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Count out the distance from origo to planet z

  1. Mar 27, 2008 #1

    I want to confirm this:

    a=8, b=5, c=7

    Count out the distance from origo to planet z = ax + by + c in three different ways.

    1) With the aid of linear algebra and geometry (no derivates!).

    Normalvector is: (a,b,-1) , the length * (8,5,1) = p ,(p=point)
    p=(x,y,z) gives 8s=x , 5s=y. -s = 8x+5y+7 => -90s=7 => s=-7/90 => x = -56/90 , y = -35/90 , z = 7/90 , Is this correct?

    2) Through solving one optimization problem in two variables without bee conditions.

    q^2 = (ax+by+c)^2 where q is the distance

    uses the df/dx = df/dy = 0 and gets:


    and then solves what x,y,z is ? Or have i done something wrong?

    3) Through using Lagranges multiplier method

    I havent done this because i dont know how to use this method, can anyone help me with this?

    I want some help to confirm if i am solving this correctly.
  2. jcsd
  3. Mar 27, 2008 #2


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    You want to find the distance from the origin, (0, 0, 0) to the plane z= 8x+ 5y+ 7?
    Yes, any vector normal to that plane is a multiple of (8, 5, -1) and the line through the origin normal to the plane has parametric equations x= 8t, y= 5t, z= -t. That line will intersect the plane where 8x+ 5y+ 7= 64t+ 25t+ 7= -t or 90t= -7 so t= -7/90. Yes, that point is x= -56/90, y= -35/90, z= 7/90. But you aren't finished- you haven't found the distance from that point to (0,0,0).

    What distance? The distance from a point (x, y, z) on the plane to the origin (squared) is [itex]x^2+ y^2+ z^2= x^2+ y^2+ (8x+ 5y+ 7)^2[/itex], not what you have. What you have is just z2

    PLEASE don't use "f" without telling us what it means! I think f(x)= q2, the distance squared.

    The first method is correct, once you actually answer the question- for the second, you have the distance wrong.

    Lagrange's multiplier method uses the fact that the gradient of the function to be minimized points directly toward the direction of fastest increase. We will be at a minimum on the surface when that is perpendicular to the surface- and so parallel to a normal vector. In this case, taking f(x,y,z) to be the square of the distance from (x, y, z) to (0, 0, 0): f(x,y,z)= x2+ y2+ z2 so the gradient is [itex]\nabla f= 2x\vec{i}+ 2y\vec{j}+ 2z\vec{k}[/itex]. Since a normal vector to the surface is [itex]8\vec{i}+ 5\vec{j}-\vec{k}[/itex], at the nearest point we must have
    [itex]\nabla f= 2x\vec{i}+ 2y\vec{j}+ 2z\vec{k}= \lambda(8\vec{i}+ 5\vec{j}-\vec{k})[/itex] where [itex]\lambda[/itex] is the "multiplier". That is, we must have [itex]2x= 8\lambda[/itex], [itex]2y= 5\lambda[/itex], and [itex]2z= -\lambda[/itex]. Those, together with 8x+ 5y+ 7= z, give 4 equations for the four "unknowns" x, y, z, and [itex]\lambda[/itex]. since we don't really need to find [itex]\lambda[/itex], one good way to eliminate it is to divide one equation by another. Dividing the first equation by the second, [itex]2x/2y= 8/lambda/5\lambda[/itex] so 8y= 5x. Dividing the first equation by the third, [itex]2x/2z= 8\lambda/-\lambda[/itex] so 8z= -x. Now y= (5/8)x and z= -x/8. Put those into the equation of the plane, 8x+ 5y+ 7= z, solve the equation for x, find y and z, and find the distance from (0,0,0) to that point.
    Last edited: Mar 27, 2008
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