1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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)

    A(-2,-1,-2)


    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
    -a2=4
    -2a3 = 3

    in augmented form:

    -2,0,0|-1
    0,2,0|-4
    0,0,2|3

    REF...

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

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    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

    Mark44

    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.
    No.
    No.

    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

    Mark44

    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

    Mark44

    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
    Woops.

    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

    Mark44

    Staff: Mentor

    NO!!
    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

    Mark44

    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

    LCKurtz

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    @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

    Mark44

    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

    Mark44

    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

    LCKurtz

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    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?
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Parallel vectors
  1. Parallel vectors (Replies: 3)

  2. Vectors are parallel (Replies: 3)

  3. Parallel Vectors (Replies: 1)

Loading...