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Vector questions

  1. Jun 13, 2006 #1
    1. Consider two lines L1: r = (2,0,0) + t(1,2,−1) and L2: r =(3,2,3) + s(p,q,1)
    Find a relationship between p and q [independent of s and t] that ensures that L1 and L2 intersect.

    i proceeded like any other interesection of line question but got stuck when i got single equations with 3 unknowns (including p and q)


    2. The position vectors of the points A and B, with respect to the origin O, are 2i−3j+3k and 5i + j+ck respectively, where c is a constant. The point C is such that OABC is a rectangle.
    (a) Find the value of c
    (b) Find the point C

    how do i find c? if i had the exact coordinates of B, im pretty sure developing an equation like OA = CB would give be the point C (which is part b))


    3. In the following system of equations, k is a real number.
    x−2y+z=4
    x−y−z=3
    x+y+kz=1

    (a) Determine the value(s) of k for which the system of equations has
    (i) exactly one solution
    (ii) infinitely many solutions

    should i solve this using a standard matrix? what happens when i get down to x = something, y = something, z = something/k. i don't know really what to do differently for i) and ii), what would each part entail?


    i know i haven't shown a tremendous amount of work but i'm not the greatest thinker and i really have a deficiency of knowledge in this area. thanks for any help given!
     
  2. jcsd
  3. Jun 13, 2006 #2
    1. for this one, note that 2 lines will intersect iff their direction vectors are not parallel, i.e. not scalar multiples of each other. so do you know of a condition you can put on (p,q,1) such that it is not a multiple of (1,2,-1)?

    3. when you put it into matrix form and reduce it, which values of k will give you a row of zeros? for those values, there will be infinitely many solutions. if there are no rows of zeros there will be one solution.
     
  4. Jun 14, 2006 #3
    hmm ill start working with what you said, thanks by the way. any further advice is welcome too!

    i also have a question about the distance between two skew lines. geometrically, what is the relationship between the line that represents the distance and the skew lines. the distance line is perpendicular to both no?
     
  5. Jun 14, 2006 #4
    For Q2, use the fact that adjacent sides of a rectangle are perpendicular to each other.
     
  6. Jun 14, 2006 #5

    HallsofIvy

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    No, that's not true. In 3 dimensions two lines may be "skew", neither parallel nor intersecting. A simple example is the x axis, r= t(1, 0, 0) and the line r= (1, 1, 0)+ t(0, 0, 1), through (1, 1, 0) parallel to the z-axis.

    masterofthewave124, you started correctly. in order for the lines, r = (2,0,0) + t(1,2,−1) and L2: r =(3,2,3) + s(p,q,1) to intesect, you must have x= 2+ t= 3+ ps, y= 2t= 2+ qs, z= -t= 3+ s for some t and s. Multiply the last, z, equation by 2 and add to the second, y, equation to eliminate t: 0= 8+ (2+ q)s so s= -8/(2+q). Now use the second equation to determine t: -t= 3+ s= 3- 8/(2+q)= (3q- 2)/(2+ q) so t= (2- 3q)/(2+ q). Finally, those two values must satisfy the first, x, equation:
    2+ t= 3+ ps or 2+ (2-3q)/(2+q)= 3- 8p/(2+q).
    (6- q)/(2+q)= (6+ 3q- 8p)/(2+q). In order for that to be true, and the two lines intersect, we must have 6- q= 6+3q- 8p or q= 2p (and, of course, q not equal to -2).

    As pizzasky said, to be a rectangle, two sides, here the vectors 2i- 3j+ 3k and 5i+ j+ ck, must be perpendicular. Set the dot product of those vectors equal to 0 and solve for c.
    I'm not sure why you say "if i had the exact coordinates of B"- you do. Being told that the "position vector of B, with respect ot the origin O, is 5i+ j+ ck" tells you that the "exact coordinates" of B are (5, 1, c) and you just found c.


    If you are familiar with matrix methods, sure, go ahead and row-reduce the augmented matrix. For what value of k is the last row (excluding the "augmented part") not 0? What does that tell you? For what value of k are all the terms of the last row, (including the "augmented part") equal to 0? What does that tell you? There is, of course, a possible (iii) NO solution. For what value of k does that happen?

    You don't have to use matrix methods of course. I you use you old algebraic methods to solve for x, y, and z, you will eventually have to divide by something involving k. For what values of k is that non-zero so you can solve the equation? For what values of k does that give a fraction in which both numerator and denominator are 0 (and what does that mean)? For what values of k (if any) does that give a fraction in which the numerator is non-zero but the denominator is zero?
     
  7. Jun 14, 2006 #6
    what i meant about the exact coordinates of B is i dont have c. i was suggesting that you needed the answer from a) to do part b).

    i have a question about your method, being that the rectangle OABC is formed by taking the vertices in order, shouldnt the dot product of A and C be 0 and not A and B? simply because isnt B a diagonal of the rectangle?
     
  8. Jun 14, 2006 #7
    ok, i tried this method and i need some help here:

    so x-2y+z = 4 (1)
    x-y-z = 3 (2)
    x+y+kz = 1 (3)

    (1) - (2) = -y + 2z = 1 (4)
    (1) - (3) = -3y + z(1-k) = 3 (5)

    Then
    3x(4) - (5) = 6z - z(1-k) = 0
    z(5+k) = 0

    it appears that for any value k, z will be zero. if k is -5, z would = 0/0? does that mean for any value k that's not -5, the lines intersect at a point and when k is -5, they intersect at a line?

    evidently, i still dont understand the implications of the values of k. (i.e. k is nonzero, the numerator and denominator are zero, the denominator is zero). this is all so confusing.
     
  9. Jun 14, 2006 #8

    HallsofIvy

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    Okay, you reduced it to z(5+k)= 0. No, it is not the case that "for any value k, z will be zero"- if k= -5, the equation reduces to 0*z= 0 and z can be anything. You have now answered your question: If z is any number except, -5, z= 0. Putting that into the previous equations
    12x- 5= 0 so x= 5/12, and y= -1. That's a single solution. If k= -5, then z can be anything so there are an infinite number of solutions.

    "does that mean for any value k that's not -5, the lines intersect at a point and when k is -5, they intersect at a line?"
    Yes, that's exactly what it means!
     
  10. Jun 14, 2006 #9

    HallsofIvy

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    My experience is, that when a single problem has "part a" and "part b", that's often the case!:rolleyes:

    Oh, I see your point. I misread the problem. I assumed that OA and OB were two adjacent sides of the rectangle but since it specifically says "2. The position vectors of the points A and B, with respect to the origin O, are 2i−3j+3k and 5i + j+ck respectively, where c is a constant. The point C is such that OABC is a rectangle." It clearly intends that the A and B be adjacent so that OB is a diagonal, not a side of the rectangle.
    (Notice that A, B, C, O are points, not vectors. It's better to say that OB is a diagonal rather than "B is a diagonal".)
    Of course, OA+ OC= OB so OC= OB- OA= 5i + j+ck - (2i−3j+3k)= 3i+ 4j+ (c-3)k. Put in the value of c you got in (a) and take the dot product of OA and OC to see if those vectors are pependicular.
     
  11. Jun 14, 2006 #10
    But at this point, we still haven't determined the value of c!

    edit: nvm, your wording was just a bit confusing. thanks for all your help though!

    for part b), should i just sub in the value for c and solve for OC using OB - OA?

    has anyone given any consideration to the question: "geometrically, what is the relationship between the line that represents the distance and the skew lines. the distance line is perpendicular to both no?"
     
    Last edited: Jun 14, 2006
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