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Math Help - Orthogonal projection and linear transformation

  1. #1
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    Orthogonal projection and linear transformation

    Let T : R^3 -> R^3 be orthogonal projection onto the plane x + y + z = 0.
    a) Show that the closest point to (1,0,0) on the plane x + y + z = - is (2/3,-1/3,-1/3).
    b) Show that the standard matrix for T is A = 1/3[2 -1 -1; -1 2 -1; -1 -1 2]

    Can anyone show me how to do (a)? I'm not sure I have done it right. I first found the basis for the plane: {(-1,1,0),(-1,0,1)} (correct?) and then changed it to an orthonormal basis. Then I went on to find the orthogonal projection of (1,0,0) using the orthonormal basis vectors.

    Any ideas about (b)?
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  2. #2
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    The shortest distance from a point to a plane is along the line perpendicular to the plane. The normal vector to x+ y+ z= 0 at any point on the plane is <1, 1, 1>. The line through (1, 0, 0) with direction vector < 1, 1, 1> is x= t+1, y= t, z= t. Where does that line intersect the plane x+ y+ z= 0?

    What (a) tells you is that (1, 0, 0) is mapped by A onto (3/2, -1/3, -1/3). Do the same thing to determine what A maps (0, 1, 0) and (0, 0, 1) to. those three vectors form the columns of A in the standard basis.
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  3. #3
    MHF Contributor FernandoRevilla's Avatar
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    Quote Originally Posted by mushroom View Post
    Any ideas about (b)?

    Another way: according to a well known theorem the projection matrix in \mathbb{R}^n over the hyperplane \pi\equiv a_1x_1+\ldots+a_nx_n=0 is P=I-\dfrac{1}{N^tN}NN^t where N^t=(a_1,\ldots,a_n) .


    P.S. Of course, I don't know if you have covered that theorem.
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