Finding the tangent planes to a torus parallel to $3x+4y-5z=20$.

calculusmultivariable-calculus

I've been tasked with finding all the points on the torus $S$ given by
$$
\left(6-\sqrt{x^2+y^2}\right)^2+z^2=2
$$
at which the tangent plane is parallel to the plane $P$ given by $3x+4y-5z=20$ (which is, itself, tangential to $S$ at $(3,4,1)$). The method that I have employed gives me another point, but since $S$ is a torus, and $P$ is not parallel to any of $x=0$, $y=0$ or $z=0$, I know there must be at least two more.

The method that I have employed is as follows. Let $f(x,y,z)=\left(6-\sqrt{x^2+y^2}\right)^2+z^2$, then we know that a vector normal to $S$ at $(x_0,y_0,z_0)$ is given by $\nabla f(x_0,y_0,z_0)$. We use this fact, in conjunction with the fact that for two planes to be parallel, their normal vectors must also be parallel, to arrive at the condition
$$
\nabla f(x_0,y_0,z_0)=\lambda\begin{pmatrix}3\\4\\-5\end{pmatrix}
$$
for some scalar, $\lambda\in\mathbb{R}$. We compare the components of these vectors to obtain expressions for $z$ and $x$ in terms of $y$, and then substitute these expressions into the equation for $S$ to find that $y=4$ or $y=28/5$. Since we already have a point at which $y=4$, we discard this value, and hence we have that another tangent to $S$ which is parallel to $P$ is given by $5x/5+8y/5-2z=16$. Graphically, these are the two tangent planes in question
enter image description here

However, as we can see, there is an entire other half of the torus, one which I'm confident has two more points at which the tangent planes are parallel to $P$. How come I have missed these points? Any help is appreciated.

Best Answer

Note that the planes parallel to $3x+4y-5z=20$ are all of the form $3x+4y-5z=k$ for some $k$.

The gradient is $$ (-\frac { 2(6-\sqrt {x^2+y^2})x}{\sqrt {x^2+y^2}},-\frac {2 (6-\sqrt {x^2+y^2})y}{\sqrt {x^2+y^2}},2z). $$

You are looking for points in the torus where the gradient is a scalar multiple of $(3,4,-5)$. This gives you four equations (three from the gradient and one from the torus):

\begin{align} z&=-5\lambda\\ (6-\sqrt {x^2+y^2})x&=-3\lambda\sqrt {x^2+y^2} \\ (6-\sqrt {x^2+y^2})y&=-4\lambda\sqrt {x^2+y^2}\\ (6-\sqrt {x^2+y^2})^2+z^2&=2 . \end{align} Dividing the second by the third you get $$ \frac{x}{y}=\frac34, $$ so $y=4x/3$. Then $$ \sqrt{x^2+y^2}=\sqrt{x^2+\frac{16x^2}{9}}=\frac{5|x|}{3}. $$ So now the system is \begin{align} z&=-5\lambda\\ (6-\frac{5|x|}{3})x&=-5\lambda{|x|} \\ (6-\frac{5|x|}{3})^2+z^2&=2 . \end{align} From the first and second equation you now get

$$ z=- {5\lambda} =5\,\frac{(6-\frac{5|x|}{3})x}{{5|x|}{}}=\frac{(6-\frac{5|x|}{3})x}{{|x|}{}}. $$ Now the last equation is $$ (6-\frac{5|x|}{3})^2+\left(\frac{(6-\frac{5|x|}{3})x}{{|x|}{}}\right)^2=2, $$ so $$ 2(6-\frac{5|x|}{3})^2=2. $$ This gives $$ x=\pm\frac{18\pm3}5,\ \ \ y=\frac{4x}3=\pm\frac{24\pm4}5,\ \ \ z=\pm1.$$

So you get

  • $(\tfrac{21}5,\tfrac{28}5,-1)$ and the plane is $3x+4y-5z=40$

  • $(3,4,1)$ and the plane is $3x+4y-5z=20$

  • $(-\tfrac{21}5,-\tfrac{28}5,1)$ and the plane is $3x+4y-5z=-40$

  • $(-3,-4,-1)$ and the plane is $3x+4y-5z=-20$

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