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Urgent+orthogonal projection

  1. Jan 17, 2009 #1
    please can u do this sum for me...really urgent situation

    find the equation of the orthogonal projection of the line x+1/1 = 2y/-1 = z+1/2 on the plane x+2y+z=12

    thanks in advance
  2. jcsd
  3. Jan 17, 2009 #2
    pleaaaaase help me guys!!!1
  4. Jan 18, 2009 #3
    is there anyone who can do this
  5. Jan 18, 2009 #4


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    You probably made a mistake putting this in the "Topology and geometry" section!

    The plane x+ 2y+ z= 12 has <1, 2, 1> as a normal vector. The line x+1/1 = 2y/-1 = z+1/2, which, setting each fraction equal to t, is the same as the parametric equations x+ t- 1, y= -(1/2)t, z= t- 1/2 have directional vector <1, -1/2, 1>. The projection of that vector onto <1, 2, 1> has length
    [tex]\frac{<1,-1/2, 1>\cdot<1, 2, 1>}{||<1, 2,1||}= \frac{1}{\sqrt{6}}[/tex]
    so the vector projection of <1, -1/2, 1> on <1, 2, 1> is that times a unit vector in the direction of <1, 2, 1>, [itex]\sqrt{6}<1, 2, 1>[/itex], and so is <1/6, -1/3, 1/6>. The projection of the vector onto the plane is <1, -1/2, 1>- <1/6, -1/3, 1/6>= <5/6, -5/6, 5/6>. That is, the parametric equations of the line are [itex]x= (5/6)t+ x_0[/itex], [itex]y= -(5/6)t+ y_0[/itex], [itex]z= (5/6)t+ z_0[/itex] where [itex](x_0, y_0, z_0)[/itex] is a single point on that projection. Putting the parametric equations of the line for x, y, and z in the plane, it see that the line intersects the plane when t= 27/2 so x= 25/2, y= -27/4, z= 26/2= 13 is a point on the plane and line and therefore on the projection of the plane into the line.

    I am going to move this to the Calculus and Beyond Homework section.
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