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Unit Vector

  1. Nov 17, 2007 #1
    Vector Directions

    Determine if the vectors u= (1,2,-3) and v = (4,8,-9) have the same direction.

    I am unsure how to determine if they are in the same direction. How would I do this?

    I can see that 4u=v, but doesn't know if that would make them in the same direction. Any help?
     
  2. jcsd
  3. Nov 17, 2007 #2
    4u is not equal to v, the z coordinate does not match. If its a typo, and it DOES match, then as the quantities have the same sign, they're in the same direction. If the signs dont match, then they are in opposite directions.
     
  4. Nov 18, 2007 #3

    rock.freak667

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    If you can write one as a scalar multiple of the other then they are parallel. if the scalar is +ve then they are in the same direction
     
  5. Nov 18, 2007 #4
    I am sorry rock.freak667, what do you mean by this

     
  6. Nov 18, 2007 #5
    What if u factored out the second vector? What would that give you? What does it mean?
     
  7. Nov 18, 2007 #6

    CompuChip

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    +ve is a shorthand for positive.
    So if [itex]\vec u = n \vec v[/itex]
    for some number n, and n is a positive number, then they are in the same direction (if you'd draw them as arrows starting at the same point, they would lie on the same line and point in the same direction). If the number n is negative (they would lie on the same line, but they would point in opposite directions) we say they are opposite.
     
  8. Nov 19, 2007 #7
    This is not a scalar multiple, but the signs are all the same, so based off of what I have read than this would be heading in the same direction. Is that correct?
     
  9. Nov 19, 2007 #8
    If the signs are the same than does that mean it is in the same direction? Does scalar multiplication matter for direction?
     
  10. Nov 19, 2007 #9

    CompuChip

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    I have the feeling you don't really get the geometric picture behind it. Did you try drawing them on a piece of paper (preferably the kind with small squares)?

    Did you also try drawing several sets of vectors lying along the same line, and then written down their coordinates?
     
  11. Nov 28, 2007 #10
    I am not quite sure if I have answered this correctly so I was trying to see my answer is correct.

    Question:
    Determine if the vectors u=(1,2,-3) and v=(4,8,-9) have the same direction.

    Answer:
    Vectors u and v are in the same direction. This is due to the signs of the coordinates being the same.
     
  12. Nov 28, 2007 #11

    robphy

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    Consider a simpler question in the plane:
    Do vectors u=(1,1) and v=(1,2) have the same direction?
    (If you gave them as directions for two airplanes, will they fly over the same path?)
    Plot on graph paper if necessary.


    (There is a sense in which they sort-of (but not strictly) point in the same direction... involving the dot-product.
    But I don't think that your question is looking for that.)

    (To make connection with the title of this thread....
    two nonzero vectors have the same direction if their respective unit-vectors (the vectors scaled-down by their magnitudes) are equal...
    that is... if [tex]\frac{\vec u}{|\vec u|} = \frac{\vec v}{|\vec v|}[/tex] ).
     
    Last edited: Nov 28, 2007
  13. Nov 28, 2007 #12

    CompuChip

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    This is the same question you asked a while ago. So I guess you didn't really understand the explanations there. I think you should help us find the problem. For example, do you know what a vector is? Do you have any geometric feeling for it? Do you understand why it makes sense to speak about direction? I think the problem is somewhere in this area, otherwise you'd probably have no trouble drawing them and answering the question.
    What level in maths are you on? Did you ever encounter linear algebra / vectors before?
     
  14. Nov 28, 2007 #13
    I am in Matrix Algebra and no I have never had Linear Algebra. I honestly don't really understand direction. To determine a length of a vector I would use this equation:

    [tex]||u||=\sqrt{1^2+2^2-3^2}=\sqrt{14}[/tex]
    [tex]||v||=\sqrt{4^2+8^2-9^2}=\sqrt{161}[/tex]

    So if I am understanding correctly these would not be in the same direction because they don't have the same length
     
  15. Nov 28, 2007 #14

    HallsofIvy

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    No, "direction" and "length" are not the same thing! In fact a vector has exactly those two properties- direction and length.

    A vector 4 feet long pointing east would have the same length as a vector 4 feet long pointing south but different directions. (If you "don't really understand direction" to the extent that you don't know east from south, I give up!)

    A vector 4 feet long pointing east would have a different length from a vector 6 feet long pointing east but the same direction.
     
  16. Nov 28, 2007 #15
    I get the concept of direction, but I am unsure if I am supposed to follow a certain equation to show this. My textbook is very unclear and I am very new to vectors. I am just trying to make sure I am doing everything correctly

    Based off the points it would be headed in the same directions if I plotted this.

    Should I use this:[tex]\frac{\vec u}{|\vec u|} = \frac{\vec v}{|\vec v|}[/tex]

    to prove that.
     
  17. Nov 29, 2007 #16

    Chris Hillman

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    In a three dimensional euclidean inner product space? You didn't really mean to write minus signs there, did you?

    Question: do [itex](1,2), \; (3,6)[/itex] point in the same direction? That is, as vectors are they scalar multiples of each other? Yes, since [itex](3,6) = 3 \, (1,2)[/itex]. Do [itex](1,2), \; (2,1)[/itex] point in the same direction? No, because (a) draw a picture (b) for no real number is it true that [itex](1,2) = \lambda \, (2,1)[/itex], so they are not scalar multiples of each other. Because if they were we would have [itex]1 = 2\, \lambda, \; 2 1= \lambda[/itex] or [itex]1=4[/itex], which isn't ([itex]1, \, 2, \, 3, \, 4 \neq 1[/itex]) true :wink:

    Caveat one: the algebraic notion I am hinting at is linear dependence of two vectors, while the geometic notion appears to distinguish between positive and negative multiples. It turns out that the first notion is so convenient, mathematically speaking, that no-one pays attention to the second notion!

    Caveat two: I also hinted that it matters what the underlying field is in our vector space, but in this case it seems safe to assume it is R, the field of real numbers.

    Sigh... I'm not helping, am I? :rolleyes:

    Title and author? Not that I take your word for its alleged inadequacies, but I am curious.
     
    Last edited: Nov 29, 2007
  18. Nov 29, 2007 #17
     
  19. Nov 29, 2007 #18

    robphy

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  20. Nov 29, 2007 #19

    HallsofIvy

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    Yes, if [tex]\frac{\vec u}{|\vec u|} = \frac{\vec v}{|\vec v|}[/tex]
    then two vectors are "in the same direction".

    Notice that that also implies [tex]\vec{u}= \frac{|\vec{u}|}{|\vec{v}|} \vec{v}[/itex]. That is, two vectors are in the same direction (parallel) if and only if one is a positive multiple of the other. That's often easier to show since it doesn't involve messing with square roots!

    (If one is a negative multiple of the other, they point in opposite directions (anti-parallel).)
     
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