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Homework Help: Vector equation of a line

  1. Jan 30, 2010 #1
    1. The problem statement, all variables and given/known data
    Using vector algebra, derive the tridimensional equation of a straight line going thru points a and b. In other words assuming that the tip of position vector A is located at point a, tip of position vector B is located at b and an arbitrary point r is located along the straight line and at the tip of the position vector R, the desired straight line equation is
    Rx(B-A)= AxB (all are vectors)

    2. Relevant equations

    3. The attempt at a solution
    I have no idea. Need some clues.
    I have attached the figure I came up with. I doubt if its correct. I thought the idea was to prove AxB is parallel to Rx(B-A). But not sure.
    1. The problem statement, all variables and given/known data

    2. Relevant equations

    3. The attempt at a solution

    Attached Files:

    Last edited: Jan 30, 2010
  2. jcsd
  3. Jan 30, 2010 #2


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    All three position vectors, of course, have their ends at the same point, the origin. One way of thinking about this is that the area of a parallelogram, with adjacent sides given by the vectors [itex]\vec{u}[/itex] and [itex]\vec{v}[/itex] is [itex]|\vec{u}\times\vec{v}|[/itex] and so the area of a triangle having those vectors as sides (the triangle is half the parallelogram) is [itex]\frac{1}{2}|\vec{u}\times\vec{v}|[/itex].

    [itex]|\vec{A}\times\vec{B}|[/itex] is twice the area of the triangle formed by a, b, and the origin. B-A is the vector from a to b so [itex]|\vec{R}\times(\vec{B}- \vec{A})|[/itex] is twice the area of the triangle formed by placing [itex]\vec{R}[/itex] at b. Those are different triangles but both have the line ab as base and the distance from the origin perpendicular to ab as altitude and so the same area.
  4. Jan 30, 2010 #3
    ok. That makes sense. But what is the general approach here. Are we saying both triangles have the same base. So its a straight line?
    How to go about if AxB=R x (B-A) wasn't given.
    I am trying to understand coz the problems after this keep building on this and go to vector equation of a plane etc.
  5. Feb 1, 2010 #4
  6. Feb 1, 2010 #5


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    Hi likephysics! :smile:

    (btw, in your diagram, it would be usual to put r between a and b, partly because then it is a positive combination of both a and b … and partly to save space!! :biggrin:)

    Sorry, but I don't like HallsofIvy's method, I think they're looking for a purely vectorial answer. :redface:
    No, because a x b is simply a vector out of the page … which in a plane isn't much of a distinction! :wink:

    The idea is to prove that AR is along AB, ie (r - a) x (b - a) = 0.

    Now expand. :smile:
  7. Feb 1, 2010 #6


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    If you are given points a and b, you can find the equation of the line in space that contains these points. I think that's what you're asking, so if not, please clarify your questions.

    Suppose the coordinates of A are (x0, y0, z0) and the coordinates of B are (x1, y1, z1). Suppose also that P(x, y, z) is an arbitrary point on this line. (I have switched from a and b to A and B, because I like to use capital letters for the names of points and lower case letters for vectors.)

    A vector with the same direction as the line is AB = <x1 - x0, y1 - y0, z1 - z0>.

    The vector AP = <x - x0, y - y0, z - z0> is parallel to AB, meaning that AP = tAB for some scalar t.

    The equation of an arbitrary point on the line is the sum of the vectors OA and AP, where OA is the vector from the origin to point A. Geometrically to get to point P, we go first from O to A, and then from A to P. Vectorially, OP = OA + AP = OA + tAB.

    So <x, y, z> = <x0, y0, z0> + t<x1 - x0, y1 - y0, z1 - z0>.

    This equation gives you the parametric representation of the line through points A and B.
  8. Feb 2, 2010 #7
    tinytim, that's great. I thought of proving vectors are parallel. But couldn't proceed further. I was thinking in terms of triangle base as hallsofivy had mentioned.
    Can you throw some light for the vector equation of a plane [(R-A)x(C-A)].(B-A)=0
    The equation makes sense and I know the cross product of 2 vectors on a plane dotted with the third will be zero as they are perpendicular. The cross products are parallel. So I tried using [(R-A)x(C-A)]x[(R-A)x(B-A)]=0
    But got nowhere. Any help?
  9. Feb 2, 2010 #8


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    Hi likephysics! :smile:

    For R (measured from a point not on the plane) to be in the same plane as ABC, the volume of the tetrahedron RABC must be zero, so the triple product of eg the three sides from A must be zero. :wink:

    (or, to put it another way, the plane RAB must contain RC, so the normal to RAC must be perpendicular to AB, so …

    in fact, can't you read that from [(R-A)x(C-A)].(B-A)=0? reading from the left, it says "the vector perpendicular to both AR and AC is also perpendicular to AB")
  10. Feb 2, 2010 #9
    That's the way I understand it. But how do I get there.
    You said the triple product of 3 sides from A should be 0, (R-A)x(R-B)x(R-C)=0
    I expanded using bac-cab. But got nowhere.
  11. Feb 2, 2010 #10


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    No, the scalar triple product, [(R-A)x(C-A)].(B-A) :smile:
  12. Feb 2, 2010 #11
    ok. Now the trapezoid volume formula makes sense.
    But how do I arrive at this result - [(R-A)x(C-A)].(B-A)=0.
    Am I missing something?
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