Let $r(t) = (t^2, t, t^4)$. Find normal plane at point t = 1.
While working on this problem I have come to point where I am not sure what to do. I originally thought to use the formula $T=\frac{r'(t)}{|r(t)|}$
The result is the vector $<\frac{2t}{\sqrt{3}}, \frac{1}{\sqrt{3}}. \frac{4t^3}{\sqrt{3}}>$
Afterwards I believe I should plug the vector and points at t=1 into the equation for a plane, is this the correct path to the solution or have a gone astray? Thanks in advanced.
(This is a study exam, not homework)
$\endgroup$ 21 Answer
$\begingroup$You want the plane to be normal to $\langle \frac{2}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{4}{\sqrt{3}}\rangle$ (or just $\langle 2,1,4\rangle$). You also want the plane to pass through $(1,1,1)$. The equation is hence $(2,1,4)\cdot((x,y,z)-(1,1,1))=0$, or $2(x-1)+(y-1)+4(z-1)=0$, which can be rearranged to $2x+y+4z=7$.
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