TITLE: A $\pm$ in a Vector Equation
QUESTION [0 upvotes]: I am following a derivation in a book by Bryson and Ho. The problem is for path calculation in a velocity field (wind). Wind is $w = \vec{w}$ position $r = \vec{r}$ and control heading $\hat{u} = \vec{\hat{u}}$ where $\hat{u} \cdot \hat{u} = 1$.
Aircraft velocity equation:
$$
\dot{r} = V \hat{u} + w
$$
Hamiltonian
$$
\mathcal{H} = \lambda \cdot (V \hat{u}  + w) + 
\mu (1 - \hat{u}\cdot\hat{u}) + 1
$$
Euler-Lagrange
$$
\dot{\lambda} = - \frac{\partial H}{\partial r} = - \nabla (\lambda \cdot w)
$$
$$
0 = \frac{\partial H}{\partial \hat{u}} = V \lambda - 2\mu\hat{u}
$$
Then, using the last equation above, the author says
$$
2\mu = \pm V |\lambda| \to \hat{u} = \pm \frac{\lambda}{|\lambda|}
$$
I understand the algebra except the $\pm$ part.. Why was that necessary? 
Thanks,

REPLY [1 votes]: We have that
$$V\lambda = 2 \mu \hat{u}$$
Which means that their direction is the same, so we can write that
$$\hat{u}=C \lambda$$
for some $C$. But because $\hat{u}$ is an unit vector, we must have that
$$|C|\, ||\lambda|| = 1$$
Which means that
$$|C| = \frac{1}{||\lambda||}$$
But $C$ can be positive and negative as well, that's why we write
$$\hat{u}=\pm |C| \lambda = \pm \frac{\lambda}{||\lambda||}$$
And we will pick the appropriate sign later (which will make the Hamiltonian zero, as I can see).
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