Use Lagrange's method to find the point on the line of intersection of the two planes $a_1x+a_2y+a_3z+a_0 = 0, b_1x+b_2y+b_3z+b_0 = 0$ which is nearest to the origin. you may also assume that the tow planes really intersect, but you should explain where this enters into the calculation.
I tried to solve this and I reached up to this point:
assuming $g=a_1x+a_2y+a_3z+a_0 = 0, h=b_1x+b_2y+b_3z+b_0 = 0$ and using Lagrange's multiplier method $L(x,y,z,\lambda,\mu) = f -\lambda g -\mu h$ and proceeding by taking the $gradient L =0$; I got the plane $2x^2+2y^2+2z^2 =0$. But the question need a point on the line of intersection of the two planes. How can I find this point?
Best Answer
The distance from a point to the origin is $\sqrt{x^2+y^2+z^2}$. Minimising this is equivalent to minimising its square. So your constrained optimisation problem is $$ \begin{align}\min x^2+y^2+z^2\\ \text{s.t. } a_1x+a_2y+a_3z+a_0&=0\\ b_1x+b_2y+b_3z+b_0&=0\end{align} $$ This has the equivalent Lagrangian formulation $$L(x,y,z,\lambda,\mu)=x^2+y^2+z^2+\lambda\left(a_1x+a_2y+a_3z+a_0\right)+\mu\left(b_1x+b_2y+b_3z+b_0\right).$$
Finding $\nabla L$ gives $$ \nabla L=\begin{bmatrix}2x+a_1\lambda+b_1\mu\\ 2y+a_2\lambda+b_2\mu\\ 2z+a_3\lambda+b_3\mu\\ a_1x+a_2y+a_3z+a_0\\ b_1x+b_2y+b_3z+b_0\end{bmatrix}$$
Setting $\nabla L=0$, equating the last two elements to zero gives the linear system which produces the line of intersection of the two planes.
Focusing on these two equations, you can get the solution as $x=s_1+s_2z$, $y=t_1+t_2z$, with $z$ a free parameter and the constants $s_1,s_2,t_1$ and $t_2$ depending on the $a_*$ and $b_*$ values in some way. This result can then be substituted into the first three equations to give you a linear system in $z,\lambda,\mu$, which can be solved for these three values. $x$ and $y$ can then be determined by substituting the appropriate value for $z$ into their respective solutions.