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Why use a Direction Vector?

  1. Apr 5, 2012 #1
    When constructing the vector equation for the position vector r (finishing at point P) in the drawing below. Why is it necessary to use a "direction vector", v ? Couldn't one just use vector a and have done with it??

    Why do we need to define this new vector r??What is the point.


    I would appreciate it if you guys could reply in simple laymans english.

    I thank you in advance.

  2. jcsd
  3. Apr 5, 2012 #2


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    Science Advisor

    You need to clarify what is known and what you are trying to find.
  4. Apr 5, 2012 #3
    Please expand on this. Can you give me an explicit reason for its purpose?
  5. Apr 5, 2012 #4


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    Staff: Mentor

    mathman is saying that we do not understand your question. Are the vectors "v" and "a" meant to have the same magnitude and direction? And what is the significance of "The Line"?
  6. Apr 5, 2012 #5
    The Line is just a line in 3D space that I wish to find the equation of. the line may represent the direction of the greatest rate of increase of the scalar field. e.g the direction of the greatest increase in electric field.

    Vectors a and v are parallel. Therefore and there is a parameter t such that:



  7. Apr 7, 2012 #6
    Two vectors are parallel if they are multiples of each other

    ie [itex]\vec{a}[/itex]= <1,2,3> is parallel to [itex]\vec{b}[/itex]=<2,4,6> and [itex]\vec{c}[/itex]=<3,6,9> and so on

    So the parameter t in [itex]\vec{a}[/itex] = [itex]\vec{v}[/itex]t just allows the vector [itex]\vec{r}[/itex] to move along the line, where [itex]\vec{r}[/itex] is the position vector of a point on the line.

    When you change the value of t, you change your position the line.
  8. Apr 7, 2012 #7

    Stephen Tashi

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    A vector whose tail (the non-arrow end) doesn't begin at the origin ( a so-called "free vector") has a head that doesn't inform us of the components of the vector. For example if your position at time t0 is (4,5) and your position at time t1 is (6,9) then the algebraic calculation of the change in position is (6-4,9-5) = (2,4). But if you look at the vector whose tail is at (4,5) and whose head is at (6,9) the head of that vector obviously can't be interpreted as indicating (2,4). To have a geometric picture of a vector whose head is (2,4), you must draw a "bound vector". This amounts to "moving" the free vector so its tail is (0,0).

    The reason for using bound vectors is to have a clear visualization of the magnitude of the components of the vector.

    You can plot a curve showing the 2D path of an object and you can indicate time by making tick marks on that curve. The object also has a vector velocity. If you want to visualize that velocity clearly, you need to draw the path of a vector whose head informs you of the velocity. To do that, you must plot the value of a "bound" vector indicating the velocity.
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