1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Given vectors v and w, prove |v*w| =< |v|*|w|

  1. Feb 5, 2009 #1
    1. The problem statement, all variables and given/known data
    Given vectors v and w, show that |v*w| =< |v|*|w|

    2. Relevant equations
    I know that |v| = sqrt((x1)^2+(x2)^2+...+(xn)^2)
    and that |w| = sqrt((y1)^2+(y2)^2+...+(yn)^2)

    also |v*w| = x1y1 + x2y2 + ... + xnyn

    We were also told
    Hint: use |v+c*w|^2
    and (v+c*w)(v+c*w)

    3. The attempt at a solution
    I am not at all sure what to do. I distributed the second expression in the hint to get
    w^2 + 2vwc + c^2*v^2, but that doesn't tell me much.

    Thanks!
     
  2. jcsd
  3. Feb 5, 2009 #2

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    (v+cw).(v+cw)>=0, right? Now pick a special value of c=(-v.w)/(w.w). It's basically a trick. It's the value of c that extremizes (v+cw).(v+cw) as a function of c.
     
  4. Feb 5, 2009 #3
    How come (v+cw).(v+cw) >=0? And after finding that value of c, I'm still not sure how that relates back to the original question. Sorry if I'm missing something obvious!
     
  5. Feb 5, 2009 #4

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    The dot product of any vector with itself is >=0. Look at your formula, it's a sum of squares. Put the special value of c in and expand it out.
     
  6. Feb 5, 2009 #5
    Hm, I think I get it. (Vectors are fairly new to me so forgive me if I misunderstand and misuse these operations.)

    knowing c = (v.w)/(w.w)
    in my expansion I had

    v.v - 2*(v.w)*c + (w.w)*c^2 >= 0

    a vector times itself must be the magnitude of itself squared, right? So knowing that and plugging in c gets me

    |v|^2 - 2(v.w)(v.w)/(w.w) + (|w|^2)*(v.w)(v.w)/((w.w)(w.w)) >= 0

    multiplying everything by |w|^2

    |w|^2*|v|^2 - 2(v.w)(v.w) + (v.w)(v.w) >= 0
    and
    |w|^2*|v|^2 >= (v.w)(v.w)

    and I suppose I can take the square root of both sides to get

    |w|*|v| >= (v.w)

    Is this correct?
     
  7. Feb 5, 2009 #6

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    Yes, but what you actually get after the square root is |w||v|>=|v.w|. When you take square roots of squares you get absolute values. Good job.
     
  8. Feb 5, 2009 #7
    Ah, that makes sense. Thank you very much!
     
  9. Feb 5, 2009 #8

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    No problem. The inequality is called Cauchy-Schwarz, if you didn't already know that.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Given vectors v and w, prove |v*w| =< |v|*|w|
Loading...