angle of inclination

I know that if

$F(x,y,z)=0$

is a surface, then the angle of inclination at the point $(x_0, y_0, z_0)$ is defined by the angle of inclination of the tangent plane at the point or

$\cos(A)=\dfrac{\nabla F(x_0,y_0,z_0)*k}{|\nabla F(x_0,y_0,z_0)|}$

my question is just what is a k? Is it about i,j,k? Which are just component of vectors? or for example what is equal to following $<1/2,1/2,1/2>*k$? and why? thanks

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Recall that we can define cos as $cos(\theta) = \dfrac{ A \cdot B}{|A||B|}$, where $\theta$ is the orthogonal to the surface (why? because the gradient is perpendicular to level curves). So taking the dot product of this orthogonal with the up direction, and then dividing by the magnitude (k is already a unit vector), will yield the cosine of the angle of inclination, just as the above formula works.
So in your question, I would presume that $< 1/2, 1/2, 1/2> * \hat{k} = < 1/2, 1/2, 1/2> \cdot \;\hat{k} = \frac{1}{2}$
@user: you're absolutely. It's just $\frac{1}{2}$. – mixedmath Jul 31 '11 at 8:44