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Vector with the same direction but a different magnitude

  1. Feb 1, 2015 #1
    1. The problem statement, all variables and given/known data
    Find a vector that has the same direction as <-2, 4, 2> but has length 6.

    2. Relevant equations
    Length of vector in 3D = |v| = (x2 + y2 + z2)0.5

    3. The attempt at a solution
    I can see that the length of the given vector is (-2)2 + 42 + 22 = 24
    36 is (1.5)24, so I tried multiplying each component of the vector by the scalar 1.5. However, this doesn't give me the desired length of six (rather, it gives me the square root of 54). I had a feeling this method would be faulty, but nothing else comes to mind.
     
  2. jcsd
  3. Feb 1, 2015 #2

    jedishrfu

    Staff: Mentor

    Compare lengths not the squares of lengths to get the right factor.
     
  4. Feb 1, 2015 #3

    Fredrik

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Not according to the formula you included under "relevant equations".
     
  5. Feb 1, 2015 #4
    I tried it again with the length of the desired vector = (F(-2)2 + F42 + F22)0.5 Where F = 1.5, and the answer seems to be correct...
     
  6. Feb 1, 2015 #5
    I just noticed my error. In my original attempt, (F * IJK)2 for each component I did when it should have been F(IJK2)
     
  7. Feb 1, 2015 #6

    Mark44

    Staff: Mentor

    Perhaps you understand what you mean, but this is very weird notation. If u = <a, b, c> and v = 2<a, b, c>, then |v| = 2|u|
     
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