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Homework Help: 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?


    2. Relevant equations

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


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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


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    Gold Member

    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


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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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