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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)

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A normal plane at a point on a curve: (1) passes through the point, and (2) has its normal vector parallel to the tangent vector to the curve. You found (2), now incorporate (1) and you're done. –  vadim123 May 4 '13 at 18:10
Would I just set t=1 and solve the vector for the point? Then plug into the formula for a plane? @vadim123 –  jrquick May 4 '13 at 18:19

1 Answer 1

up vote 5 down vote accepted

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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