Calculus – Maximum Angle Between Vectors

calculusvectors

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.

Best Answer

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 $|u|=|v|\cos\theta$.

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