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Vector question in 3d

  1. May 26, 2016 #1
    1. The problem statement, all variables and given/known data
    The points A,B and C have position vectors , relative to the origin O given by ## OA= i+2j+3k, OB=4j+k , OC=2i+5j-k.## A fourth point D is such that the quadrilateral ABCD is a parallelogram.
    i) Find the position vector of D and verify that the parallelogram is a Rhombus.
    ii)The plane p is parallel to OA and the line BC lies in p. Find the equation of p,giving your answer in the form ##ax+by+cz=d##


    2. Relevant equations


    3. The attempt at a solution
    ##AB= (4j+k)-(i+2j+3k)=-i+2j-2k, AC= (2i+5j-k)-(i+2j+3k)=i+3j-4k, BC= (2i+5j-k)-(4j+k)=2i+j-2k## i am just groping in the dark here, but i know vectors on same line and parallel vectors should have a scalar or something...relating them
     
  2. jcsd
  3. May 26, 2016 #2

    SteamKing

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    Let's take things one at a time.

    What do you know about the sides of a general quadrilateral figure which would make it a rhombus?

    https://en.wikipedia.org/wiki/Rhombus

    Hint: it's OK to make a sketch, if that helps.
     
  4. May 26, 2016 #3
    i am unable to make a sketch on this application, ok for a quadrilateral the opposite sides are equal in terms of distance ##d##
    In Euclidean geometry, a rhombus(◊), plural rhombi or rhombuses, is a simple (non-self-intersecting) quadrilateral whose four sides all have the same length. Another name is equilateral quadrilateral, since equilateral means that all of its sides are equal in length. The rhombus is often called a diamond, after the diamonds suit in playing cards which resembles the projection of an octahedral diamond, or a lozenge, though the former sometimes refers specifically to a rhombus with a 60° angle (see Polyiamond), and the latter sometimes refers specifically to a rhombus with a 45° angle.

    Every rhombus is a parallelogram and a kite. A rhombus with right angles is a square.
    220px-Rhombus.svg.png
     
  5. May 26, 2016 #4

    SteamKing

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    You don't have to make a sketch here in the edit box at PF, but just for yourself.

    For a rhombus, is it just the opposite sides which are equal in length, or ...?

    In the article referenced above, there is a whole list of geometric properties which must be satisfied for a quadrilateral to be called a rhombus.
     
  6. May 26, 2016 #5
    agreed , a square is a rhombus too point noted....all sides equal also qualifies to be a rhombus. In that case the two sides opposite to each other in our problem should be equal in terms of distance i.e parallelogram
     
  7. May 26, 2016 #6

    SteamKing

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    Good. This should help you find the position vector for point D. Some of the other properties may also be used to check your result.
     
  8. May 26, 2016 #7
    ok, ##AD=BC, BC=2i+j-2k, → ##DA=d-a= 2i+j-2k,
    ##D=(i+2j+3k)+(2i+j-2k)= 3i+3j+k##
     
  9. May 26, 2016 #8

    SteamKing

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    OK, that looks good for OD.

    Now, what about part ii)?
     
  10. May 27, 2016 #9
    now what do they mean by part i) of the question, that verify that the parallelogram is a Rhombus? am i supposed to show using ##L=√(x2-x1)^2+(y2-y1)^2+(z2-z1)^2## that the opposite sides are same in terms of length(distance) ie ##AD=BC, AB=DC?##
     
  11. May 27, 2016 #10

    SteamKing

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    That would be one of the ways to show that the figure is a rhombus, according to the wiki article. There are several other methods which are equally valid.
     
  12. May 27, 2016 #11
    like finding mid point?......
     
  13. May 27, 2016 #12
    for part ii) ##BC= (2i+5j-k)-(4j+k)=2i+j-2k,→2x+y-2z, OA=i+2j+3k→x+2y+3z## what next?
     
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