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Two vectors in 3D, always form a plane?

  1. Nov 5, 2007 #1
    If you have two vectors in space (so 3d) and they don't intersect, am I right in saying that it doesn't make sense to say they form a plane? Because there's no single plain we could be talking about. Whereas if they do intersect, they will always form a plane where those two lines are part of right?

    And is it right to say if you have 3 vectors in space, a, b, and c, and three constants p, q and r, if you can write pa + qb = rc then they all lie on a plane? Thanks.
     
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  3. Nov 5, 2007 #2

    HallsofIvy

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    If two lines do not intersect, they might be parallel- and still in a single plane. Vectors, however have finite length and might just be too short to cross.

    Yes, the three vectors are not independent and so lie in a plane.
     
  4. Nov 5, 2007 #3

    JasonRox

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    Let v and w be two vectors. Let v = w, hence they both intersect. Do they form a plane?
     
  5. Nov 5, 2007 #4
    All vectors intersect at the origin.
     
  6. Nov 5, 2007 #5

    JasonRox

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    Are you positive?
     
  7. Nov 5, 2007 #6
    You can have a vector as long as it begins from a point and end at a another point right? Thats what I get from vector formulas. It is not required for a vector to start at the an origin right? Is that what you mean?
     
  8. Nov 5, 2007 #7

    JasonRox

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    You're right. That's what meant. :smile:
     
  9. Nov 6, 2007 #8
    In most pure texts vectors are considered to be starting from the origin. We're dancing around the definitions of directed line segments and vectors. I've seen some texts go as far as saying that directed lines are vectors that are located at arbitrary points in space.

    How do you draw vectors anyways? Given the coordinates of some vector you plot the point and then connect the origin to this point. Hence, every vector intersects at the origin. I was just playing devil's advocate. Had he said line segment I wouldn't have said anything.
     
    Last edited: Nov 6, 2007
  10. Nov 6, 2007 #9

    JasonRox

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    In it's pure form, vectors don't all connect at the origin. I don't see where they define vectors as anything else but that.
     
  11. Nov 6, 2007 #10
    Not in any text I've read. The type of vector you're referring to is usually called a position vector (it defines the position of a point in a coordinate system), to differentiate it from a general vector.
     
  12. Nov 8, 2007 #11
    I don't get it, if they intersect at a point and after that point they will be travelling away so they must form a plane. Right?
     
  13. Nov 8, 2007 #12

    HallsofIvy

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    Yes, you don't get it: he said "Let v= w". The don't "travel away" (from each other).

    Now, let's go back to an issue of terminology- two vectors don't "form a plane" to begin with! They may well "span" a plane or, if the are dependent, only a single line.
     
  14. Nov 10, 2007 #13
    Ok Ithink I understand a bit more now, I was getting confused with terminology etc.

    But one final question. If you have a plane with equation say px + qy + rz = d where p, q and r are some numbers, why is it that the vector k(pi + qj + rk) (where k is a number and i, j, k are unit vectors in the usual directions) is perpendicular to the plane? Is there any easy way to see/understand why? It isn't obvious to me why but I feel as if it should be.
     
  15. Nov 10, 2007 #14

    HallsofIvy

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    I can see why you would be confused! Your notation is confusing me. I think anyone would be confused when you use "k" to rpresent both a number and a unit vector!

    First, pick a fixed point in the plane px+ qy+ rz= d. Taking x= y= 0, z= d/r will suffice. Let (x, y, z) represent any point in that plane. Then [itex](x-0)\vec{i}+ (y-0)\vec{j}+ (z- d/r)\vec{k}[/itex] represents an arbitrary vector in that plane.

    The dot product of [itex](x-0)\vec{i}+ (y-0)\vec{j}+ (z- d/r)\vec{k}[/itex] with [itex]ap\vec{i}+ aq\vec{j}+ ar\vec{k}[/itex], for a any number, is apx+ aqy+ ar(z-d/r)= a(px+ qy+ rz- d) The term in parentheses is 0 because (x, y, z) satifies the equation of the plane so the dot product is 0 and the two vectors are pependicular. Since (x, y, z) could be any point in the plane, [itex](x-0)\vec{i}+ (y-0)\vec{j}+ (z- d/r)\vec{k}[/itex] could be any vector in the plane, [itex]ap\vec{i}+ aq\vec{j}+ qr\vec{k}[/itex] is perpendicular to every vector in the plane an so to the plane itself.
     
  16. Nov 10, 2007 #15
    lol sorry about the notation. I don't know what I was thinking when I wrote it. But thanks for the post it makes perfect sense!!
     
  17. Dec 18, 2007 #16
    One question. If you have this:

    |PQ|^2 = (b-a) • (c × d) then why is |PQ| not the square root of the RHS? If |PQ| = d then it's easier to see why it should be. But apparently it isn't.

    b, a, c and d are vectors. PQ is meant to have an arrow above it (so it's a vector too).
     
  18. Dec 19, 2007 #17
    Vector is an equivalence class. So you can pick a representative from any place in a space.
    Two vectors don't form a space, you need a whole bunch of their linear combinations to form a plane unless they are equal or opposite in direction (zerovector has whichever direction). Vectors don't intersect. Lines do.
    So you are right in saying: it doesn't make sense... .

    But with a quite lots of good will i would say this is just what you meant.

    (0) Startingpoint and endpoint define the vector representative. (so there's nothing between!)
    (1) We are talkin about nonzero vectors u,v in R^n.
    (2) Fix a point O in space.
    (3) take the representatives, which start from that point O.
    (4) Be sure that O is the only point in common.
    Then probably {au + bv| a,b from R} is a "plane"
     
    Last edited: Dec 19, 2007
  19. Dec 20, 2007 #18
    What is the context of this equation? Without context, I don't see any reason for what you're saying to be false.
     
  20. Dec 29, 2007 #19
    Hence we can define a flat plane by a single vector. In physics classes I have used the unit normal vector to find the stresses acting on a plane. Of course, to do this, I also needed a tensor - "the stress tensor".
     
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