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Math Help - Finding the equation of a plane

  1. #1
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    Finding the equation of a plane

    Could somebody please explain me how I should solve the following?

    Determine the vector equation of the plane P(a) containing the line AB and perpendicular to the vector v(a).

    Given

    A(1,0,1) B(0,-1,1) and v(a) = i - j + (a)k
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  2. #2
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    Let \vec u = (1,1,0) \parallel \vec {AB}

    Then \vec w = \vec u\times \vec v=(a,-a,-2) will be perpendicular to both \vec u and \vec v.


    \left(\begin{array}{ccc}x\\y\\z\end{array}\right)=  \left(\begin{array}{ccc}1\\0\\1\end{array}\right)+  s\left(\begin{array}{ccc}1\\1\\0\end{array}\right) +t\left(\begin{array}{ccc}a\\-a\\-2\end{array}\right)<br />
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  3. #3
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    A plane perpendicular to vector <A, B, C> and containing point (x_0,y_0, z_0) has equation A(x- x_0)+ B(y- y_0)+ C(z- z_0)= 0

    Here (1, 0, 1) is a point in the plane and <1, -1, a> is perpendicular to the plane so (x-1)- y+ a(z-1)= 0. We can write that as x- y+ az= a+1. We can solve for x: x= y- az+ a+ 1 and, using y and z as parameters write
    \begin{bmatrix}x \\ y \\ z\end{bmatrix}= \begin{bmatrix}a+1 \\ 0 \\ 0\end{bmatrix}+ \begin{bmatrix} 1 \\ 1 \\ 0\end{bmatrix}y+ \begin{bmatrix}-a \\ 0 \\ 1\end{bmatrix}z.
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