Originally Posted by audioxbliss
So I had a math midterm today. One problem I couldn't get:
A plane is given by 3x+4y+12z=26. Find the point closest to the origin.
I got as far as f(x,y,z)=3x+4y+12z-26
Then I gave up. Thoughts?
Yes, you need to optimize using Lagrange multis with boundary condition 3x + 4y +12z - 26 = 0
Using Lagrange, f(x, y, z, λ) = x^2 + y^2 + z^2 + λ(3x + 4y + 12z - 26)
Then finding extremes with respect to all directions in Eucl. space:
∂f/∂x = 2x + 3λ = 0
∂f/∂y = 2y + 4λ = 0
∂f/∂z = 2z + 12λ = 0
Solving this linear system gives 2x = -3λ, 2y = -4λ, 2z = -12λ => y = (4/3)x , z = 4x
Thus using our boundary condition, 3x + 4 * (4/3)x + 12 * 4x = 26 => (144/3 + 16/3 + 9/3) x = 26 => x = 26/169
Then y = (4/3) * 26/169 = (34+2/3)/169, z = 4 * 26/169 = 104/169.
∴p = <26/169, (34+2/3)/169, 104/169>