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Finding closest point from a parabola to a plane

  1. Dec 6, 2013 #1
    The problem statement, all variables and given/known data.

    Find the point in the parabola ##y^2=x##, ##z=0## closest to the plane ##z=x+2y+8##


    The attempt at a solution.

    I've solved some problems where I had to find the closest point from a given surface to another point on the space. In this case, I have to find the distance between a curve and a plane so to speak. Maybe I could use Lagrange multipliers, but I am not so sure how to apply it in this particular problem.
     
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  3. Dec 6, 2013 #2

    ShayanJ

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    Gold Member

    You can use Lagrange multipliers but the tricky part is finding the function which should be minimized.That can be done by recalling the definition of the distance of a point to a plane. That's just the length of the line from the point normal to the plane.The normal vector of your plane is [itex] \hat{n}=\frac{1}{\sqrt{5}}(1,2,-1) [/itex]. Now consider a point [itex] P(x,y,z) [/itex] in space.But I want P to be on the parabola you mentioned so it becomes [itex] P(y^2,y,0) [/itex].I make a line with P and [itex] -\hat{n} [/itex].[itex] L:(y^2-tn_x,y-tn_y,-tn_z) [/itex] t being a parameter.Now lets see in what t,my line intersects your pane!
    [itex]
    -tn_z=y^2-tn_x+2(y-tn_y)+8 \Rightarrow (n_x+2n_y-n_z)t=y^2+2y+8 \Rightarrow t=\frac{y(y+2)+8}{n_x+2n_y-n_z}=t_p [/itex]
    Now we can find the distance of a point on your parabola to your plane by [itex] d=\sqrt{(y^2-t_p n_x-y^2)^2+(y-t_p n_y-y)^2+(-t_pn_z)^2}=t_p \sqrt{n_x^2+n_y^2+n_z^2} [/itex] But [itex] \hat{n} [/itex] is a unit vector so [itex] d=t_p [/itex].The function you should minimize,is [itex] t_p [/itex].
    But wait...We now have a function of one variable with no constraints...so you can just differentiate with respect to [itex] y [/itex] and set it equal to zero and that gives you the answer.
     
    Last edited: Dec 6, 2013
  4. Dec 6, 2013 #3

    HallsofIvy

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    Another way to do it is to use the geometric fact that the "shortest distance" to a plane is always along a line perpendicular to the plane. We can write this plane as x+ 2y- z= -8 so it has <1, 2, -1> as perpendicular vector. A line through [itex](x_0, y_0, 0)[/itex] in the direction of that vector is [itex]x= x_0+ t[/itex], [itex]y= y_0+ 2t[/itex], [itex]z= -t[/itex]. Further we have [itex]y_0= x_0^2[/itex] so we can write that line as [itex]x= x_0+ t[/itex], [itex]y= x_0^2+ 2t[/itex], [itex]z= -t[/itex].

    That line will cross the plane when [itex]x_0+ 2(x_0^2+ 2t)- (-t)= 5t+ x_0+ 2x_0^2= -8[/itex]. You can solve that for t in terms of [itex]x_0[/itex], so determine the point in terms of [itex]x_0[/itex]. Then minimize the distance (more easily, the distance squared) to find [itex]x_0[/itex].
     
  5. Dec 6, 2013 #4

    Ray Vickson

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    Besides what others have said, you can (as you wanted) use the Lagrange multiplier method: minimize the squared distance from (t^2,t,0) to (x,y,z), subject to x+2y-z+8=0. The variables are t,x,y,z. This works out very nicely.
     
    Last edited: Dec 6, 2013
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