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Distance from plane to origin

  • Thread starter misterau
  • Start date
  • #1
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Homework Statement


Show that the distance D from the origin of any point (x , y , z) lying on the plane x + y + z= 1 satisfies
D^2 + x^2 + y^2 +(x + y -1)^2 .
By considering partial derivatives, find the point that is closest to the origin. Prove that this distance is genuinely minimal.

Homework Equations





The Attempt at a Solution


I did the first part. Knowing Distance is D^2 = x^2 + y^2 + z^2 then by re-arranging x + y + z =1 to -z = x + y - 1.
z^2 = (-z)^2 = (x + y - 1)^2
D^2 = x^2 + y^2 + (x + y - 1)^2
Not sure how to do the next part. How does considering partial derivatives help me find a point closest to the origin?
Any help would be great.
 

Answers and Replies

  • #2
33,152
4,838

Homework Statement


Show that the distance D from the origin of any point (x , y , z) lying on the plane x + y + z= 1 satisfies
D^2 + x^2 + y^2 +(x + y -1)^2 .
Shouldn't this be an equation? Namely, D^2 = x^2 + y^2 +(x + y -1)^2.

By considering partial derivatives, find the point that is closest to the origin. Prove that this distance is genuinely minimal.

Homework Equations





The Attempt at a Solution


I did the first part. Knowing Distance is D^2 = x^2 + y^2 + z^2 then by re-arranging x + y + z =1 to -z = x + y - 1.
z^2 = (-z)^2 = (x + y - 1)^2
D^2 = x^2 + y^2 + (x + y - 1)^2
Not sure how to do the next part. How does considering partial derivatives help me find a point closest to the origin?
Any help would be great.
You have D2 as a function of x and y. You're looking for the smallest value of D2 from the origin to the plane. You can use partial derivatives to find the minimum point in a way analogous to finding the minimum value of a function of one variable (i.e., finding the derivative, and setting it to zero to get the critical points, and then checking with the second derivative to see if you have a local maximum or local minimum).
 
  • #3
chiro
Science Advisor
4,790
131

Homework Statement


Show that the distance D from the origin of any point (x , y , z) lying on the plane x + y + z= 1 satisfies
D^2 + x^2 + y^2 +(x + y -1)^2 .
By considering partial derivatives, find the point that is closest to the origin. Prove that this distance is genuinely minimal.

Homework Equations





The Attempt at a Solution


I did the first part. Knowing Distance is D^2 = x^2 + y^2 + z^2 then by re-arranging x + y + z =1 to -z = x + y - 1.
z^2 = (-z)^2 = (x + y - 1)^2
D^2 = x^2 + y^2 + (x + y - 1)^2
Not sure how to do the next part. How does considering partial derivatives help me find a point closest to the origin?
Any help would be great.
The distance from a plane P to the origin is equal to a.x + b.y + c.z = d where <a,b,c> is
a normal vector and d is the distance to the origin.

We do this through the notion of a Euclidean Inner Product. Basically what happens here is
that in three dimensional geometry we express the inner product in the form of
<a,b> = |a||b|cos(a,b) where |a| is the length vector of a and |b| is the lebgth vector of
b. cos(a,b) is the cosine of the angle between a and b is denoted in that way.
 

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