Can someone help with Pset 4, Part II, problem #2? On their answer for (b.), I can't work out where they're getting their Z coordinate for the normal vectors to both planes. For both, it is -1, but I have no idea why! Can anyone explain?

Question

Can someone help with Pset 4, Part II, problem #2?  On their answer for (b.), I can't work out where they're getting their Z coordinate for the normal vectors to both planes.  For both, it is -1, but I have no idea why!  Can anyone explain?

joshdanziger23 · Answer

jamesond, I didn't see this explained in the course but what I think is going on is this:

You have $z=f(x,y)$ so $dz=f_x dx+f_y dy$, and so the tangent at $P=(x_0,y_0)$ satisfies $$z-z_0=f_x(x_0,y_0)(x-x_0)+f_y(x_0,y_0)(y-y_0).$$Rearranging and collecting all the constant terms into $k$ gives $$f_x(x_0,y_0)x+f_y(x_0,y_0)y-z=k.$$This is now the standard form for the equation for a plane (ie, $ax+by+cz=k$) with normal vector $<f_x(x_0,y_0),f_y(x_0,y_0),-1>$, which is the result being relied on the Pset.

Best wishes

Josh

anonymous · Answer

Oh I see now!  Thanks for pointing me to it!