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Linear algebra normalising a vector?

  1. Oct 29, 2009 #1
    1. The problem statement, all variables and given/known data

    Let the vector V = (x1,y1,z1)

    x1' = x1/sqrt(x1^2 + y1^2 + z1^2)
    y1' = x1/sqrt(x1^2 + y1^2 + z1^2)
    z1' = x1/sqrt(x1^2 + y1^2 + z1^2)

    What do you call x1', y1' and z1' in mathematics terms?

    is x1' the norm of x1 and y1' the norm of y1?

    Thanx



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Oct 29, 2009 #2

    HallsofIvy

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    No. x1', y1', and z1' are the x, y, and z components, respectively, of the normalized vector- the unit vector parallel to V.
     
  4. Oct 31, 2009 #3

    LCKurtz

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    If [itex]\vec V = \langle a,b,c \rangle[/itex] is a vector and
    [tex]|\vec V| = \sqrt{a^2 + b^2 + c^2}[/tex]
    is its length, or norm, then the vector
    [tex]\hat V = \frac 1 {|\vec V|}\ \vec V = \langle \frac a {\sqrt{a^2 + b^2 + c^2}},\frac b {\sqrt{a^2 + b^2 + c^2}},\frac c {\sqrt{a^2 + b^2 + c^2}}\rangle[/tex]
    is called a unit vector. The components of this unit vector are sometimes called the direction cosines of [itex]\vec V[/itex].
     
  5. Oct 31, 2009 #4

    HallsofIvy

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    I assume you mean
    y1' = y1/sqrt(x1^2 + y1^2 + z1^2)
    z1' = z1/sqrt(x1^2 + y1^2 + z1^2)
    rather than having "x1" as every numerator.

     
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