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Math Help - find the equation of the plane.

  1. #1
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    find the equation of the plane.

    Find the equation, in the form of ax + by + cz = d, of the plane going through a point
    P = (-3, 4, 2) to a line L: (x, y, z) = (3, -2, -1) + (1, -2, 2)t.


    Can you solve this with brief explanation ?
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  2. #2
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    Quote Originally Posted by eunse16 View Post
    Find the equation, in the form of ax + by + cz = d, of the plane going through a point
    P = (-3, 4, 2) to a line L: (x, y, z) = (3, -2, -1) + (1, -2, 2)t.


    Can you solve this with brief explanation ?
    Hi eunse16.

    The plane ax+by+cz+d passes through (-3,4,2) so

    -3a+4b+3c\ =\ d\quad\ldots\quad\fbox1

    And every point on the line (x,y,z)=(3+t,-2-2t,-1+2t) lies on the plane. Take any two points on the line, say (2,0,-3) and (0,4,-7), and plug the values into the plane equation:

    2a-3c\ =\ d\quad\ldots\quad\fbox2

    4b-7c\ =\ d\quad\ldots\quad\fbox3

    Solve the simultaneous equations \fbox1-\fbox3 for a,\,b,\,c in terms of d.

    You should get a=d,\ b=\frac56d,\ c=\frac13d; hence the equation of the plane is x+\frac56y+\frac13z=1.
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  3. #3
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    Another way to do this: Choose two points on the line. For example, if t= 0 then (x, y, z) = (3, -2, -1) + (1, -2, 2)0= (3, -2, -1) and if t= 1, (x, y, z) = (3, -2, -1) + (1, -2, 2)1= (4, -4, 1). That gives (-3, 4, 2), (3, -2, -1), and (4, -4, 1) as three points in the plane and we can use the standard method of finding the equation of the plane that includes three given points.

    Now, the vector from (-3, 4, 2) to (3, -2, -1) is 6i- 6j- 3k and the vector from (-3, 4, 2) to (4, -4, 1) is 7i- 8j- k. The cross product of those two vectors is a normal vector to the plane and its equation can be written A(x-x0)+ B(y-y0)+ C(z-z0)= 0 where Ai+ Bj+ Ck is a normal vector and (x0, y0, z0) is a point in the plane.
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  4. #4
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    Here is a general way.
    If \ell :Q + tD\;\& \,P \notin \ell then the plane containing \ell~\&~P is:
    \left( {\overrightarrow {QP}  \times D} \right) \cdot \left( {\left\langle {x,y,z} \right\rangle  - Q} \right) = 0
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