# Intersection Point of a Line and four Planes

Let's assume a helicopter crashes into a wall after flying in a straight line:

$$g : \overrightarrow {OX} = \begin{pmatrix}2\\5\\28 \end{pmatrix}+ \lambda*\begin{pmatrix}1\\\frac{1}{3}\\\frac{-1}{11} \end{pmatrix}$$

There are four walls which form a valley (it is formed by 1,2; 2,3; 3,4 and 4,1): $$\epsilon_1 : 9922x+6716y+539z = 95033$$ $$\epsilon_2 : 20314x-6708y+4543z = 262821$$ $$\epsilon_3 : 67179x+7766y+12803z = 4737245606$$ $$\epsilon_4 : 15414x+135576y+22540z = 1213884$$

How can I determine in which of the these planes the helicopter crashes? I was able to calculate four different $\lambda$ (after substituting $g$ into $\epsilon$) and the four intersection points. I understand why there are four intersection points since the planes have an infinity x,y,z range and as long as the line is not parallel there will always be an intersection point...

But there has to be a way to determine the actual wall.

-

Calculate the lambdas for each of the planes so that the helicopter intersects the particular plane (I guess lambda would be infinity if the helicopter is going parallel to the plane), and choose the "earliest" (i.e. smallest positive) lambda. The associated plane must be the one the helicopter crashes into.

-