Reducing nonlinear equations to obtain simplified form

Question

Apologies in advance if my question has been posed in a nonstandard way.

I am currently reading a journal article that contains the following two equations:

$\lambda_{u}\cos \gamma_{T}-(\lambda_{v}\cos\psi_{T}+\lambda_{w}\sin\psi_{T})\sin\gamma_{T}=0$

$-\lambda_{v}\sin\psi_{T}+\lambda_{w}\cos\psi_{T}=0$

Using the above two equations, the author states that the following three results can be obtained:

$\sin\gamma_{T}=\lambda_{u}/\lambda_{V}\\ \cos\gamma_{T}\cos\psi_{T}=\lambda_{v}/\lambda_{V}\\ \cos\gamma_{T}\sin\psi_{T}=\lambda_{w}/\lambda_{V}$

where $\ \lambda_{V}=\sqrt{\lambda_{u}^{2}+\lambda_{v}^{2}+\lambda_{w}^{2}}$. After several attempts at deriving these results, I have unfortunately not been able to obtain these results. I would be extremely grateful if someone could kindly take a look at the above relationships and advise whether these results can indeed be obtained.

Many thanks.

Answer 1

Write your two equations in the following form:

cosψ·λv + sinψ·λw = cotγ·λu
-sinψ·λv + cosψ·λw = 0

and solve them to get

λv = cotγ·λu·cosψ
λw = cotγ·λu·sinψ

It follows that λV = λu·√(1+cot²γ) = λu/sinγ,
whence the values of the the ratios λu/λV, λv/λV, λw/λV follow readily.

Answer 2

Let $\theta=\gamma_T$ be the azimuthal angle and $\phi=\psi_T$ the axial angle in a spherical coordinate system $(r,\theta,\phi)$. Let $\boldsymbol{\lambda}=-\lambda_v\mathbf{\hat{x}}-\lambda_w\mathbf{\hat{y}}+\lambda_u\mathbf{\hat{z}}$ where $\{\mathbf{\hat{x}},\mathbf{\hat{y}},\mathbf{\hat{z}}\}$ are the Cartesian unit vectors. Then the two given equations are

$$\begin{align}\boldsymbol{\lambda}\cdot\mathbf{\hat{r}}&=0\\ \boldsymbol{\lambda}\cdot\boldsymbol{\hat{\phi}}&=0 \end{align}$$ which implies $$\begin{align} \boldsymbol{\hat{\lambda}}=\pm\boldsymbol{\hat{\theta}} \end{align}$$

where the $\{\mathbf{\hat{r}},\boldsymbol{\hat{\theta}},\boldsymbol{\hat{\phi}}\}$ are the spherical-coordinate unit vectors and $\boldsymbol{\hat{\lambda}}$ is the unit vector in the $\boldsymbol{\lambda}$ direction. The results you give correspond to $\boldsymbol{\hat{\lambda}}=-\boldsymbol{\hat{\theta}}\text{.}$