*Calculus 3* Find the points of intersection (if there are any) of the tangent line to the curve r(t)= (t is a real number) at the point (3,-1,-3) with the plane x-y+z-1=0 *How would I solve this problem?

Question

*Calculus 3* Find the points of intersection (if there are any) of the tangent line to the curve r(t)=<3t^2,t^3,-3t^4> (t is a real number) at the point (3,-1,-3) with the plane x-y+z-1=0
*How would I solve this problem?

nikvist · Answer

$$r(t)=(3t^2,t^3,-3t^4)\quad,\quad (3,-1,-3)=r(-1)$$$$r'(t)=(6t,3t^2,-12t^3)\quad,\quad r'(-1)=(-6,3,12)$$Tangent line$$l(t)=r(-1)+(t-(-1))r'(-1)=(3,-1,-3)+(t+1)\cdot(-6,3,12)=$$$$=(3-6(t+1),-1+3(t+1),-3+12(t+1))=(-6t-3,3t+2,12t+9)$$Plane$$x-y+z-1=0$$$$-6t-3-(3t+2)+12t+9-1=0$$$$-6t-3-3t-2+12t+9-1=0$$$$3t+3=0\quad,\quad t=-1$$Point  of intesection $$l(-1)=r(-1)=(3,-1,-3)\quad start\,\,point\,\,:))$$

anonymous · Answer

Thanks that helped out alot :)