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Parallel unit vector ?

  1. Sep 6, 2009 #1
    What PARALLEL UNIT VECTORS are and Why i gotta divide a vector with its value to get a PARALLEL UNIT VECTOR ?
    This way
    [tex]\hat{a}[/tex] = [tex]\stackrel{\rightarrow}{A}[/tex]
    /|[tex]\stackrel{\rightarrow}{A}[/tex]
    | ?

    I am sure that the answere would be very easy but i am such a . . . ! Please help!
     
  2. jcsd
  3. Sep 6, 2009 #2
    First, we usually don't use the word "value" to denote the magnitude or length of a vector. Just use the phrase "magnitude of a vector" and people will know what you mean.

    So your question is why does dividing a vector [tex]\vec{V}[/tex] by its magnitude [tex]|\vec V|[/tex] produce a unit vector [tex]\hat V[/tex] parallel to the original vector?

    First understand the parallel part. Whenever your multiply or divide a vector by a number you are simply scaling the vector leaving its direction unchanged. Thus a two-dimensional vector with components (6, 8) is parallel to the vector (3, 4) and has twice the magnitude.

    Next understand the unit part. Consider the vector with components (3, 4). Its magnitude is
    [tex](3^2 + 4^2)^{1/2} = 5[/tex]. Hence the vector divided by its magnitude is (3/5, 4/5). The claim is that this vector has unit magnitude. Just check it: [tex] [(3/5)^2 + (4/5)^2]^{1/2} = [9/25 + 16/25]^{1/2} = (25/25)^{1/2} = 1.[/tex]
     
  4. Sep 6, 2009 #3
    For two vectors to be "parallel" just means they point in the same direction. That's all.

    Multiplying by a scalar (such as the magnitude of the vector) does not change the vector's direction, only its magnitude. So there you go.
     
  5. Sep 6, 2009 #4
    Yes, if P (3,4) then[tex]\stackrel{\rightarrow}{OP}[/tex] = 3[tex]\hat{i}[/tex] + 4[tex]\hat{j}[/tex]
    so parallel unit vector of [tex]\stackrel{\rightarrow}{OP}[/tex], would be [tex]\stackrel{\rightarrow}{OR}[/tex] = [tex]\stackrel{\rightarrow}{OP}[/tex]/|[tex]\stackrel{\rightarrow}{OP}[/tex]|
    =(3[tex]\hat{i}[/tex] + 4[tex]\hat{j}[/tex])/[tex]\sqrt[]{3^2 + 4^2}[/tex]
    =(3[tex]\hat{i}[/tex] + 4[tex]\hat{j}[/tex])/5
    =3/5[tex]\hat{i}[/tex] + 4/5[tex]\hat{j}[/tex]

    equation of OP is (x - 3) / (3 - 0) = (y - 4) / (4 - 0)
    =>4x = 3y
    equation of OR is (x - 3/5) / (3/5 - 0) = (y - 4/5) / (4/5 - 0)
    =>20x = 15y
    =>4x = 3y
    so OP and OR is parallel without any doubt but why we are dividing with its magnitude ? If we divide with 2, 3, 4 or any other number that time also there will be a parallel vector visible, not?
     
  6. Sep 6, 2009 #5
    Thanks thanks thanks i have solved the problem !
    When we divide one vector with its magnitude we get a UNIT VECTOR so [tex]\hat{n}[/tex] = [tex]\stackrel{\rightarrow}{A}[/tex]/|[tex]\stackrel{\rightarrow}{A}[/tex]| and this unit vector must be parallel to [tex]\stackrel{\rightarrow}{A}[/tex].

    So we divide one vector with its magnitude is to calculate a unit vector and automatically they are parallel!
     
  7. Sep 6, 2009 #6
    is it possible to prove algebraically that dividing any vector by its magnitude results in a parallel unit vector?
     
  8. Sep 6, 2009 #7
    what does it mean for two vectors to be parallel? one of them can be written as a multible of the other. or in other words, they are linearly dependent, so there exists a nontrivial linear combination equal to zero.

    let a be such a vector, and let u be the unit vector acquired as follows [tex]u=\frac{a}{|a|}[/tex]

    then it is clear that:


    na+mu=0, where n and m are different from zero.

    or a=ku, with k=|a|.

    so a and u are linearly dependent, or parallel.

    it is also clear that u is the unit vector, since |u|=1.
     
  9. Sep 6, 2009 #8

    LCKurtz

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    Assuming you know that for scalar a and vector A that

    | kA | = |k| |A|

    apply that to:

    ( 1/|A|)A
     
  10. Sep 7, 2009 #9

    rock.freak667

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    Homework Helper

    Any vector parallel to xi+yj can be written as α(xi+yj)=αxi+αyj where α= scalar

    Unit vector means the magnitude of the vector you want is 1

    so


    [tex]\left | \alpha x \hat{i}+ \alpha \hat{j} \right |= \sqrt{(\alpha x)^2 + (\alpha y)^2}=1[/tex]


    So your scalar α works out to be equal to

    [tex]\alpha = \frac{1}{\sqrt{x^2+y^2}}[/tex]

    this means to obtain a unit vector, you divide the vector by its magnitude
     
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