How can I deduce the shape of these vector equations? I would like to know how I should visualize and intuitively imagine the shapes these vector equations represent:


*

*$ \textbf{a} \cdot \hat{\textbf{b}} = m|\textbf{a}| $

*$|\textbf{a} - (\textbf{a} \cdot \hat{\textbf{b}}) \hat{\textbf{b}}| = k$
Shouldn't 1. be a plane that is tilted at some angle? I have completely no idea what 2. represents. 
 A: If ${\bf a}$ is the variable vector and ${\bf b}$ is a constant vector, then
$${\bf a}\cdot \hat{{\bf b}}=m|{\bf a}| \iff \left({\bf a}={\bf0}\right) \lor \left(\hat{{\bf a}}\cdot \hat{{\bf b}}=m\right)$$
which is a conical surface with axis in the direction of $\hat{{\bf b}}$ and angle $\cos^{-1}m$, assuming $|m| \le 1$.
The second one is a cylindrical surface with axis in the direction of $\hat{\bf b}$ and radius $k$.
For analytical proof, since the equations are in vector form, we can WLOG choosing our coordinate system so that $\hat{\bf b}=\hat{\bf z}$, and let ${\bf a}=(x,y,z)$.
Then for (1):
$$z=m\sqrt{x^2+y^2+z^2}$$
In cylindrical coordinates
$$z=z$$
$$x=\rho \cos\phi$$
$$y=\rho \sin\phi$$
it is
$$z^2=m^2(x^2+y^2+z^2)$$
$$(1-m^2)z^2=m^2(x^2+y^2)=m^2(\rho^2\cos^2\phi+\rho^2\sin^2\phi)=m^2\rho^2$$
$$(1-m^2)z^2=m^2\rho^2$$
or
$$\rho=\sqrt{\frac{1-m^2}{m^2}}|z|$$
which is clearly a conical surface.
For (2):
$$|(x,y,z)-(0,0,z)|=k$$
$$\sqrt{x^2+y^2}=k$$
or in cylindrical coordinates
$$\rho = k$$
which is clearly a cylindrical surface.
A: As far as intuition goes, maybe this will help. For any two vectors $\bf a$ and $\bf b$, the dot product $\bf a \cdot \hat b$ is the component of $\bf a$ in the direction of $\bf b$. If you "remove" that component from $\bf a$ you would be left with the part of it which is perpendicular to $\bf b$:
$$
\begin{align}
{\bf a}_{\parallel} &= ({\bf a} \cdot \hat {\bf b}) \hat {\bf b} \\
{\bf a}_\perp &= {\bf a} - ({\bf a} \cdot \hat {\bf b}) \hat {\bf b}
\end{align}
$$
[Note that to do any subtraction, first we need to make that "component" into an actual vector, hence the $\bf \hat b$'s multiplying the dot products in the above.]
Thus we have obtained a decomposition of the original vector $\bf a$ as
$$ {\bf a} = {\bf a}_\parallel + {\bf a}_\perp $$
where the two "parts" are vectors that are parallel and perpendicular to $\bf b$, respectively. Now you can see quite easily that your second vector equation is equivalent to $|{\bf a}_\perp| = k$, which is exactly what a cylinder is: a set of points that are a given perpendicular distance from an axis (defined here by the vector $\hat {\bf b}$). 
The first one is not that straightforward, but you can see it is equivalent to $|{\bf a}_\parallel| = m|\bf a|$ which, if you try to visualize all possible vectors $\bf a$, tells you that when each of them is "joined" to $\bf b$ (tail-to-tail at the origin), all the triangles formed are similar; that is what a cone is.
