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Perpendicular Lines in 3 Space

  1. Aug 16, 2006 #1
    i am given parametric equations of 2 lines L1 and L2 and i am asked to find the vector, parametric and symmetric equation of a line that intersects these 2 lines at 90 degrees so a line that is perpendicular to those 2 lines. how do i go about this? at first i tried converting the parametric equation to a vector equation to get the value of vector a cause p vector = a vector + t(b vector) and then cross multiplying the two a vectors. but then i realized that the a vectors dont have the same direction and that the line and the a vector are totally different things so ... how do i go about this?
     
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  3. Aug 16, 2006 #2

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    If your vector equation will read [itex]\vec r(t) = \vec a+ \vec b t [/itex], you can take [itex]\vec a[/tex] as any point on the line and [itex]\vec b[/tex] as the difference between any two distinct points on the line.
     
  4. Aug 16, 2006 #3
    i know how to get values of vector b but i don't understand how to do the calculations. i dont know how to find the a line that intersects 2 other lines and at a 90 degree angle...
     
  5. Aug 16, 2006 #4

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    Oh, I'm sorry, I misread the question. You want to find the corresponding vectors [itex]\vec b[/itex] for these two lines and then take their cross product. As for a, it's unclear where they want you to put this line, but if the lines interect and they want it to pass through this point, just take that point as your [itex]\vec a[/itex].
     
  6. Aug 16, 2006 #5
    okay if i do a cross product of the b vectors i will get a line that is perpendicular to both of the line but only at the origin. the line does not have to necessarily come across the origin right? so this would be the like B value of my 3rd line right? now then they want me to put the line so it intersects the given 2 lines at a 90 degree angle. so how do i do that?
     
  7. Aug 16, 2006 #6

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    I see, so you want the line to intersect both given lines, at two different points if these lines don't themselves intersect, and in addition for it to be perpendicular to both lines. Sorry I repeatedly didn't read the question properly. I've been a little distracted.

    So, one thing you could do is look at the set of all vectors between the two lines. That is, if the equation for L1 is [itex] \vec a_1 + \vec b_1 t_1[/itex] and similarly for L2, then look at the set [itex] \vec a_1 -\vec a_2+ \vec b_1 t_1-\vec b_2 t_2[/itex]. Now you want to solve for [itex]t_1[/itex] and [itex]t_2[/itex] at which this vector is perpendicular to both [itex] \vec b_1 [/itex] and [itex] \vec b_2 [/itex], that is [itex] \vec b_1 \cdot (\vec a_1 -\vec a_2+ \vec b_1 t_1-\vec b_2 t_2) =\vec b_2 \cdot (\vec a_1 -\vec a_2+ \vec b_1 t_1-\vec b_2 t_2) =0[/itex]. Then you can just plug in these values to get [itex] \vec a[/itex] (the point on L1 at t1 or L2 ar t2) and [itex] \vec b[/itex] (the diference between these two points).
     
    Last edited: Aug 16, 2006
  8. Aug 16, 2006 #7
    its aliright but i still have one more question. i do not understand how you got this:
    [itex] \vec a_1 -\vec a_2+ \vec b_1 t_1-\vec b_2 t_2[/itex]
     
  9. Aug 17, 2006 #8

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    I just subtracted the vector from the point on L1 at t1 to the point on L2 at t2. This is then the vector between the two points, or equivalently, the b vector for a line that intersects both points.
     
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