Finding the vectorial expression for the mutual slant of two cones with a common vertex

Solution 1:

Start by parameterizing $\mathbf{k_2}$ in terms of $\mathbf{k_1}$. Let $\mathbf{u_1}, \mathbf{u_2}$ be two mutually orthogonal unit vectors that are also orthogonal to $\mathbf{k_1}$, then any vector $\mathbf{k_2}$ on the cone with axis $\mathbf{k_1}$ and semi-vertical angle $\alpha_{12} $ can be written as

$\mathbf{k_2} = \cos \alpha_{12} \mathbf{k_1} + \sin \alpha_{12} (cos \phi \mathbf{u_1} + \sin \phi \mathbf{u_2}) $

Now, you want $\mathbf{k_2}$ to lie on the cone that has axis $\mathbf{k_3}$, and semi-vertical angle $\alpha_{23}$, so the angle between $\mathbf{k_2}$ and $\mathbf{k_3}$ will be $\alpha_{23} $

So you need to solve the equation

$ \mathbf{k_2} \cdot \mathbf{k_3} = \cos \alpha_{23} $

for $\phi$. Upon plugging in the expression of $\mathbf{k_2}$ the above equation becomes

$ \cos \alpha_{12} (\mathbf{k_1} \cdot \mathbf{k_3}) + \sin \alpha_{12} \left(cos \phi (\mathbf{u_1} \cdot \mathbf{k_3} ) + \sin \phi (\mathbf{u_2} \cdot \mathbf{k_3})\right) = \cos \alpha_{23} $

which is of the form

$ a \cos \phi + b \sin \phi = c $

and can be solved easily (see below). It will result in two solutions if $\alpha_{12} + \alpha_{23} \gt \beta $, one solution if $\alpha_{12} + \alpha_{23} = \beta $, and no solutions if $\alpha_{12} + \alpha_{23} \lt \beta $.

Now for the step-by-step solution of $ a \cos \phi + b \sin \phi = c $, divide through by $\sqrt{a^2 + b^2} $, resulting in

$\dfrac{a}{\sqrt{a^2 + b^2}} \cos \phi + \dfrac{b}{\sqrt{a^2 + b^2}} \sin \phi = \dfrac{c}{\sqrt{a^2 + b^2}} $

Define angle $\psi$ such that $\cos \psi = \dfrac{a}{\sqrt{a^2 + b^2}} $ and $\sin \psi = \dfrac{b}{\sqrt{ a^2 + b^2}} $, then

$ \cos \phi \cos \psi + \sin \phi \sin \psi = \dfrac{c}{\sqrt{a^2 + b^2}} $

The left hand side is simply $\cos(\phi - \psi) $, hence

$ \phi = \psi \pm \cos^{-1} \dfrac{c}{\sqrt{a^2 + b^2}} $

There will two solutions if $| c | \lt \sqrt{a^2 + b^2} $ , one solution if $ |c| = \sqrt{a^2 + b^2} $ and no solutions if $|c| \gt \sqrt{a^2 + b^2} $