In SVM Algorithm, why vector w is orthogonal to the separating hyperplane?
Geometrically, the vector w is directed orthogonal to the line defined by $w^{T} x = b$. This can be understood as follows: 

First take $b = 0$. Now it is clear that all vectors, $x$, with vanishing inner product with $w$ satisfy this equation, i.e. all vectors orthogonal to w satisfy this equation. 

Now translate the hyperplane away from the origin over a vector a. The equation for the plane now becomes: $(x − a)^{T} w = 0$, i.e. we find that for the offset $b = a^{T} w$, which is the projection of the vector $a$ onto the vector $w$. 

Without loss of generality we may thus choose a perpendicular to the plane, in which case the length $\vert\vert a \vert\vert = \vert b \vert /\vert\vert w\vert\vert$ which represents the shortest, orthogonal distance between the origin and the hyperplane.

Hence the vector $w$ is said to be orthogonal to the separating hyperplane.