Edit 2: I have solved it, in the sense that I found one critical point, but it is in a pretty nasty form, so I wouldn't know how to justify that it is indeed closest to the origin. Please feel free to comment on this – thanks for reading.
EDIT: I feel that I am pretty close to a solution, using the Lagrange Multiplier method. I am seeking a way to proceed from where I am currently stuck, rather than use another method to solve this minimization problem. Thanks,
The problem statement is:
Find the point on the intersection of the two planes $a_0+a_1x+a_2y+a_3z=0$ and $b_0+b_1x+b_2y+b_3z=0$ which is nearest to the origin $(0,0,0)$.
My work:
The objective function, thinking of distance to the origin, should be the tricky $\sqrt{x^2 + y^2 +z^2}$, but since the (positive) square root is a monotone increasing function, minimizing this function is the same as minimizing the easier $f(x,y,z) = x^2 + y^2 + z^2$. We use this as our objective function, subject to two constraints,
$g_1(x,y,z) = a_1x+a_2y+a_3z=-a_0$,
$g_2(x,y,z) =b_1x+b_2y+b_3z= -b_0$.
So, by the method of Lagrange Multipliers, I want to solve the equations
$$\nabla f(x,y,z) = \lambda_1 \nabla g_1(x,y,z) + \lambda_2 \nabla g_2(x,y,z)$$
$$\implies (2x,2y,2z) = \lambda_1 (a_1,a_2,a_3) + \lambda_2 (b_1,b_2,b_3)$$
$$\implies (2x) = \lambda_1 (a_1) + \lambda_2 (b_1)$$
$$\implies (2y) = \lambda_1 (a_2) + \lambda_2 (b_2)$$
$$\implies (2z) = \lambda_1 (a_3) + \lambda_2 (b_3)$$
now, multiplying by x to the first equation, y to the second equation, z to the third equation, and then adding all three equations and using the two constraints, gives me
$$2(x^2 + y^2 + z^2) = \lambda_1(-a_0) + \lambda_2(-b_0)$$
I am currently stuck here. How could I proceed?
Any hints or suggestions are welcome.
Thanks,
Best Answer
I would like to give you a hint because it is pretty easy.
First of all, you need to find the intersection of the two planes (a line). Because you have 2 equations with 3 variables, you can reduce it to 1 equation with 2 variables. Do not forget to impose conditions to make sure two planes do not parallel each other.
Secondly, when you have a line in which the nearest point of the origin belong to, then, you need to calculate the distant between the origin and a point on the calculated line and find the minimum distant to get the solution.
Finally, using this solution to find the last point by replacing it to one of the two given planes.