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Homework Help: Finding distance for a given value between vectors

  1. Oct 30, 2015 #1
    1. The problem statement, all variables and given/known data
    Let u = (2,k,1,-4) and v = (3,-1,6,-3)

    For which values of k is the distance d(u,v) = 6?
    2. Relevant equations

    3. The attempt at a solution

    6 = √((3-2)2 + (-1-k)2 + (6-1)2 + (-3-4)2)
    6 = √(1 + 1 + k2 + 25 + 21)
    6 = √(k2 + 48)
    36 = k2 + 48
    k = √-12

    The answers seem to be 6k, -16k and -k or k= 2 and -4 (the answer sheet makes no sense and is mislabeled and unorganized horrifically) and I don't even have a remote idea how to get multiple solutions. I suppose my whole approach is wrong and I can't for the life of me understand what to do (I'm working without any textbooks)

    Kind regards
  2. jcsd
  3. Oct 30, 2015 #2


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    Gold Member

    Remember that ##d(u,v) = ||u-v|| = \sqrt{(u_1-v_1)^2+(u_2-v_2)^2 ... + (u_n - v_n)^2}##, so you're on the right track. Recheck your math.
  4. Oct 30, 2015 #3
    I redid the working (both d(u,v) and d(v,u) which turn out to be the same so it seems correct this time) and get k = √8

    I really have no idea what I'm doing wrong, can you specify where my mistake is? I have a lot more work to catch up on and I'm struggling with the basics...

    and by ||u−v|| does this mean the two magnitudes?

    So I could solve it by 6 = √(k2+21) - √(55) which gives me k = √2 which is also obviously wrong.

    What exactly is my mistake? Am I messing up +/- signs somewhere?
  5. Oct 30, 2015 #4


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    Your first expansion in the original problem description was wrong, you will get a quadratic which has two solutions.

    ||u-v|| means the length of some new vector ##\vec{r}## that is the resultant from the vector operation ##\vec{u}-\vec{v}##

    My hint is that the expansion of say ##(k+1)^2## is ##(k^2+2k+1)##, not what you're doing. Or ##(k+1)(k+1)=(k+1)^2##
  6. Oct 30, 2015 #5
    Ah I haven't done any maths in so long I had even forgotten basic factorization, I also thought that (k+1)2 = k2 + 12 for some reason, what a sin of a mistake! Thank you very much my friend.
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