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    The plane of euqation

    r = (3,1,2) + lamda(2,1,2) + mew(1,-2,-1)

    goes though the points (3,1,2) , (5,2,4), and (4,-1,1)

    how would i then find the cartesian equation of the plane?


    [I know its somewhere along the lines of eliminated the lambda and mew parameters but im not too familiar with this.]

    any help would be appreciated..thanks
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    r = (3,1,2) + lamda(2,1,2) + mew(1,-2,-1)
    goes though the points (3,1,2) , (5,2,4), and (4,-1,1)
    how would i then find the cartesian equation of the plane?
    One way would be to write the plane in the form r.n=a.n, where r=\vec{xi}+\vec{yj}+\vec{zk}, \vec{n} is a vector perpendicular to the plane(and so perpendicular to the two lines) and \vec{a} is any vector in the plane.
    Note r.n=a.n comes from\vec{(r-a)}.\vec{n}=0 as \vec{(r-a)} represents any vector in the line and it's scalar product with \vec{n} is, of course, .
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    (Original post by TerrorSquad)
    The plane of euqation

    r = (3,1,2) + lamda(2,1,2) + mew(1,-2,-1)

    goes though the points (3,1,2) , (5,2,4), and (4,-1,1)

    how would i then find the cartesian equation of the plane?


    [I know its somewhere along the lines of eliminated the lambda and mew parameters but im not too familiar with this.]

    any help would be appreciated..thanks
    what exam board is this for , aqa ?

    Edit: do u have the answers for the question?
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    exam board is AQA maths spec a

    i have the answer but this is not the whole question . the question i am stuck on requires you to find the angle between 2 planes [ 1 is given in hte final part in cartesian form] and the other plane which you have to work out before you gotta convert it into catesian form to calculate the angle
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    what past paper is this from?
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    i hope you check this thread b4 2morow, cos i wanna get this sorted out what - if we get a similar question in the exam, worth 6 marks :eek:
    ill be screwed !!!
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    june03 its from if i remember correctly

    yeah its a dody concept , this mite help

    parametric equation of plane is x = 6lamda + 3 mew

    y = 6 - 3mew

    z = 2mew

    hte answer for the cartesian eqn is 2y + 3z =12
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    (Original post by TerrorSquad)
    june03 its from if i remember correctly

    yeah its a dody concept , this mite help

    parametric equation of plane is x = 6lamda + 3 mew

    y = 6 - 3mew

    z = 2mew

    hte answer for the cartesian eqn is 2y + 3z =12
    where did you get those parametrics from?
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    sounds more like p6 to me
    edexcel neway
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    (Original post by TerrorSquad)
    The plane of euqation
    r = (3,1,2) + lamda(2,1,2) + mew(1,-2,-1)
    goes though the points (3,1,2) , (5,2,4), and (4,-1,1)
    -> (x, y, z) = (3, 1, 2) + lm(2, 1, 2) + mw(1, -2, -1)
    -> (x, y, z) = [(3 + 2lm + mw), (1 + lm - 2mw), 2 + 2lm - mw)]
    -> x = 3 + 2lm + mw, y = 1 + lm - 2mw, z = 2 + 2lm - mw

    Need to eliminate parameters lm and mw.

    x + z = 3 + 2lm + mw + 2 + 2lm - mw = 3 + 2lm + 2 + 2lm = 5 + 4lm
    x + y - z = 3 + 2lm + mw + 1 + lm - 2mw - 2 - 2lm + mw = 2 + lm

    Hence: x + z = 5 + 4lm and x + y - z = 2 + lm
    -> (x + z - 5)/4 = lm and x + y - z - 2 = lm

    -> Equate expressions for lm: (x + z - 5)/4 = x + y - z - 2
    -> x + z - 5 = 4x + 4y - 4z - 8

    -> Cartesian Equation: 3x + 4y - 5z = 3
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    Here's the method Gaz suggested:
    n = (5-3, 2-1, 4-2) x (5-4, 2+1, 4-1) = (-3, -4, 5)

    Let r = (x, y, z), then:
    (x, y, z) . (-3, -4, 5) = (3, 1, 2) . (-3, -4, 5)
    - 3x - 4y + 5z = - 9 - 4 + 10
    3x + 4y - 5z = 3, which agrees with Nima's answer.
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    (Original post by Nima)
    -> (x, y, z) = (3, 1, 2) + lm(2, 1, 2) + mw(1, -2, -1)
    -> (x, y, z) = [(3 + 2lm + mw), (1 + lm - 2mw), 2 + 2lm - mw)]
    -> x = 3 + 2lm + mw, y = 1 + lm - 2mw, z = 2 + 2lm - mw

    Need to eliminate parameters lm and mw.

    x + z = 3 + 2lm + mw + 2 + 2lm - mw = 3 + 2lm + 2 + 2lm = 5 + 4lm
    x + y - z = 3 + 2lm + mw + 1 + lm - 2mw - 2 - 2lm + mw = 2 + lm

    Hence: x + z = 5 + 4lm and x + y - z = 2 + lm
    -> (x + z - 5)/4 = lm and x + y - z - 2 = lm

    -> Equate expressions for lm: (x + z - 5)/4 = x + y - z - 2
    -> x + z - 5 = 4x + 4y - 4z - 8

    -> Cartesian Equation: 3x + 4y - 5z = 3
    cheers, does actually all make sense!
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    (Original post by dvs)
    Here's the method Gaz suggested:
    n = (5-3, 2-1, 4-2) x (5-4, 2+1, 4-1) = (-3, -4, 5)

    Let r = (x, y, z), then:
    (x, y, z) . (-3, -4, 5) = (3, 1, 2) . (-3, -4, 5)
    - 3x - 4y + 5z = - 9 - 4 + 10
    3x + 4y - 5z = 3, which agrees with Nima's answer.

    that is so wrong, look at your first line:

    n = (5-3, 2-1, 4-2) x (5-4, 2+1, 4-1) = (-3, -4, 5)

    (5-3) x (5-4) does not equal -3
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    (Original post by ram)
    that is so wrong, look at your first line:
    n = (5-3, 2-1, 4-2) x (5-4, 2+1, 4-1) = (-3, -4, 5)
    (5-3) x (5-4) does not equal -3
    It isn't wrong, it's perfectly correct.
    He has used the vector product - not the scalar product (hence the 'x' rather than '.').
    The vector product \vec{a}x\vec{b} gives a vector perpendicular to both \vec{a} and \vec{b}.
    Alternatively you could find a perpendicular vector by setting it to be i+aj+bk and solving the simultaneous equations you get from setting the scalar product with the two vectors in the plane to zero.
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    (Original post by ram)
    that is so wrong, look at your first line:

    n = (5-3, 2-1, 4-2) x (5-4, 2+1, 4-1) = (-3, -4, 5)

    (5-3) x (5-4) does not equal -3
    This x is the cross or vector product - not multiplication - the calculation is correct
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    from what paper is this question from.....i can always get the answers to it from thw aqa website
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    its from june 03.......in the context of the question you dont need to convert into cartesian form. they give a vector that is perpendicular to plane 1. and the normal of plane 2 you can find out just by inspecting the eqn. then u just dot product the two perpendiculars....and find the angle.
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    Equation of plane is r . n = a . n

    Find n, by crossing the multiples of lamda and mew
    n = (2,1,2) x (1,-2,-1)
    n = (3, 4, -5)

    a is just a point on the plane. Just use (3,1,2)

    so
    r . n = a . n
    r . (3, 4, -5) = (3, 1, 2) . (3, 4, -5)
    r . (3, 4, -5) = 3
    3x + 4y -5z = 3
 
 
 
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