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Math Help - Cartesian equation of a plane

  1. #1
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    Cartesian equation of a plane

    I am stuck on this problem!

    (a) Find the cartesian equation of the plane P through the points
    P(1, 2, 1), Q(2, 1, -1), R(0, 3, 1).

    (b) Find the equation of the line which is perpendicular to P and passes through the point S(2, -2, 3).

    (c) Find the coordinates of the point T on P which is closest to S and hence find the distance from S to P.

    Any help is much appreciated,

    Dranalion
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  2. #2
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    For (a), I found the vectors PQ and PR.

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

    Cross product of PQ and PR is: 2i +2j +2k = (2, 2, 2)

    Therefore the equation of the plane is:

    P = 2x + 2y + 2z? (Is this correct?)

    Or should I substitute the point P(1, 2, 1) into the equation 'P' to get:
    2(1) + 2(2) + 2(1) = 8

    Therefore, 2x + 2y + 2z = 8

    Which is correct? I am completely stuck for parts (B) and (C)!
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  3. #3
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    Hi.

    Quote Originally Posted by Dranalion View Post
    For (a), I found the vectors PQ and PR.

    PQ = (1, -1, 2)
    PR = (-1, 1, 0)
    Erm, no?

    I get PQ = (1,-1,-2). Not '+2'

    Quote Originally Posted by Dranalion View Post
    Cross product of PQ and PR is: 2i +2j +2k = (2, 2, 2)
    It's a good idea to calulate the cross product,

    I calculated

    (1,-1,-2)x(-1, 1, 0) = (2,2,0)


    Quote Originally Posted by Dranalion View Post
    Therefore the equation of the plane is:

    P = 2x + 2y + 2z? (Is this correct?)
    No, because ...


    Quote Originally Posted by Dranalion View Post
    Or should I substitute the point P(1, 2, 1) into the equation 'P' to get:
    2(1) + 2(2) + 2(1) = 8

    Therefore, 2x + 2y + 2z = 8
    Yea, exactly. Now you have the information, that the point P is in the plane.

    If you substitute the points Q and R as well, the left side should be equal to 8.

    As you can see it is wrong if you substitute R in this equation

    Quote Originally Posted by Dranalion View Post
    Which is correct? I am completely stuck for parts (B) and (C)!
    (B) You already know the euclidean vector of the line - it is (2,2,0) (you calculated (2,2,2) instead...) . This is the vector you should use, because you know it is orthogonal/perpendicular to the plane p

    So the line you need is

    L: OP + t*(2,2,0)

    If you substitute t = 0 the line L goes through P.

    (C) Of course the distance got to be 0, because the line is not parallel to the plane.
    So you can calculate the intersection point of L and P or you can use the 'hesse normal form'.

    Whatever you like.

    Rapha
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  4. #4
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    Quote Originally Posted by Rapha View Post
    (B) You already know the euclidean vector of the line - it is (2,2,0) (you calculated (2,2,2) instead...) . This is the vector you should use, because you know it is orthogonal/perpendicular to the plane p

    So the line you need is

    L: OP + t*(2,2,0)

    If you substitute t = 0 the line L goes through P.

    (C) Of course the distance got to be 0, because the line is not parallel to the plane.
    So you can calculate the intersection point of L and P or you can use the 'hesse normal form'.

    Whatever you like.

    Rapha
    Sorry, I don't understand!
    What do you mean?
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  5. #5
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    (b) Find the equation of the line which is perpendicular to P and passes through the point .
    Quote Originally Posted by Dranalion View Post
    Sorry, I don't understand!
    What do you mean?
    In (b) you want to find a line which is orthogonal to P and goes through the point S(2,-2,3)

    A line has the equation

    L: 0X + t XY, where X and Y are coordinates and t is a real number.

    So

    L: 0S + t XY = \begin{pmatrix} 2-0 \\ -2-0 \\3-0 \end{pmatrix}+ t XY

    If you substitute t = 0 in this equation, then

     \begin{pmatrix} 2 \\ -2 \\3 \end{pmatrix} + 0* XY = \begin{pmatrix} 2 \\ -2 \\3 \end{pmatrix} =S

    Hence every line with this position vector 0S apparently goes through the point S.

    And what you now want to do is finding a vector XY such that L is orthogonal to the plane P.
    To find an orthogonal vector you need to calculate the cross product.

    so if XY = (2,2,0) we get

    L: (2,-2,3) + t*(2,2,0)

    so this line is perpendicularto the plane P, because the vector (2,2,0) is orthogonal to the plane P. (This is what the normal vector [of the plane P] is about)

    Are there still any questions on (b)?
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