Given the surface $ xyz = 16 $ I have to find the tangent plane that is perpendicular to
$$
(x,y,z)^T = (1,2,3)^T + t(1,2,4)^T, \quad t \in \mathbb{R}.
$$
I have tried to find the gradient of the surface equation and equate it to $(1,2,3)$, an then finding the tangent point(s) and then finding the intercept. However, I get the wrong answer.
Best Answer
Assuming you mean your tangent plane should be normal to the given line, you are looking for a point where the normal to the surface (i.e., the gradient) is parallel to the direction vector of the line. That is, your point $(x,y,z)$ satisfies
\begin{align*} yz &= 1\lambda \\ xz &= 2\lambda \\ xy &= 4\lambda \end{align*} for some nonzero $\lambda$.
The point also satisfies $xyz=16$, so you get
\begin{align*} 16 &= \lambda x \\ 16 &= 2\lambda y \\ 16 &= 4\lambda z \end{align*} Solving this yields $x=4z$ and $y=2z$, and since $xyz=16$ we get \begin{align*} 8z^3&=16\\ \end{align*} So the point must be $(4\sqrt[3]{2},2\sqrt[3]{2},\sqrt[3]{2})$