Find the scalar equation of the plane that is orthogonal to the curve parameterised by $x(t)=t^3,\ y(t)=t^2,\ z(t) = t$ at the point where $t=-1$
Write your equation in the form $ax+by+cz=k$ where $(a, b, c)$ is a unit vector.
I began by finding the coordinate on the curve at the point $t = -1$, which is $(-1, 1, -1)$.
Don't we need to be given a normal vector? which I could dot with the point $(-1, 1, -1)$ to get my constant?
A tangent vector to the curve at $t=-1$ is $\mathbf{T}=\left(x'(-1),y'(-1),z'(-1))\right)=\ldots =(3,-2,1)$ or $\mathbf{t}=\dfrac{1}{\sqrt{14}}(3,-2,1)$ (unitary). Now, our plane is $$\frac{3}{\sqrt{14}}(x+1)+\frac{-2}{\sqrt{14}}(y-1)+\frac{1}{\sqrt{14}}(z+1)=0\Leftrightarrow\ldots$$