Outward pointing normal vector to a $k$ manifold in $\mathbb{R}^n$

What do we mean when we say that $\mathbf{N}$ is normal to the the manifold:$$\mathbf{X}:\mathbb{R}^k\to\mathbb{R}^n$$? How do we determine it? How to verify that it's an outward pointing normal?

I assume that $X:\mathbb R^k\to\mathbb R^n$ is a differentiable map and $N:\mathbb R^n\to\mathbb R^n$ is a function (a vector field). In this case $N$ is normal to the manifold $X(\mathbb R^k)$ (which is reasonably a manifold if the derivative of $X$ is injective everywhere) if for every $x\in\mathbb R^k$ and $v\in\mathbb R^k$ the vectors $N(X(x))\in\mathbb R^n$ and $DX(x)v\in\mathbb R^n$ are orthogonal. This is because the tangent plane of $X(\mathbb R^k)$ at $X(x)$ is $X(x)+DX(\mathbb R^k)$.
If you have been given explicit functions $X$ and $N$, the condition is (in principle at least) easy to verify by calculation. (It is enough if $N$ is defined on $X(\mathbb R^k)$.)
• @user215258, it would be best to know the maps $X$ and $N$ explicitly. Choosing which way is inward and which is outward is often quite arbitrary (and doesn't make any sense if $n-k\geq2$), although in some cases it is natural. – Joonas Ilmavirta Feb 13 '15 at 19:17