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If I have the space curve $r(t) = \langle t, t^2, t^3 \rangle$, how would I find an equation of the normal plane to $r(t)$ at the point $P(2,4,8)$?

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Think how to find normal vector of this plane –  userNaN Feb 8 '12 at 10:24

2 Answers 2

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  1. A plane containing the origin with normal vector $\mathbf{n}$ is given by $\mathbf{n}\cdot\mathbf{x}=0$ (this is simply a state-ment of orthogonality). As corollary, a plane containing the point $\mathbf{p}$ is given by $\mathbf{n}\cdot(\mathbf{x}-\mathbf{p})=0$.
  2. A plane normal to a curve $\gamma$ at a point $\mathbf{r}(t)=\mathbf{p}$ is exactly the plane containing the point $\mathbf{p}$ with normal vector given by the curve's tangent vector at this very point, $\mathbf{n}=\mathbf{T}(t)=\mathbf{r}\,'(t)$.
  3. For what $t$ is $\mathbf{r}(t)=(2,4,8)$? What is the derivative $\mathbf{r}\,'(t)$ evaluated at this particular time $t$?
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N.B. $\mathbf{n}\cdot(\mathbf{x}-\mathbf{p})=0$ is what is termed as the Hessian normal form of a plane. –  J. M. Feb 8 '12 at 10:57
    
I got x+8y+48z=162 Is this correct? –  penu Feb 8 '12 at 15:47
    
@user1008134: No. What's $\mathbf{n}$ and what's $\mathbf{p}$ for the plane here? –  anon Feb 8 '12 at 19:55

The normal vector at the point P is (1,4,12) and the normal plane is given by $x+4y+12z=114$ using Hessian normal form of a plane.

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