# Find greatest value of an angle

I am not sure how to go about the last bit of this problem.

With respect to an origin $O$, the points $P$ and $Q$ have variable position vectors $p$ and $q$ respectively, given by, \begin{align} p &= (\cos t)i + (\sin t)j - k \\ q &= (\cos 2t)i - (sin 2t)j + \frac12k \end{align} where $t$ is a real parameter such that $0 \le t \le 2\pi$.

Show that $p \cdot q = \cos 3t - \frac12$ and hence, or otherwise, find the greatest value of the $\angle{POQ}$.

I have worked out the first part of this problem like below,

\begin{align} p &= (\cos t, \sin t, - 1) \\ q &= (\cos 2t, -\sin 2t, \frac12) \end{align}

\begin{align} p.q &= \cos t \cos 2t - \sin t \sin 2t - \frac12 \\ &= \cos (t + 2t) - \frac12 \\ &= \cos 3t - \frac12 \end{align}

For finding the angle,

\begin{align} |p| &= \sqrt{\cos^2 t + \sin^2 t} = 1 \\ |q| &= \sqrt{\cos^2 2t + \sin^2 2t} = 1 \end{align}

\begin{align} \cos \angle{POQ} &= \dfrac{p \cdot q}{|p| |q|} \\ &= \cos 3t - \frac12 \end{align}

This is the part I am unsure about. The problem asks for the maximum value of the angle and not the cosine of the angle. I could have gotten max cosine like,

Max $\cos 3t = 1$

Max $\cos POQ = 1 - \frac12 = 1/2$

Which would give $\frac\pi3$. But I don't think that's quite right, and doesn't match the given answer which is $2.82$ radians.

How do I get the max of $\angle{POQ}$.

The norms of $p$ and $q$ aren't right because you didn't include the last component in the sum of the squares of the components.