Question

Find the equation of the plane perpendicular to the curve x=t^5, y=2t^2, z=3t at t=1

Answer 1

First find the point at t=1
(1,2,2)

Answer 2

then find the tangent vector
\[
(x'(t), y'(t), z'(t))= ( 5 t^4, 4 t, 3) \\
at \quad t=1 \\
(5,4,3)
\]

Answer 3

The equation of the plane is\[
5(x-1) + 4(y-2) + 3(z-2)=0
\]

Answer 4

THANK YOU!