Normal vector to surfaces in 3D space verification

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Book A claims that for $ z=f(x,y)$ , the standard normal to the surface of the graph z is given the vector $<-\frac{\partial{f}}{\partial{x}},-\frac{\partial{f}}{\partial{y}},1>$. In what direction does $ <-\frac{\partial{f}}{\partial{x}},-\frac{\partial{f}}{\partial{y}},1>$ point? Does $ <-\frac{\partial{f}}{\partial{x}},-\frac{\partial{f}}{\partial{y}},1>$ point in the positive $ z$ -axis or negative z-axis?

Book B claims that for the graph of z, $<\frac{\partial{f}}{\partial{x}},\frac{\partial{f}}{\partial{y}},1>$ is the standard normal to the graph $ z=f(x,y)$. In what direction does the vector $<\frac{\partial{f}}{\partial{x}},\frac{\partial{f}}{\partial{y}},1>$ which is similar to a ray point?

Book A claims that the outward normal to a sphere is $ (-a^2sin(\phi)) (sin(\phi)cos(\theta),sin(\phi)sin(\theta),cos(\phi))$

Book B claims that the outward normal to a sphere is $ (a^2sin(\phi)) (sin(\phi)cos(\theta),sin(\phi)sin(\theta),cos(\phi))$

Is book A or B right?

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A normal vector can be found by taking a cross product of a pair of vectors that are tangent to the surface. Any pair will do as long as they are not parallel/anti-parallel to each other.

For a small $\delta$, one such pair of vectors is $$ u = \begin{pmatrix} \delta \\0 \\ \delta \frac{\partial f}{\partial x} \end{pmatrix} \qquad \textrm{and} \qquad v = \begin{pmatrix} 0 \\\delta \\ \delta \frac{\partial f}{\partial y} \end{pmatrix} $$

We can take cross product of these two vectors to be either $u \times v$ or from $v \times u$. Either of these will lead to a valid normal. The rule for cross product is that $v \times u = - (u \times v)$ so the two normals will point in opposite directions. This makes sense because we are considering the surface to be locally like a flat plane.

$$ u \times v = \begin{pmatrix} - \delta^2 \frac{\partial f}{\partial x} \\ - \delta^2 \frac{\partial f}{\partial y} \\ \delta^2 \end{pmatrix} = \delta^2 \begin{pmatrix} - \frac{\partial f}{\partial x} \\ - \frac{\partial f}{\partial y} \\ 1 \end{pmatrix} $$ which gives the vector $\left(- \frac{\partial f}{\partial x}, - \frac{\partial f}{\partial y}, 1\right)^T$ as pointing in the normal direction.

The rule for cross product that $v \times u = - u \times v$ which means that $\left(\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, -1\right)^T$ is also in the normal direction.

If the surface is closed, some texts describe these normals as 'inward' and 'outward'. For a function depending on $z$, just use the dot product to see which of these directions is closer to the direction of the positive $z$ axis.

Based on the above, book B looks incorrect to me.

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Hint for part 1: What's the dot product with a unit vector in the positive $z$ direction? (Note: you've written the normal wrong -- check those variables!)

Hint for part 2: What you've written is certainly incorrect. If you changed the final "$1$" to a "$-1$", it'd be better.

As to which is "standard", it depends on your book, mostly, but I'd say that the first is favored, for if you write $$ F(x, y) = (x, y, f(x, y)), $$ which is a pretty natural thing to do, then $F$ is a natural parameterization of the surface, and the standard orientation of the $xy$-plane ($x$ first, $y$ second) then transforms, via the differential of $F$, to the BookA version.

(This is an answer to the version of the question that reads as shown below (or, to be honest, the previous one). It's changed 4 times as I was writing the answer, and I'm not willing to wait for it to stabilize.)

Book A claims that for $ z=f(x,y)$ , the standard normal to the surface of the graph z is given the vector $<-\frac{\partial{f}}{\partial{x}},-\frac{\partial{f}}{\partial{y}},1>$. In what direction does $ <-\frac{\partial{f}}{\partial{y}},-\frac{\partial{f}}{\partial{x}},1>$ point? Does $ <-\frac{\partial{f}}{\partial{x}},-\frac{\partial{f}}{\partial{y}},1>$ point in the positive $ z$ -axis or negative z-axis?

Book B claims that for the graph of z, $<\frac{\partial{f}}{\partial{x}},\frac{\partial{f}}{\partial{x}},1>$ is the standard normal to the graph $ z=f(x,y)$. In what direction does the vector $<\frac{\partial{f}}{\partial{x}},\frac{\partial{f}}{\partial{y}},1>$ which is similar to a ray point?

Book A claims that the outward normal to a sphere is $ (-a^2sin(\phi)) (sin(\phi)cos(\theta),sin(\phi)sin(\theta),cos(\phi))$

Book B claims that the outward normal to a sphere is $ (a^2sin(\phi)) (sin(\phi)cos(\theta),sin(\phi)sin(\theta),cos(\phi))$

Is book A or B right?