When is parametric line parallel to planes contained?

Question

Given the system of parametric equations

$$x + y + kz = 1$$ $$x + y + kz = 3$$ $$2x + 2y + z(1+k) = 1$$ $$x + z = 0$$

I interpret this as two always (for each k) parallel planes:

$$x + y + kz = 1$$ $$x + y + kz = 3$$

And a line:

$$2x + 2y + z(1+k) = 1$$ $$x + z = 0$$

If my calculations are correct, I find that the line can never intersect the planes, but when is it only parallel and when does it actually lie on one plane? If it lies on a plane, how can I decide on which one?

Answer 1

You’ll easily find out that the line’s equation is $$(0,1/2,0)^t+\lambda(2,k-1,-2)^t.$$

Now that line is parallel to the planes if their normal vector, namely $(1,1,k)$, is orthogonal to the line’s direction vector $(2,k-1,-2)^t$, i.e., if $k=1$. But in this case $(0,1/2,0)^t$ doesn’t belong to either plane so line and plane are parallel and disjoint.

If $k\neq1$, line and plane will intersect, so your calculations seem to be defective.

Answer 2

(I was wrong, there can be points of intersection)

We can put the line in parametric form by choosing two arbitrary pairs of values for $x$ and $y$, calculating $z$ by putting them in the line cartesian equation, then calculating the difference vector between them, in formulas:

$$x = x_0 + t (x_1-x_0)$$ $$y = y_0 + t (y_1-y_0)$$ $$z = z_0 + t (z_1-z_0)$$

In order to find the points of intersection, after the line has been put into parametric form, we can substitute the values of x, y and z into the equation of the plane(s).

That finds the value of $t$ for which the intersection happens, we can then substitute that value in the parametric line equation we found before to find the actual point of intersection.