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Another simple vector question

  1. Jan 15, 2009 #1
    1. The problem statement, all variables and given/known data
    Find a unit vector from the point P = (1, 2) and toward the point Q = (4, 6).


    2. Relevant equations



    3. The attempt at a solution

    The answer at the back of the book says 3/5i + 4/5j, shouldn't it just be 3i + 4j?
     
  2. jcsd
  3. Jan 15, 2009 #2
    A unit vector has length 1.
     
  4. Jan 15, 2009 #3
    so what do you mean? If I am asked to find another vector that has the same direction with length 10, how do I do that? if I changed the question to (3,6) instead of (4,6) what would be the answer then
     
  5. Jan 15, 2009 #4
    Multiplying a vector by a nonzero scalar does not change its direction. If the length of a nonzero vector v is a, then (1/a)v has length 1.

    The following is then an exercise for you.
     
  6. Jan 15, 2009 #5
    ok so if that's so then finding another vector that has the same direction as 3/5i + 4/5j, will be something like

    1/10 (3/5i + 4/5j) am I right?

    the answer at the back of the book is 6i + 8j, I don't where that came from...
     
    Last edited: Jan 15, 2009
  7. Jan 15, 2009 #6
    Come on, Equi, what is the length of that vector?

    You have [tex]\textstyle\mathbf{v} = (\frac{3}{5}, \frac{4}{5})[/tex], which has length

    [tex].\; \; \; \; \; \textstyle ||\mathbf{v}|| = \left\|\frac{1}{5}(3, 4)\right\| = \frac{1}{5}||(3, 4)|| = \frac{1}{5}\sqrt{3^2+4^2} = \frac{1}{5}5 = 1[/tex].

    Hence [tex]\textstyle 10\mathbf{v} = 10(\frac{3}{5}, \frac{4}{5}) = (6,8)[/tex] has length 10 since

    [tex].\; \; \; \; \; ||10\mathbf{v}|| = 10||\mathbf{v}|| = 10[/tex].
     
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