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Parallel vectors

  1. Apr 1, 2014 #1
    Mod note: Corrected a typo in the problem statement.
    1. The problem statement, all variables and given/known data

    [STRIKE]If[/STRIKE]Is the vector AP parallel to u?

    u= (1,4,-3)


    3. The attempt at a solution

    I do know that if AP is parallel to u then there must be some scalar multiple such that Ax = u.
    I think it's a matter of setting up the question that I'm muddled.

    Anyway, my attempt.

    Let A = a1,a2,a3

    (a1,a2,a3)(-2,-1,-2) = (1,4,3)

    -2a1= 1
    -2a3 = 3

    in augmented form:



    a1= -1/2
    a2= -2
    a3 = 3/2
    Last edited by a moderator: Apr 1, 2014
  2. jcsd
  3. Apr 1, 2014 #2


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    What do you mean by "AP"? Normally, a notation like that means "A" and "P" are points and "AP" is the vector from A to P. But there is no "P" given here.

    You are being very confusing here. AP is parallel to u if and only if there exist sum scalar multiple x such that xAP= u. Is "A" here a point or a vector?

    Didn't you just tell us that A was (-2, -1, -2)?

    ??what kind of multiplication is this? You should be trying to find a single real number, x, such that x(-2, -1, -2)= (-2x, -1x, -2x)= (1, 4, 3) (That is, assuming your "AP" is just the vector A).

    I still have no clue what you are doing. Assuming that your "AP" is just the vector (-2, -1, -2) the you should be trying to find a single number x such that x(-2, -1, -2)= (-2x, -x, -2x)= (1, 4, 3). Yes, that gives -2x= 1 so that x= -1/2. But then -x= 1/2, not 4 so we can see immediately that this cannot be done.

    In fact, the question "Is (-2, -1, -2) parallel to (1, 4, 2) can be answered quickly by noting that -2 divided by 1 is -2 but that -1 divided by 4 is -1/4. Those are not the same so one is not the multiple of the other.
  4. Apr 1, 2014 #3


    Staff: Mentor

    Due to your unclear problem statement, it took me a while to figure out what the problem is asking. I think it should read something like this:
    Find the coordinates of a point P so that the vector ##\vec{AP}## is parallel to u, where u and A are shown above.
    You are given A. Why are you trying to solve for its coordinates? What you need to find are the coordinates of a point P.

    The main concept to understand here is how two vectors can be equal. The dot product doesn't enter into things in this problem
  5. Apr 1, 2014 #4
    I know it's unclear. I was trying to infer from what the problem was implying.
    Now that you said, it make sense that A = (-1,-2,-2). When I typed A(-1,-2,-2), it was NOT a typo but a direct copy from the source.
  6. Apr 1, 2014 #5

    After Mark44's clarification, I understand what the question is asking.
    I believe the question meant to suggest A = (-2,-1,-2) instead of A(-2,-1,-2).
  7. Apr 1, 2014 #6


    Staff: Mentor

    That was not my concern. Either way, A is a point. ##\vec{AP}## is a vector. You need to find the coordinates of P (not A) so that ##\vec{AP}## is parallel to your given vector u. Do you understand how to determine whether two vectors are parallel?
  8. Apr 1, 2014 #7
    I think I do. Both vectors are parallel if every coordinates in u is a scalar multiple of A.

    A = (-2,-1,-2) u= (1,4,-3)

    if AP is parallel to u then AP cross u = zero vector
    This requires the operation of determinant.

    But as Hallsofivy suggested, and, what I initially thought so, was through the use of scalar multiples.
    If AP is parallel to u then AP is a scalar multiple of u such that
    γAP = u or AP 1/γ (u)
  9. Apr 1, 2014 #8


    Staff: Mentor

    No. The only possible interpretation of A is that it is a point. Otherwise ##\vec{AP}## wouldn't be meaningful.
    To distinguish between points and vectors, I usually write vectors the angle brackets; e.g., u = <1, 4, -3>.
    There is no need whatsoever to do it this way.
    Yes, although you are missing an = in the second equation.
  10. Apr 1, 2014 #9

    So....γ(-2,-1,-2) = <1,4,-3>
    -2γ = 1
    -γ = 4
    -2γ = -3

    as can be seen, for all γ, γ differ in values. Therefore, the vector AP is not parallel to the point <u>.
  11. Apr 1, 2014 #10


    Staff: Mentor

    You are still treating A as if it were a vector, which it most emphatically is not! Since the question is asking about ##\vec{AP}##, you need to come up with an expression for ##\vec{AP}##. Since you're not given the coordinates for P (a point), how can you represent this point? Once you have P, then you can write an expression for the vector ##\vec{AP}##.

    You have misunderstood the question in this thread from the very start.
  12. Apr 1, 2014 #11
    Just do a dot product and solve for the angle. It has to be zero degrees to be parallel.

    [itex]\vec{A}[/itex][itex]\cdot[/itex][itex]\vec{u}[/itex] = |A|*|u|cosθ
  13. Apr 1, 2014 #12


    Staff: Mentor

    No. A is not a vector. The question clearly asks about ##\vec{AP}##, the displacement vector between the points A and P.
  14. Apr 1, 2014 #13
    Ah, I misread it then, my apologies. Can use that to check the solution though :).
  15. Apr 1, 2014 #14


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    @negation: Do you know how to write the equation of the line through A with direction vector u?
  16. Apr 1, 2014 #15

    I must have forgotten. Could I have some assistance?
    edit: is the displacement AP expressed as AX =P?
    Last edited: Apr 1, 2014
  17. Apr 1, 2014 #16


    Staff: Mentor

    What does that even mean? A is a point.

    If P1(x1, y1, z1) and P2(x2, y2, z2) are points in R3, then the displacement vector between these points is ##\vec{P_1P_2}## = ##<x_2 - x_1, y_2 - y_1, z_2 - z_1>##. I hope that looks familiar.
  18. Apr 1, 2014 #17


    Staff: Mentor

    There's a much simpler way to check that two vectors are parallel. The dot product is not needed.
  19. Apr 1, 2014 #18


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    Don't you have a text book that gives the equation of a line through a point P with direction vector D?
  20. Apr 2, 2014 #19
    I don't. My course book covers only convuluted theorem and proof which makes it rather painful to understand for someone relatively new to vector and subspace.
    I would really appreciate a leg-up
  21. Apr 2, 2014 #20
    In order the find the displacement, it is p2 - p1?
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