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1 vectors and 1 limits problem

  1. Nov 14, 2011 #1
    Hi guys,

    Starting revision for my exams and I can't seem to solve these problems. ):

    1. Given that three non zero vectors A, B and C are perpendicular to each other and that the vector R = pA + qB + rC is a zero vector, show that p = q = r = 0

    2. In another question, given that f(x) is differenciatable at x = 0 and [itex]\frac{Lim}{x->0}[/itex] [itex]\frac{f(x) - f(kx)}{x}[/itex] = a, wjere a and k are constants.

    b. Show that f'(0) = a/1-k

    What I did was take the defination of differenciation:
    [itex]\frac{Lim}{h->0}[/itex] [itex]\frac{f(x+h) - f(x)}{h}[/itex]

    and I let x = kx, and h = (1-k)x,
    [itex]\frac{Lim}{(1-k)x->0}[/itex] [itex]\frac{f(x) - f(kx)}{(1-k)x}[/itex]

    and I did this (assuming that I can write [itex]\frac{Lim}{(1-k)x->0}[/itex] as [itex]\frac{Lim}{x->0}[/itex]

    1/(1-k)[itex]\frac{Lim}{x->0}[/itex] [itex]\frac{f(x) - f(kx)}{x}[/itex] = a/1-k

    Is this correct?

    Thanks in advance for your help guys!

    *Edit* Just realized that I misread the forum title and posted in the wrong one. ):
    Last edited: Nov 14, 2011
  2. jcsd
  3. Nov 14, 2011 #2
    What can you tell us about the inner products R.A, R.B and R.C??


    No, you can't do that. You took

    [itex]\frac{Lim}{h->0}[/itex] [itex]\frac{f(x+h) - f(x)}{h}[/itex]

    but the x in there is a constant!! But later

    [itex]\frac{Lim}{(1-k)x->0}[/itex] [itex]\frac{f(x) - f(kx)}{(1-k)x}[/itex]

    your x is suddenly a limit variable.

    I'll illustrate it by taking x=2, then you write

    [itex]\frac{Lim}{h->0}[/itex] [itex]\frac{f(2+h) - f(2)}{h}[/itex]

    this is perfectly valid and it equals f'(2). But then you write

    [itex]\frac{Lim}{(1-k)2->0}[/itex] [itex]\frac{f(2) - f(k2)}{(1-k)2}[/itex]

    which obviously makes no sense. (1-k)2 -> 0 makes no sense!!

    I'll give you a hint on how to solve it:

    &= \lim_{h\rightarrow 0}{\frac{f(h)-f(0)}{h}}\\
    &= \lim_{h\rightarrow 0}{ \frac{\frac{f(h)-f(kh)}{h}}{ \frac{f(h)-f(kh)}{f(h)-f(0)}}}\\
    &= \lim_{h\rightarrow 0}{ \frac{\frac{f(h)-f(kh)}{h}}{ \frac{(f(h)-f(0)) - (f(kh)-f(0))}{f(h)-f(0)}}}\\
  4. Nov 14, 2011 #3

    R.R = R.A + R.B + R.C
    Since R.R = 0,
    R= p = q = r?

    How did you get from the first to the second step? )(Actually I'm lost from that step onwards also)

    Anyway thanks for your help so far micromass!
  5. Nov 15, 2011 #4


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    R.A = pA.A Since R=0 then R.A=0, however A.A > 0, therefore p=0. Similarly for q and r.
  6. Nov 16, 2011 #5


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    Why do you say "R.R= 0"?? And "R= p= q= r" makes no sense. R is a vector while p, q, and r are numbers.
  7. Nov 16, 2011 #6


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    R=0, therefore R.R=0

    The second line is a typo - I meant p=q=r=0.
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