At what point on the curve $x = t ^ { 3 } , y = 9 t , z = t ^ { 4 }$ is the normal plane parallel to the plane $3 x + 9 y - 4 z = 4 ?$
A) $( - 1 , - 3,9 )$
B) $( 9,18 , - 2 )$
C) $( - 1 , - 9,1 )$
D) $( - 18,9,1 )$
E) $( - 9,1,1 )$

Question

Quizplus · Accepted Answer

The normal plane is parallel to the given plane when their normal vectors are scalar multiples. The normal vector to the curve is $(3t^2, 9, 4t^3)$. Setting it proportional to $(3, 9, -4)$ gives $t = -1$, leading to the point $(-1, -9, 1)$.

At What Point on the Curve x=t3,y=9t,z=t4x = t ^ { 3 } , y = 9 t , z = t ^ { 4 }x=t3,y=9t,z=t4

At What Point on the Curve $x = t ^ { 3 } , y = 9 t , z = t ^ { 4 }$