How do you find the intersection of a line and a plane?

Question

Find intersection of line that goes through (2,-1,6) & is perpendicular to the plane r = (1,2,3) + s(1,3,5) + t(2,5,-1)? I'm not quite sure how to start solving this question.

Answer 1

Clearly you need a vector perpendicular to the plane. One way to get this is to convert the given parametric form for the plane to the form

Ax + By + Cz = D

because we know that the vector <A, B, C> is perpendicular to the plane.

Another way to write the parametric representation is

x = 1 + s + 2t

y = 2 + 3s + 5t

z = 3 + 5s - t

Solve, say, the y and z equations for s and t. I get

s = ((y-2) + 5(z-3))/28

t = ((5(y-2) - 3(z-3))/28

Put these into the x-equation. After simplifying, I get

28x - 11y + z = 9

So the vector <28, -11, 1> is perpendicular to the plane.

Now we can write a parametric of the line through (2,-1,6) and perpendicular to the plane:

x = 2 + 28u

y = -1 - 11u

z = 6 + u

Put these into the plane equation and solve for u, then put that into the parametric line equations. I get

x = 10/453, y = -101/453, z = 2686/453

which you may confirm satisfy the parametric equations for the plane with s = 201/453 and t = -322/453