Thread: normal plane to the curve intersecting a plane

Hey everyone,
This question is driving me mental. Sorry for not using latex, I swear I will figure it out soon

ok here it is.

At what point on the curve r(t) = (t^3,3t,t^4) is the normal plane parallel to the plane 3x+3y-4z = 9 (the normal plane is the plane through the point r(t) which is normal to r(t))

I usually explain what I have tried already, but none of it is making any sense, so I decided to spare myself the humiliation.

Thanks ahead of time

Originally Posted by ubique

Hey everyone,
This question is driving me mental. Sorry for not using latex, I swear I will figure it out soon

ok here it is.

At what point on the curve r(t) = (t^3,3t,t^4) is the normal plane parallel to the plane 3x+3y-4z = 9 (the normal plane is the plane through the point r(t) which is normal to r(t))

The normal plane is perpendicular to the tangent. So you want to find the value(s) of t for which the tangent to r(t) is a scalar multiple of the vector (3,3,–4) (because that vector is perpendicular to the plane 3x+3y-4z = 9.

I was on the right track, I just didnt fit it all together. Thanks a lot