An Euclidean geometry problem

Question

Consider two sets of three unit, coplanar vectors $(\mathbf{T}_1,\mathbf{N}_1,\mathbf{S}_1)$ and $(\mathbf{T}_2,\mathbf{N}_2,\mathbf{S}_2)$. Such that $\mathbf{T}_1\cdot\mathbf{N}_1=\mathbf{T}_2\cdot\mathbf{N}_2=0$, and $\mathbf{S}_1$ makes an angle $0<\sigma<\pi$ with $\mathbf{N}_1$, and similarly $\mathbf{S}_2$ makes an angle $0<\rho<\pi$ with $\mathbf{N}_2$. We align $\mathbf{S}_1$ and $\mathbf{S}_2$ such that $\mathbf{S}_1=-\mathbf{S}_2$, and impose some $\gamma\neq 0$ or $\pi$ such that $\mathbf{T}_1\cdot\mathbf{T}_2=\cos\gamma$.

The question is, what is $\mathbf{N}_1\cdot\left(\mathbf{T}_1\times\mathbf{T}_2 \right)$ in terms of $\sigma$, $\rho$ and $\gamma$ ?

I approached the problem by writing

\begin{align} 0=\mathbf{S}_1+\mathbf{S}_2=\sin\sigma\: \mathbf{T}_1+\cos\sigma\: \mathbf{N}_1+\sin\rho\: \mathbf{T}_2+\cos\rho\: \mathbf{N}_2 \end{align}

Then crossing with, say, $\mathbf{T}_2$ and dotting with, say, $\mathbf{N}_1$ we get an equation involving $\mathbf{N}_1\cdot\left(\mathbf{T}_1\times\mathbf{T}_2 \right)$ and $\mathbf{N}_1\cdot\left(\mathbf{N}_2\times\mathbf{T}_2 \right)$. Crossing with, say, $\mathbf{T}_2$ and dotting with, say, $\mathbf{N}_2$ we get another equation and so on, until we have a homogeneous system of 6 equations involving the 6 mixed products. On top of that we can write $\mathbf{T}_1$ in terms of the basis $(\mathbf{T}_2,\mathbf{N}_2,\mathbf{T}_2\times\mathbf{N}_2)$ and get another equation from which we can extract a non-trivial solution. The answer is

\begin{align} \left(\mathbf{N}_1\cdot\left(\mathbf{T}_1\times\mathbf{T}_2 \right)\right)^2=1-\sec^2\sigma\left(\cos^2\gamma+\sin^2\rho+2\cos\gamma\sin\rho\sin\sigma \right) \end{align}

I'm looking for a more direct, geometrical way to the answer.

Answer 1

We can chose a coordinate system such that: $$ \mathbf{S}_1=(0,0,1),\quad \mathbf{N}_1=(\sin\sigma, 0,\cos\sigma),\quad \mathbf{T}_1=(\cos\sigma, 0,-\sin\sigma),\quad \mathbf{S}_2=(0,0,-1). $$ Vectors $\mathbf{N}_2$ and $\mathbf{T}_2$ lie then on a plane through $z$-axis, making an angle $\alpha$ with $x$-axis. Hence:

$$ \mathbf{N}_2=(\sin\rho\cos\alpha, \sin\rho\sin\alpha,-\cos\rho),\quad \mathbf{T}_2=(\cos\rho\cos\alpha, \cos\rho\sin\alpha,\sin\rho). $$ The condition $\mathbf{T}_1\cdot\mathbf{T}_2=\cos\gamma$ gives: $$ \cos\alpha={\cos\gamma+\sin\rho\sin\sigma \over\cos\rho\cos\sigma}. $$ We can then compute: $$ \mathbf{N}_1\cdot(\mathbf{T}_1\times\mathbf{T}_2)=\cos\rho\sin\alpha $$ and finally: $$ \big(\mathbf{N}_1\cdot(\mathbf{T}_1\times\mathbf{T}_2)\big)^2 =\cos^2\rho-{(\cos\gamma+\sin\rho\sin\sigma)^2 \over\cos^2\sigma}. $$

Answer 2

Since $S_1 = - S_2$, we can initially take

$ S_1 = j $

where $j$ the unit vector along the $Y$ axis

and

$S_2 = - j $

Now the angle between $N_1$ and $S_1$ is $\sigma$, therefore

$N_1 = - \sin \sigma \ i + \cos \sigma \ j $

($i$ is the unit vector along the $X$ axis)

and

$ T_1 = \cos \sigma \ i + \sin \sigma \ j $

In a similar fashion, we have

$ N_2 = \sin \rho \ i - cos \rho \ j $

and

$ T_2 = - \cos \rho \ i - \sin \rho \ j $

Now we'll rotate $T_2$ about the $ Y $ axis (to preserve the relation $S_1 = - S_2$). The rotate $T_2$ is given by

$ T'_2 = R_Y( \phi ) T_2 = R_Y(\phi) ( - \cos \rho \ i - \sin \rho \ j ) $

And this equals

$ T'_2 = - \cos \rho ( \cos \phi , 0, - \sin \phi ) - \sin \rho (0, 1, 0) $

which can be written as

$ T'_2 = ( - \cos \rho \cos \phi, - \sin \rho, \cos \rho \sin \phi ) $

Since $ T'_2 \cdot T_1 = \cos \gamma $

Therefore,

$\cos \gamma = -\cos \sigma \cos \rho \cos \phi - \sin \sigma \sin \rho $

It follows that

$ \cos \phi = - \dfrac{ \cos \gamma + \sin \sigma \sin \rho }{\cos \sigma \cos \rho } $

Now, from the triple product identity,

$ N_1 \cdot (T_1 \times T'_2 ) = T'_2 \cdot (N_1 \times T_1)$

and

$N_1 \times T_1 = - k $

where $k$ is the unit vector along the $Z$ direction.

Hence

$ N_1 \cdot (T_1 \times T'_2) = - \cos \rho \sin \phi $

From which it follows that

$ \bigg( N_1 \cdot (T_1 \times T'_2) \bigg)^2 = \cos^2 \rho \left( 1 - \left( \dfrac{\cos \gamma + \sin \sigma \sin \rho }{ \cos \sigma \cos \rho } \right)^2 \right)$

And this equals

$ \bigg( N_1 \cdot (T_1 \times T'_2) \bigg)^2 = \sec^2 \sigma \left( \cos^2 \sigma \cos^2 \rho - \cos^2 \gamma - \sin^2 \sigma \sin^2 \rho - 2 \cos \gamma \sin \sigma \sin \rho \right) $

Replacing $\sin^2 \sigma $ with $1 - \cos^2 \sigma $ gives

$\bigg( N_1 \cdot (T_1 \times T'_2) \bigg)^2 = \sec^2 \sigma \left( \cos^2 \sigma - \cos^2 \gamma - \sin^2 \rho - 2 \cos \gamma \sin \sigma \sin \rho \right) $

And this equals

$\bigg( N_1 \cdot (T_1 \times T'_2) \bigg)^2 = 1 - \sec^2 \sigma \left( \cos^2 \gamma + \sin^2 \rho + 2 \cos \gamma \sin \sigma \sin \rho \right) $