Orthogonality of the gradient to level sets of the function

I know that gradient is orthogonal to the tangent plane of any point in the level set,but is it true claiming that it is orthogonal to the level set?I think they are different in fact.

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What does "orthogonal to the level set" mean, otherwise? –  Dylan Moreland Mar 2 '12 at 2:39
i am not sure,look at en.wikipedia.org/wiki/Gradient the part about the level set and gradient and you will see –  Mathematics Mar 2 '12 at 2:50
Dylan's point is that "orthogonal to the level set" means "orthogonal to the tangent plane of the the level set". Claiming that one of these could be true without the other one also being true is self-contradictory. If you want to convince anyone that, no, they actually mean different things, the onus is on you to explain which different things you want them to mean. –  Henning Makholm Mar 2 '12 at 3:00

We have to assume that the gradient is nonzero at the point of consideration. Otherwise the level set may be rather nonsmooth and admit no tangent plane. Although it's true that zero vector is orthogonal to any subspace, it is not clear to me whether $\vec 0\perp \text{undefined}$ is a true statement.
Anyway, a common way to define $v\perp M$ at point $p$ (where $M$ is some subset of Euclidean space) is to require that $v\perp \gamma'(0)$ for every curve $\gamma:(-1,1)\to M$ such that $\gamma(0)=p$. Suppose $M=\{f(x)=c\}$. Then for every $\gamma$ as above we have $f\circ \gamma\equiv c$, hence (by the chain rule) $$0=\frac{d}{dt}_{|t=0}f(\gamma(t))=\nabla f(p)\cdot \gamma'(0)$$ as required.
As a matter of fact, the set of all vectors that can arise as $\gamma'(0)$ is exactly the tangent space of $M$ at $p$.