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Math Help - Vector and parametric equations of a plane

  1. #1
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    Unhappy Vector and parametric equations of a plane

    Hey!
    I am having trouble with a question. I have NO IDEA how to go about finding the solution, so any help is greatly appreciated!

    Question
    Find a vector equation and parametric equations of the plane that contains the points M(2, 1, 3) and N(-1, 5, 7) and is perpendicular to the plane x + 2y + 3z + 4 = 0.

    What I know
    So I know how to find the vector equation of a plane if I am given 3 points (or 2 direction vectors and a point) on the plane. I also know how to find the parametric equation given the same thing.

    My theory on how to solve this is using the two points given to get a direction vector (3, -4, -4).

    I know that the Cartesian equation of a plane is in the format Ax + By + Cz + D = 0. The capital letters (A, B, C) are the normal to the plane the Cartesian vector is referring to (so, in this case, it would be the second direction vector I need). The x y z are a point on that plane. I do not know how to combine this information in order to solve the problem though!

    Thank you for reading, and SUPER thank you if you help
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  2. #2
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    Quote Originally Posted by Kakariki View Post
    Hey!
    I am having trouble with a question. I have NO IDEA how to go about finding the solution, so any help is greatly appreciated!

    Question
    Find a vector equation and parametric equations of the plane that contains the points M(2, 1, 3) and N(-1, 5, 7) and is perpendicular to the plane x + 2y + 3z + 4 = 0.

    What I know
    So I know how to find the vector equation of a plane if I am given 3 points (or 2 direction vectors and a point) on the plane. I also know how to find the parametric equation given the same thing.

    My theory on how to solve this is using the two points given to get a direction vector (3, -4, -4).

    I know that the Cartesian equation of a plane is in the format Ax + By + Cz + D = 0. The capital letters (A, B, C) are the normal to the plane the Cartesian vector is referring to (so, in this case, it would be the second direction vector I need). The x y z are a point on that plane. I do not know how to combine this information in order to solve the problem though!

    Thank you for reading, and SUPER thank you if you help
    \begin{vmatrix}<br />
i & j & k\\ <br />
3 & -4 & -4\\ <br />
x & y & z<br />
\end{vmatrix}=(-4z+4y)\mathbf{i}-(3z+3x)\mathbf{j}+(3y+4x)\mathbf{k}=<1,2,3>
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  3. #3
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    Quote Originally Posted by dwsmith View Post
    \begin{vmatrix}<br />
i & j & k\\ <br />
3 & -4 & -4\\ <br />
x & y & z<br />
\end{vmatrix}=(-4z+4y)\mathbf{i}-(3z+3x)\mathbf{j}+(3y+4x)\mathbf{k}=<1,2,3>
    Thank you for responding! I do not understand what you put though. Can someone explain, please?

    Thanks!
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  4. #4
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    Quote Originally Posted by Kakariki View Post
    Thank you for responding! I do not understand what you put though. Can someone explain, please?

    Thanks!
    The cross product of two vectors will produce a normal vector to the plane. You know two vectors in your problem and we know one of the vectors is normal to the plane; therefore, if you take the cross product of an arbitrary vector <x,y,z> with your vector you know, you will obtain a normal vector. However, you want you normal vector to be <1,2,3>
    Last edited by dwsmith; May 24th 2010 at 07:46 PM.
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  5. #5
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    Quote Originally Posted by dwsmith View Post
    The cross product of two vectors will produce a normal vector to the plane. You know two vectors in your problem and we know one of the vectors is normal to the plane; therefore, if you take the cross product of an arbitrary vector <x,y,z> with your vector you know, you will obtain a normal vector. However, you want you normal vector to be <1,2,3>
    Took me a few minutes to understand, but I get it now. So I write out the cross product and solve for x y and z. I think you wrote out the cross product wrong though, it isn't following the formula I have.

    Using the formula I have it writes out as:
     (-4z + 4y), (-4x - 3z), (3y + 4x) = (1,2,3)



    Which makes the system of equations to solve:
     -4z + 4y = 1
     -4x - 3z = 2
     3y + 4x = 3

    I can't seem to get a straight answer solving this system of equations!

    Btw, I am using:  a2b3 - a3b2, a3b1 - a1b3, a1b2 - a2b1

    HELP please!
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