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Math Help - projection of point and plan

  1. #1
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    projection of point and plan

    Finding a projection from point A (2,-1,3) in the x-3y-z-13 = 0



    Answer:
    (3,-4,2)
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  2. #2
    Super Member flyingsquirrel's Avatar
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    Hello
    Quote Originally Posted by Apprentice123 View Post
    Finding a projection from point A (2,-1,3) in the x-3y-z-13 = 0

    Answer:
    (3,-4,2)
    Let H be the orthogonal projection of A on the plane which equation is x-3y-z=13

    We know two things about H :
    • As the projection is orthogonal, \vec{AH} is orthogonal to the plane.
    • H(x_h,y_h,z_h) lies in the plane : x_h-3y_h-z_h=13


    Luckily, the equation of the plane {\color{red}1}\cdot x{\color{red}-3}\cdot y{\color{red}-1}\cdot z=13 gives us a vector which is orthogonal to the plane : \vec{n}=\begin{pmatrix}1\\-3\\-1\end{pmatrix}. As \vec{n} and \vec{AH} are both orthogonal to the plane, they share the same direction that is to say that there exists a real number t such that \vec{AH}=t\vec{n}.

    As \vec{AH}=\begin{pmatrix}x_h-2\\y_h-(-1)\\z_h-3 \end{pmatrix} and t\vec{n}=\begin{pmatrix}t\\-3t\\-t \end{pmatrix} we have four equations :

    <br />
\begin{cases}<br />
t=x_h-2 \\<br />
-3t=y_h+1\\<br />
-t=z_h-3\\<br />
13=x_h-3y_H-z_h<br />
\end{cases}\implies \begin{cases}<br />
x_h=t+2 & (1)\\<br />
y_h=-3t-1& (2)\\<br />
z_h=-t+3&(3)\\<br />
13=x_h-3y_H-z_h&(4)<br />
\end{cases}

    Substituting (1),\,(2) and (3) in (4) will give you the value of t from which you'll get H(x_h,y_h,z_h). Good luck
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  3. #3
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    Thank you . I do AH (x-2,y+1,z-3) = (1,-3,-1). And I find x, y and z.


    Thank you very much
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