# Find normal lines which pass via origin

1. Oct 18, 2011

### athrun200

1. The problem statement, all variables and given/known data
1. Find all the points $P_{0}$ on the surface z = 2 − xy at which the normal line passes
through the origin.

2. Relevant equations

3. The attempt at a solution
See the photo.
It seems as long as I can find a,b and c, then the question is done.
However I don't know how to do that

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2. Oct 18, 2011

### HallsofIvy

Yes, writing the surface as $f(x,y,z)= z+xy= 2$ gives the normal at any point $(x_0, y_0, z_0)$ as $\nabla f= y_0\vec{i}+ x_0\vec{j}+ \vec{k}$. In order to give a line that passes through the origin, that vector must be parallel to the "position" vector of the point: $x_0\vec{i}+ y_0\vec{j}+ z_0\vec{k}$ and so must satisfy $y_0= x_0t$, $x_0= y_0t$, and $z_0= t$ for some t. The two equations $y_0= x_0t$ and $x_0= y_0t$ give $x_0^2= y_0^2$ so that $y_0=\pm x_0$. With the last, $z_0= t$ we must have $t= z_0$ and so $y_0= x_0t= x_0z_0$. Putting those together, $y_0= x_0= x_0z_0$ would give $z_0= 1$ so that $1= 2- x_0y_0$ $x_0y_0= 1$ so that $x_0= y_0= 1$ or $x_0= y_0= -1$. That is, two such points are (1, 1, 1) and (-1, -1, 1). Or $y_0= -x_0= x_0z_0$ so we must have $z_0= -1$ and then $x_0y_0= 3$ which is not possible since $x_0$ and $y_0$ are of different signs.

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