Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 $P(2,4,8)$?

share|cite|improve this question
Think how to find normal vector of this plane – Norbert Feb 8 '12 at 10:24
up vote 3 down vote accepted
  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 $t$?
share|cite|improve this answer
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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.