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I'm given the surface $ x^3-2y^2+z^2=27 $ and have to find where the tangent plane is perpendicular to the line described by \begin{align*} x &= 3t-5 \\ y &= 2t+7\\z&=1-t\sqrt2\end{align*}

I started with letting $f(x,y,z) =x^3 -2y^2 +z^2$

I know that the equation for the tangent plane at some vector $\vec{a} =(x_0, y_0, z_0)$ is

$$ \begin{align*} 0 &= f_x(\vec{a}) (x-x_0)+ f_y(\vec{a}) (y-y_0) + f_z( \vec{a}) (z-z_0) \\ 0 &= 3x_0^2(x-x_0)-2y_0(y-y_0)+2z_0(z-z_0)\end{align*} $$

and I'm not quite sure where to take it from here. I've tried plugging in the parametrized equations of the aforementioned line into the equation above, but it gave me an equation with more unknowns.

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Clearly one direction of the line is given by ${\bf n} = (3,2,-\sqrt{2})$. If the plane is perpendicular to the line, then the normal direction to the plane, the gradient, must be parallel to our $\bf n$.

You want $(x_0,y_0,z_0)$ in the surface such that $\nabla f (x_0, y_0, z_0) = \lambda {\bf n}$ for some $\lambda \in \Bbb R$. You want to solve $$\begin{cases} 3x_0^2 = 3\lambda \\ -4y_0 = 2\lambda \\ 2z_0 = -\lambda \sqrt{2} \\ x_0^3 - 2y_0^2 + z_0^2 = 27 \end{cases}$$ Solve for $x_0, y_0 $ and $z_0$ in terms of $\lambda $ in the first three equations. Substitute in the fourth. Find $\lambda $. Go back and find $x_0, y_0$ and $z_0$. Now find the tangent plane at this point.

(It is possible that more than one point verify these relations, so that the problem might have more than one solution.)

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  • $\begingroup$ I clearly over thought this $\endgroup$ – rgarci0959 Jun 28 '15 at 4:48

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