1. ## Tangent Planes

The question:

Show that the curve $\displaystyle r(t) = t^2\underline{i} + t\underline{j} + (5t - 4)\underline{k}$ is normal to the surface $\displaystyle 2x^2 + y^2 + 5z^2 = 8$ at the point (1, 1, 1).

My attempt:
I found the tangent plane for the second surface, which is:
$\displaystyle 2x + y + 5z = 8$

I then noticed that the curve is at point (1, 1, 1) when t = 1. I differentiated the curve, substituted t = 1, and got:

$\displaystyle 2\underline{i} + \underline{j} + 5\underline{k}$

Which is the same as the tangent plane for the second surface. I'm confused, how can it be normal to the surface if it has the same tangent? Or have I calculated it wrong?

2. Originally Posted by Glitch
Which is the same as the tangent plane for the second surface. I'm confused, how can it be normal to the surface if it has the same tangent? Or have I calculated it wrong?

Take into account that $\displaystyle (2,1,5)$ is a perpendicular vector to $\displaystyle 2x+y+5z=8$ .

3. Huh, now that I think about it, it is isn't it? >_<

Thanks.

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