Recently, I've been trying to make a program that calculates the intersection of a line and a plane. To do so, I need a universal equation. Here's the question.
Let's say there's a plane in 3d space, with a normal vector n of $<x_1,y_1,z_1>$. The point $(x_2,y_2,z_2)$ lies on the plane as well. There is also a line that passes through points $(x_3,y_3,z_3)$ and $(x_4,y_4,z_4)$. What, in terms of those variables, are the co-ordiantes of the intersection between the line and the plane?
I've tried solving it myself, but I got the following, which doesn't seem to work at all.
$$x= x_3 + \frac{x_1x_2+y_1y_2+z_1z_2}{(x_3-x_1)(y_3-y_1)(z_3-z_1)(z_1)(x_1)(y_1)}(x_3-x_1)$$
$$y= y_3 + \frac{x_1x_2+y_1y_2+z_1z_2}{(x_3-x_1)(y_3-y_1)(z_3-z_1)(z_1)(x_1)(y_1)}(y_3-y_1)$$
$$z= z_3 + \frac{x_1x_2+y_1y_2+z_1z_2}{(x_3-x_1)(y_3-y_1)(z_3-z_1)(z_1)(x_1)(y_1)}(z_3-z_1)$$
Best Answer
Simplify notation a bit by calling the point on the plane $P,$ the first point on the line $Q,$ and the difference between the second point and first point on the line $v$ (so $v$ is the vector from the first point to the second point).
The equation of the plane is $n\cdot x + d = 0.$ Since $P$ is on the plane, $d$ must equal $-n\cdot P.$
The equation of the line is $Q + tv$, where $t$ is any real number. The value of $t$ where the line and plane intersect must satisfy $$ n\cdot (Q + tv) + d = 0. $$
Solving this equation for $t$ yields $$ t = \frac{-d - n\cdot Q}{n\cdot v}. $$