Originally Posted by

**scorpion007** The question is:

Find the Cartesian form of the equation for the plane that is perpendicular to the space curve

$\displaystyle \mathbf{r}(t)=16t^{\frac{1}{4}}\mathbf{i} - 4t^{\frac{3}{4}}\mathbf{j}+t\mathbf{k}$

at the point P(32, -32, 16).

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Now the simple part of course, once I have the normal (n) to the curve at P, I can construct a plane equation via $\displaystyle \mathbf{n} \cdot \mathbf{v} = 0$ where $\displaystyle \mathbf{v} = \mathbf{Q}-\mathbf{P}$ (the vector from the plane's origin (P) to any point Q).

Of course, I can find the derivative of r(t), which is:

$\displaystyle \dot{\mathbf{r}}(t) = 4t^{-3/4}\mathbf{i}-3t^{-1/4}\mathbf{j}+\mathbf{k}$

That will give me the tangent vector at any time **t**.

I know that t = 16, from the 'k' component, so I can evaluate $\displaystyle \dot{\mathbf{r}}(16) = (\frac{1}{2}, -\frac{3}{2}, 1)$

But I need the normal vector (perpendicular to the curve).

Any tips?