Increasing the angle between two vectors Having two vectors $a$ and $b$ and with $b$ fixed, how would I go about increasing the angle between the two vectors by a specified $\Delta \theta$?
The dot product equation:
$$ a \cdot b = |a| \cdot |b| \ \cos \theta .$$
gives multiple solutions in the form of a cone for $3$-dimensional vectors.  I want the new vector $a'$ to be in the same plane spanned by $a$ and $b$. 
Also can this case be extended to arbitrary number of dimensions? 
 A: You can decompose $a$ into a part parallel to $b$ and a part perpendicular to $b$, then increase the perpendicular part.  The parallel part of $a$ is $|a| \cos \theta \hat{b}=\frac{a\cdot b}{b^2}b$.  The perpendicular part is the rest: $a-\frac{a\cdot b}{b^2}b$.  $\tan \theta$ is the ratio of the magnitudes of these: $\frac{|a-\frac{a\cdot b}{b^2}b|}{|\frac{a\cdot b}{b^2}b|}$, so you can increase the perpendicular part enough to give the desired $\theta$
A: I thought more about your problem in general, without the restriction. and I think it is an interesting one. Here's what I came up with, intuitively.
Given two vectors, $a$ and $b$ in $\mathbb R ^n$, choose your coordinate axes to allow $b$ to be on one of the axes$^1$. Now spin $a$ around the axis which $b$ is on. This will create the cone you describe in $\mathbb R ^3$, and some other "surface" in other dimensions. Points on this surface have the same angle with $b$ as $a$. Vectors on the same side of the surface as $b$ have a lesser angle with $b$, and points on the side of the surface without $b$ have a greater angle with $b$.
I don't know if something interesting might come from algebraically trying this out in, say, $\mathbb R ^4$, but it should not be too difficult. Actually, in $\mathbb R ^3$, the cone of vectors with angle greater than that between $a$ and $b$ exists only if the angle between $a$ and $b$ is obtuse. If it is acute, you get the complement of a cone (all of  $\mathbb R ^3$ minus the cone), and if the angle is $\frac \pi 2$, you get half of $\mathbb R ^3$.
If we take a two-dimensional "slice" of $\mathbb R ^3$, we would get either a triangle of infinite height, the complement of such a triangle, or a half-plane. In the higher dimensions, I would conjecture that a three-dimensional "slice" would always yield a cone, the complement of a cone, or a half-space, and a two-dimensional "slice" would therefore be similar to that already described.
$^1$ The purpose of this is just to make the algebraic manipulations easier. Obviously the choice of axes plays no direct role, and such a result is congruent to one in which $b$ is not placed on the axis.
