If $f$ and $\vec G$ are point functions finding the components of $\vec G$ normal and tangential to $f=0$ 
If $f\;and\;\vec G$ are point functions prove that the components of
  $\vec G$ normal and tangential to the surface $f=0$ are $\frac{(\vec
 G.\nabla f)\nabla f}{(\nabla f)^2}$ and $\frac{\nabla f \times(\vec G \times\nabla f)}{(\nabla f)^2}$

Now $\nabla f$ is normal to $f$ and $(\vec G .\nabla f)\nabla f$ projection  $G$ on $\nabla f$ therefore unit normal would be given by $\frac{(\vec
 G.\nabla f)\nabla f}{\nabla f}$ but the question is saying that the denominator would have a square term , I got a similar result for the tangential component also 
Please help : And also can someone explian the geometric significance of this problem , as to what is happening between the surfaces.
 A: The vector $\nabla f$ is normal to the surface
$f = 0, \tag 1$
but is not in general orthonormal to it, being as we typically will not have
$\vert \nabla f \vert = 1 \; \text{on} \; f = 0; \tag 2$
nevertheless, wherever 
$\nabla f \ne 0, \tag 3$
we may define the unit vector field
$\dfrac{\nabla f}{\vert \nabla f \vert } \tag 4$
which is orthonormal to $f = 0$, since
$\left \vert \dfrac{\nabla f}{\vert \nabla f \vert} \right \vert = \dfrac{\vert \nabla f \vert}{\vert \nabla f \vert} = 1; \tag 5$
it follows that $\vec G^\bot$, the component of $\vec G$ normal to $f = 0$, is given by
$\vec G^\bot = \left ( \vec G \cdot \dfrac{\nabla f}{\vert \nabla f \vert } \right ) \dfrac{\nabla f}{\vert \nabla f \vert } = \dfrac {\left ( \vec G \cdot \nabla f \right ) \nabla f}{\vert \nabla f \vert^2 }.  \tag 6$
From this equation we see that the component of $\vec G$ parallel to $f = 0$ is
$\vec G^\parallel = \vec G - \vec G^\bot = \vec G -\dfrac {\left ( \vec G \cdot \nabla f \right ) \nabla f}{\vert \nabla f \vert^2 } = 
\left ( \dfrac{\nabla f}{\vert \nabla f \vert } \cdot \dfrac{\nabla f}{\vert \nabla f \vert } \right ) \vec G - \left ( \vec G \cdot \dfrac{\nabla f}{\vert \nabla f \vert } \right ) \dfrac{\nabla f}{\vert \nabla f \vert }.  \tag 7$
We now apply the well-known vector identity
$\vec A \times (\vec B \times \vec C) = (\vec A \cdot \vec C) \vec B - (\vec A \cdot \vec B)\vec C \tag 8$
to the right-hand side of (7), yielding
$\vec G^\parallel = \left ( \dfrac{\nabla f}{\vert \nabla f \vert } \right ) \times \left (\vec G \times \dfrac{\nabla f}{\vert \nabla f \vert } \right ) = \dfrac{\nabla f \times (\vec G \times \nabla f)}{\vert \nabla f \vert ^2}, \tag 9$
as desired.
