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Homework Help: Vector Problem

  1. Nov 24, 2007 #1
    [SOLVED] Vector Problem


    Suppose a and b are vectors in Vn and theta is the angle between them. If a=<1,1,...,1> and b=<1,2,...,n>, find the limiting value of theta as n-->infinity

    Relevent equations:
    a*b = |a|*|b| * cos(theta)
    theta = arccos ((a*b)/(|a|*|b|)

    I know:

    1) as n --> infinity, a*b = 1*1 + 1*2 + 1*3 + ... + 1*infinity
    Therefore a*b equals infinity as n approaches infinity

    2) as n --> infinity, |a|*|b| = sqrt(1 + 1 +...+ 1) * sqrt(1 + 4 + 9 + 16 +...+ (infinity)^2)

    I can make all kinds of assumptions from here but I don't know where to go with this problem. Just looking at what I have so far it looks as if |a|*|b| goes to infinity faster then a*b, but I dont know how to show that and I am stuck. Thanks for the help
  2. jcsd
  3. Nov 24, 2007 #2
    nono you got the scalar product all wrong :) |a|*|b| -> max(a_i*b_i) = n2 because n*n is the biggest summand in the scalar product.

    so the numerator goes to n^2.

    Now you gotta take a better approch to calculating a*b=1+2+3+...=n*(n+1)/2

    so you have arccos((n^2+n/2)/n^2)) goes to arccos(1) = 0, Pi , 2Pi,... so on :)

    i guess it goes that way doesn´t it ?
  4. Nov 24, 2007 #3

    D H

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    Find expressions for [itex]\vec a \cdot \vec b[/itex], [itex]|\vec a|[/itex], and [itex]|\vec b|[/itex] for finite n, form [itex]\cos\theta[/itex], and finally take the limit as [itex]n\to\infty[/itex].

    Good so far. How about the rest?

    Nope. Not that way.
  5. Nov 27, 2007 #4
    I agree with the [itex]\vec a \cdot \vec b[/itex] = [tex]\frac{n*(n+1)}{2}[/tex]

    I don't agree with the part where you say that [itex]|\vec a|[/itex] [itex]|\vec b|[/itex]
    = n[tex]^{2}[/tex]

    My reasoning is [itex]|\vec a|[/itex] = [tex]\sqrt{1 + 1 +...+ 1}[/tex] and as [itex]n\to\infty[/itex], the limit goes to [tex]\sqrt{n}[/tex]

    -- Also shouldn't the sum of the series: [tex]\sqrt{1 + 4 + 9 + 16 +...+ \infty^{2}[/tex] be soemthing like [tex]\sqrt{ \frac{(2n+1)*(n+1)}{2}}[/tex] as [itex]n\to\infty[/itex] ???

    I am not sure if the [tex]\sqrt{ \frac{(2n+1)*(n+1)}{2}}[/tex] is correct or not and it is the last thing holding me up. I looked through my calculus book and I can't find it in the series section. Can anyone confirm this or tell me what it auctually is? Thanks in advance.
  6. Nov 27, 2007 #5


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    Sum of the first n squares is n*(n+1)*(2n+1)/6. How did you arrive at your guess??
  7. Nov 27, 2007 #6
    I was just trying to remember from previous classes that I took. Simply a guess.
  8. Nov 27, 2007 #7
    I think that means that as [itex]n\to\infty[/itex], then [tex]\frac{\vec a \cdot \vec b
    }{|\vec a| |\vec b|}\to\frac{\sqrt{3}}{2}[/tex]

    That means that [itex]\arccos\frac{\sqrt{3}}{2} \to \frac{\pi}{6}[/itex], which is the angle between the two vectors

    Thanks for the help
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