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Homework Help: Vectors and Planes

  1. Mar 23, 2010 #1
    1. The problem statement, all variables and given/known data
    Question 1.
    Given position vectors OA : i+2j-k and OB: 2i+j+3k in a 3D Cartesian space with origin O of the points A and B.

    a) Find the scalar equation of the plane which contains A and which is perpendicular to vector AB.

    b) Find the shortest distance from the point (1,-1,1) to the plane obtained in a (a)


    2. Relevant equations



    3. The attempt at a solution

    Here is what i did
    (a) Equation of line AB
    r= (2,1,3)+ t(1,-1,4)

    AP.n = 0
    OP.n = OA.n
    (x,y,z).(1,-1,4) = (2,1,3).(1,-1,4)

    x-y+4z = 13 [equation of plane]?

    (b) vector form of plane => r.(1,-1,4) = 13
    line equation => r= (1,-1,1) + t(1,-1,4)

    Using r=r for both equations i get t= 2/9

    Substitute t=2/9 into line equation giving me (11/9, -2/9, 17/9)

    I'm stuck here then for part (b)
    I'm pretty weak in this chapter, couldn't seem to grab the concept here.

    1. The problem statement, all variables and given/known data
    Question 2.
    Obtain the scalar equation of the plane which passes through point P(1,2,3) and contain a straight line
    x(t)= 3t
    y(t)= 1+t
    z(t)= 2-t
    t is the parameter of the line


    2. Relevant equations



    3. The attempt at a solution

    Equation of the line= (0,1,2) + t(3,1,-1)

    Let Q and R be the points on the line
    t=0 Q=(0,1,2)
    t=1 R= (3,2,1)

    Therefore PQ= (-1,-1,-1) PR= (2,0,2)

    PQxPR= (-2,0,-2)

    Thus the equation of the plane
    a(x-x1) + b(y-y1) + c(z-z1)= 0
    -2(x-1) + 0 -2(z-3) = 0
    -2x - 2z + 8 = 0
    is this correct?
     
    Last edited: Mar 23, 2010
  2. jcsd
  3. Mar 24, 2010 #2

    LCKurtz

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    You will find that (1,2,-1) doesn't work in that equation so the point is not on the plane.
    For R = <x,y,z>, the form on the equation of a plane is

    [tex](\vec R - \vec A)\cdot \vec N = 0[/tex]

    The correct equation for the plane might get you started.
     
  4. Mar 24, 2010 #3
    so what does it mean when it says the plane contains A?

    so is it like this?
    [(x,y,z)-(1,2,-1)].(1,-4,4)=0

    1(x-1)-4(y-2)+4(z+1)=0

    x-4y+4z=11
     
    Last edited: Mar 24, 2010
  5. Mar 24, 2010 #4

    LCKurtz

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    Yes, except check the sign on the 11.
     
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