# Moving a 1 long segment in order to reach the highest point on a plane

We are standing in point $(0,0,0)$ on the plane $2x-2y-3z=0$. In order to achieve maximum height by walking a segment which length is $1$, on the plane, what direction we should move?

I was thinking about that the gradiant is the plane's normal, and it is obviously connected to it's projection on the plane.

How do I continue from this point? thanks.

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It might be clearer if you think about the plane as the graph of the function $f(x,y) = \frac{2}{3}x - \frac{2}{3}y$. What does the gradient, $\nabla f$, say then? – Antonio Vargas May 20 '12 at 7:12
The fastest climb would be in the direction $(0, 0, 1)$ but that is not along the plane. However, you can project this direction back onto the plane using the normal. – WimC May 20 '12 at 7:35

To get the direction of steepest ascent, you just take the vector $Z = (0,0,1)$ and project it onto your plane (in a direction along the plane normal). One way to do this is as follows. The unit upward plane normal is $N = (-2, 2, 3)/\sqrt{17}$, and the projection of Z onto this vector is $V = (Z \cdot N)N = (-6,6,9)/17$. So, the projection $U$ of $Z$ onto the plane is $Z-V$, which is $(6, -6,8)/17$. So, in tidier terms, you should move in the direction of the vector $(3,-3,4)$. With a step of length $1$, the highest point you can reach is $(3,-3,4)/\sqrt{34}$. This is at a height of $4/\sqrt{34}$, which is roughly $0.685994341$.
The downward normal to the plane is $(2,-2,-3)$. So the direction of maximum slope is just $(2,-2,a)$, where you choose $a$ so that $(2,-2,a)$ lies in the plane.