_1$ + c$v_1$ = 0

a + b$u_2$ + c$v_2$ = 0

a + b$u_3$ + c$v_3$ = 0

And now, I just have to find the values for u and v which would imply that all the scalars are nul. It sounds like it should be really simple, but I'm not sure how to go about this. Any help is greatly appreciated. Thank you.

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    $\begingroup$ "...whether they are referring to the dimensions of the vectors..." there is no such thing: as you guessed, they want a basis of $\mathbb{R}^3$, hence $3$ vectors. You might find this useful. What you say is right, the only problem is that there are infinitely many solutions, so you just need to choose an approach. $\endgroup$ – 57Jimmy Apr 2 '18 at 14:24
  • $\begingroup$ @57Jimmy Thank you very much for your reply. I see I have some gaps in my knowledge that I need filled; I'll do some research and then post again if I'm still unsure. $\endgroup$ – iaskdumbstuff Apr 2 '18 at 15:06
  • $\begingroup$ No worries. Just to clarify what I meant at the beginning: we speak of the dimension of a vector space, hence you can also talk about the dimension of the linear span of some vectors inside a vector space, but the dimension of the span of any single non-zero vector will be one. $\endgroup$ – 57Jimmy Apr 2 '18 at 19:48
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We can find a linearly independent orthogonal vector to the given vectors according to one of the two following ways

  • find the third vector by cross product $v_3=v_1\times v_2$

  • find an orthogonal vector by dot product solving the system $(a,b,c)\cdot v_1=0$ and $(a,b,c)\cdot v_2=0$

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  • $\begingroup$ Thanks for your reply; is there any way to use the cross product method when I'm only given one vector? $\endgroup$ – iaskdumbstuff Apr 3 '18 at 16:29
  • $\begingroup$ @user494405 Yes of course, it suffices to find by simple inspection one second vector orthogonal to the given one and then use cross product! $\endgroup$ – user Apr 3 '18 at 16:30
  • $\begingroup$ Oh of course, that should've been obvious. I'm just afraid that I'll get a problem where intuitively finding for an orthogonal vector would be too difficult. I can't really imagine what such a problem would look like, but I assume it's possible? Anyway, thank you for the help! $\endgroup$ – iaskdumbstuff Apr 3 '18 at 16:35
  • $\begingroup$ @user494405 For example if $v_1=(a,b,c)$ is given it suffices to take $v_2=(-b,a,0)$ which is orthogonal to $v_1$ and then calculate $v_3=v_1\times v_2$. $\endgroup$ – user Apr 3 '18 at 16:37

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