Consider the line and plane defined by

`vec r_"L" = vec a + tvec m qquad and qquad vec r_Pi = vec b + rvec u + svec v`.

They may or may not intersect. In `RR^3`, there are three cases:

- coplanar (`oo`)
- The line lies on the plane. Formally, `AA t\ EE r EE s` such that `vec r_"L" = vec r_Pi`. This implies that `vec m`, `vec u`, and `vec v` are coplanar, therefore `vec m * vec u xx vec v = 0`,
*and*that the position `vec a` lies on the plane. There are `oo` points of intersection. - parallel (0)
- The line and the plane are parallel, so the three direction vectors are coplanar as before, but the position vector `vec a` does
*not*lie on the plane. For all combinations of values of*t*,*r*, and*s*, we have `vec r_"L" != vec r_Pi`. There are zero points of intersection. - nonparallel (1)
- All other systems fall into this category. The line and the plane are not parallel, so the three direction vectors are not coplanar. There is exactly one point of intersection.

It is possible to find the point of intersection of a line and a plane this way by converting to parametric form, because you will have three variables (free parameters) and three equations. But this is a lot of work—there is an easier way. Two easier ways, in fact.

The first method works uses the parametric equations of a line and the standard equation of a plane. Given the latter,

`ax + by + cz + d = 0`,

you can substitute the values of *x*, *y*, and *z* for the right-hand sides of the parametric equations of the line:

`a(x_0 + tm_1) + b(y_0+tm_2) + c(z_0+tm_3) + d = 0`.

You can then solve for *t*. If you get a value for *t*, the system is nonparallel, and the single point of intersection is found by evaluating the parametric equations of the line at *t*. If all the *t* terms and constant terms cancel and you are left with `0=0`, then the value of *t* is irrelevant (the equation holds for all values `t in RR`) and the system is coplanar—there are `oo` points of intersection. If only the *t* terms cancel and you are left with `0t=a` where `a != 0`, then there is no solution and the system is parallel but not coplanar (zero points of intersection).

The second method uses the vector equation of a line and the point-normal equation of a plane. Given the line and the plane defined by

`vec r_"L" = vec a + tvec m qquad and qquad vec n * (vec r_Pi - vec b) = 0`,

we can substitute the first into the second because we are looking for the point of intersection, which occurs when `vec r_"L" = vec r_Pi`:

`vec n * (vec a + tvec m - vec b) = 0.`

Rearranging this to solve for *t* gives us

`t = ((vec b - vec a) * vec n)/(vec n * vec m)`.

If this equation gives you a value for *t*, there is one point of intersection. If you get the indeterminate case, zero over zero, there are `oo` points of intersection. And if the numerator isn’t zero, then there are none. It isn’t hard to see why: if `vec m` is perpendicular to the normal of the plane, then it is *parallel* to the plane; and if `vec b - vec a` is parallel to the plane, then the position vector `vec a` lies on the plane.

Sometimes, in addition to the point of intersection, we want to find the *angle* of intersection. As with the angle between two planes, we only care about the acute angle. Let `phi` represent the angle between `vec m` and `vec n`, and let `theta` represent the angle of intersection. They are complementary: `phi + theta = 90º`. Since `cos phi = cos(90º-theta) = sin theta`, we can find the angle with

`theta_"acute" = sin^-1((|vec m * vec n|)/(|vec m||vec n|))`.