Suppose I have a vector $\vec{a}$ in 3 dimensions. I realise that there is no unique solution for vectors a fixed angle $\alpha$ away from vector $\vec{a}$ - instead there is a cone of possible solutions.

Is there a way to parameterise the solution by some parameter than ranges over $0-2\pi$?

Any vector $\vec{b}$ an angle $\alpha$ away must satisfy,

$$ \text{cos } \alpha = \frac{\vec{a} \cdot \vec{b}}{|a||b|}$$

If we are dealing with unit vectors, then this becomes,

$$ \text{cos } \alpha = a_1b_1 +a_2b_2 +a_3b_3$$

Is it then possible to parameterise this expression for all possible vectors?


  • $\begingroup$ Suppose $a$ is a unit vector. Find two unit vectors $a_2,a_3$ such that $a,a_2,a_3$ are orthogonal. Then the cone is generated by the vectors $t(a+ \tan \alpha ( \cos \theta a_2 + \sin \theta a_3))$, with $t \in \mathbb{R}$ and $\theta \in [0,2 \pi]$. $\endgroup$
    – copper.hat
    Dec 13, 2016 at 17:51
  • $\begingroup$ Can you explain how you got this? $\endgroup$ Dec 13, 2016 at 18:37
  • $\begingroup$ I added an answer below. $\endgroup$
    – copper.hat
    Dec 13, 2016 at 18:48

1 Answer 1


Suppose $a$ is a unit vector.

Take a plane perpendicular to $a$ and passing through $a$. The cone intersects this plane at a circle of radius $r=\tan \alpha$ centered at $a$.

That is, the cone passes through points $a + r v$, where $v \bot a$ is a unit vector, so the whole cone is described by the points $t(a+r v)$, where $t$ is real and $v$ is a unit vector perpendicular to $a$.

If $a_2,a_3$ are such that $a,a_2,a_3$ are orthonormal, then $v$ must have the form $v=\cos\theta a_2 + \sin \theta a_3$ for some $\theta$.

Hence the points are given by $\{ t (a + \tan \alpha(\cos\theta a_2 + \sin \theta a_3)) | t\in \mathbb{R} , \theta \in [0,2 \pi] \}$.

The cone is unique, but different choices of $a_2,a_3$ will give different parameterisations.


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