Maximum angle between vectors

Question

Consider 3 vectors $\textbf{v},\textbf{v}',\textbf{u}$ related by

$$\textbf{v}=\textbf{u}+\textbf{v}'$$

Let $\theta$ be the angle between $\textbf{v}$ and $\textbf{u}$ and let $\phi$ be the angle between $\textbf{v}$ and $-\textbf{v}'$.

For what angle $\theta$ is the angle $\phi$ a maximum? The magnitudes $v,u$ are given and you can express the angle in terms of them.

I started by doing the dot product of the equation with itself getting

$$v^{2}=vu\cos\theta+vv'\cos\phi=u^{2}+v^{'2}+2uv'\cos (\pi/2-\phi-\theta)$$

I thought I would then find the derivative of this expression with respect to $\theta$ and set $d\phi/d\theta=0$. That gives

$$-vu\sin\theta=2uv'\sin(\pi/2-\phi-\theta)$$ or

$$v\sin\theta=2v'\sin(\phi+\theta-\pi/2)=-2v'\cos(\phi+\theta)$$

which seems overdetermined.

Answer 1

Unless you're working under peculiar constraints, doing this algebraically is almost certainly a tactical error. Instead, draw a diagram:

      A
     / \
   u/   \v
  |/     \|
  B------>C
      v'

The angles are $\phi$ at C and $\theta$ at A.

$\phi$ will certainly be a maximum if we can make it $\pi$ which happens if $|u|>|v|$ and $\theta=0$.

Then consider the case $|u|<|v|$. If we have already decided on the positions of A and C, the possible positions of B is a circle with center A. Among those, the ones that maximize the angle at C are the ones where BC is tangent to the circle -- which are the ones where the angle is B is right!

Basic trigonometry then gives us $|u|=|v|\cos\theta$.

Answer 2

$$ \vec v = \vec u -(-\vec v') = \vec v = \vec u -\vec w $$

so now

$$ \vec u\cdot\vec v = ||\vec u||^2-\vec u\cdot\vec w $$

or

$$ ||\vec u||||\vec v||\cos\theta=||\vec u|| - ||\vec u||||\vec w||\cos\phi $$

and then assuming that $\vec u\cdot\vec w \ne 0$

$$ \phi = \arccos\left(a+b\cos\theta\right) $$

and now deriving

$$ \frac{d\phi}{d\theta} = \frac{b \sin (\theta )}{\sqrt{1-(a+b \cos (\theta ))^2}} = 0 $$

which gives $\theta = 0 + k\pi$