- #1
B.Cantarelli
- 4
- 0
Question:
Find the point in the plane x+2y-3z=1 with minimum distance to point (1,1,1).
Attempt at resolution:
Well, techinically, I already have the means to solve this.
I can find the distance between the point and the plane to be 1/sqrt(14) and then I can either solve for the sphere (x-1)^2+(y-1)^2+(z-1)^2=1/14 and the plane x+2y-3z=1, or I can use Calculus (Lagrange's method) to get L=x^2+y^2+z^2-2x-2y-2z+3+l(x+2y-3z-1) and take all the partial derivatives.
Both yield (15/14, 8/7, 11/14), but I'm quite sure I'm missing some way more efficient method for doing this using vectors or something.
Find the point in the plane x+2y-3z=1 with minimum distance to point (1,1,1).
Attempt at resolution:
Well, techinically, I already have the means to solve this.
I can find the distance between the point and the plane to be 1/sqrt(14) and then I can either solve for the sphere (x-1)^2+(y-1)^2+(z-1)^2=1/14 and the plane x+2y-3z=1, or I can use Calculus (Lagrange's method) to get L=x^2+y^2+z^2-2x-2y-2z+3+l(x+2y-3z-1) and take all the partial derivatives.
Both yield (15/14, 8/7, 11/14), but I'm quite sure I'm missing some way more efficient method for doing this using vectors or something.
Last edited: