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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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up vote 1 down vote accepted

Yes, k is the 'up' unit vector (the same up that give the sense of inclination). And I will assume that by * you mean the dot product. We can think of this in a few ways.

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 $< 1/2, 1/2, 1/2> * \hat{k} = < 1/2, 1/2, 1/2> \cdot \;\hat{k} = \frac{1}{2}$

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one question @mixedmath from here simple 1/2k is 1/2 yes?because we need number to define angle or arcoss(something) and also if instead of all 1/2 if we would have like this <1/2,2/3,1/4> then dot product with k will be simple 1/4*k yes? – dato datuashvili Jul 31 '11 at 7:44
thanks for answer first of all – dato datuashvili Jul 31 '11 at 7:44
@user: you're absolutely. It's just $\frac{1}{2}$. – mixedmath Jul 31 '11 at 8:44

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