# 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 $\angle{POQ}$.

Thanks for all your help.

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## 1 Answer

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.

Also, to maximize the angle, you need to minimize, not maximize the cosine.

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A Doh! moment if there ever was one. It makes sense now. Thanks for your help. – mathguy80 Aug 6 '11 at 10:59
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