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Calc 3, vectors and length

  1. Sep 11, 2010 #1
    1. The problem statement, all variables and given/known data

    Find a vector that has the same direction as the given vector but has length 6.
    ‹4, 2, -2›

    2. Relevant equations

    L = ((x-i)^2+(y-j)^2+(z-k)^2)^.5


    3. The attempt at a solution

    For some reason I am just stuck. I don't know how to start. I assume L = ((x-i)^2+(y-j)^2+(z-k)^2)^.5 becomes L = (x^2+y^2+z^2)^.5 because of the 2 points (0,0,0) and (x,y,z) but how does one find something with the length of 6 in 3 dimensions? All I really have is this

    36=x^2+y^2+z^2

    I guess I could eliminate z by finding an equation for a line on points (0,0,0) and (4,2,-2). I found that equation to be z=x-3y but after plugging that in it seems that I have 2 independent variables(x and y). How do I deal with this?



    Thanks!
     
  2. jcsd
  3. Sep 11, 2010 #2

    cronxeh

    User Avatar
    Gold Member

    Ok so lets say a is scalar and v is your vector. The a*v has same direction but different magnitude than a, and so the magnitude of a*v is sqrt((ax)^2 + (ay)^2 + (az)^2)

    sqrt((4a)^2 + (2a)^2 + (-2a)^2) = 6, and so a=sqrt(3/2). so a*v should equal to 6

    Please refer to the vector calculus notes here http://tutorial.math.lamar.edu/Classes/CalcII/VectorArithmetic.aspx you need to know difference between addition, substraction, and multiplication of vectors and what it looks like
     
  4. Sep 11, 2010 #3
    Ah yes yes yes. Ok now I see how you go about this. Simply find a scalar to multiply the original vector by. I found out that I multiply by (3/2)^.5 and the new vector is <4(3/2)^.5,2(3/2)^.5,-2(3/2)^.5> and this has a length 6.

    Thanks!
     
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