1. PF Contest - Win "Conquering the Physics GRE" book! Click Here to Enter
    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!

Scalar projection of b onto a (vectors)

  1. Sep 9, 2010 #1
    1. The problem statement, all variables and given/known data

    If a = <3,0,-1> find the vector b such that compaB = 2

    2. Relevant equations


    3. The attempt at a solution

    [tex]|a| =\sqrt{3^2 + 1^2} = \sqrt{10}[/tex]

    compaB = [tex]\frac{ a\cdot b}{|a|}[/tex]

    [tex]2 = \frac{3(b1) - 1(b3)}{\sqrt{10}}[/tex]

    [tex]2\sqrt{10} = 3(b1) - 1(b3)[/tex]

    I dont know what to do from here
  2. jcsd
  3. Sep 9, 2010 #2


    User Avatar
    Science Advisor
    Homework Helper

    You said, find the vector b whose projection on a is twice as long as a itself, but of course there are infinitely many such vectors. What you have shown here is that there is that a vector of the form b = (r, s, 3r - 2√10) for any real numbers r and s satisfies compab = 2.
  4. Sep 9, 2010 #3
    Sorry but I dont understand.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook