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I am currently having issues with calculating plane intersection of a ray.

I start with the following equation $P = P_0 +tR_t$

$R_t$ is the Unit Vector of the Trajectory.

Now we have a plane which is given by

$Ax+Bx+Cx +D = 0$

Where the normal vector is given by $N = \langle A,B,C\rangle$

To calculate the Plane Intersection you find $t$

$t = \frac{R_0\cdot N + D}{R_t \cdot N}$

Then once you have that you substitude and find intersection.

Assuming I want to find the ray intersection to the y=0 plane, what would be my D value in this case. I know that is should be $D=-Ax-By-Cz$ . But how to determine x,y,and z?

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help!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! – Anonymous Nov 29 '12 at 18:09

The plane should be $Ax+By+Cz+D=0$ (not all vars $x$ as you wrote). Then the plane $y=0$ fits this using $A=0,B=1,C=0,D=0$.

So your $N=(0,1,0)$ and your formula should work.

However your formula $P=P_0+tR_t$ makes it look like the coordinates of the unit vector $R_t$ depend on the parameter $t$. In this case the formula for $P$ does not represent a ray, but rather a curve in space, and finding its intersection with a plane will be in general difficult. If instead $R_t$ is a constant vector, say $P=P_0+tR_0$, then this lies on the plane iff $N \cdot P+D=0$ and the formula for $t$ may be obtained on substituting $P=P_0+tR_0$ for $P$ in the plane equation $N \cdot P+D=0.$

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+1 for the remark on $R_t$. – user1551 Nov 30 '12 at 10:14

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