Finding the directional derivative parallel to an intersection of planes

Question

We must find the directional derivative of $ f(x,y,z) = x^2 + 2xyz -y^2 $ at $ (1,1,2) $ in a direction parallel to the straight line $ \frac{x-1}{2} = y-1 = \frac{z-2}{-3} $

The straight line is an intersection of 2 planes but in 3 variables, how do I write this in the form of a vector so I can find the directional derivative?

Answer 1

The direction vector of the line is $(2, 1, -3)$, the denominators of the three parts.

The parameterised form of the line is

$$x=1+2t, y=1+t, z=2-3t$$

which is

$$\begin{pmatrix}x\\y\\z\end{pmatrix}=\begin{pmatrix}1\\1\\2\end{pmatrix}+t\begin{pmatrix}2\\1\\-3\end{pmatrix}$$

Answer 2

That equation is not an intersection of planes, but just the standard form of a line in 3D space. You can see by setting

$$ \frac{x-1}{2} = y - 1 = \frac{z-2}{-3} = t $$

Solving for $x$, $y$, $z$ each in terms of $t$ yield

$$ x = 1 + 2t $$ $$ y = 1 + t $$ $$ z = 2 - 3t $$

which is, of course, the parametric form