<s>[INST] Find the unknown vectors of a parallelogram and equilateral triangle I am not looking for the answers, could someone help break the questions down into simpler terms.  I can find out what the answers are so that is not the goal.
Vectors $p$, $q$ and $r$ are represented on the diagram as shown:


*

*$BCDE$ is a parallelogram

*$ABE$ is an equilateral triangle

*$\left|p\right| = 3$

*Angle $ABC = 90^\circ$



(a) Evaluate $p\cdot(q + r)$
I am not sure what this means.  Is it saying evaluate Multiplying vector $p$ by the sum of $q$ plus $r$?
(b) Express ${\vec {AC}}$ in terms of $p$, $q$ and $r$
For this question, is it saying the path to $EC$ using the vectors.
Would it be $q + p + r$?
(c) Given that ${\vec{AE}\cdot\vec{EC} = 9{\sqrt{3}-{9\over2}}}$, find $\left|r\right|$
I don't understand this question, I am not sure how I find the length of $r$ from the values given.
I am specifically looking for the explanations of $a$ and $c$.  I don't understand how I can deduce $\left|r\right|$ from the values.
 [/INST] (a) First distribute the dot product over addition as in
$$p\cdot(q+r) = p\cdot q + p\cdot r$$
and notice that $p$ and $r$ are perpendicular and so $p\cdot r = 0$. And so we are left with $p\cdot q$. We know that the angle between $p$ and $q$ is $\frac{\pi}{3}$ and that $\|p\|=\|q\|=3$ because we are told $ABE$ is an equilateral triangle. But we also know that for any vectors $a,b$ that $a\cdot b=\|a\|\|b\|\cos{\theta}$ where $\theta$ is the angle between $a$ and $b$. Thus 
\begin{align}
p\cdot q &= \|p\|\|q\|\cos{\frac{\pi}{3}} \\
         &= (3)(3)\frac{1}{2} \\
         &= \frac{9}{2}
\end{align}
and so
$$p\cdot(q+r) = p\cdot q + p\cdot r = \frac{9}{2} + 0 = \frac{9}{2}$$
(b) Notice two things. First that, because $BCDE$ is a parallelogram, that $\vec{EB}=\vec{DC}$. And, second, that $\vec{EB}=p-q$. Now to get $\vec{AC}$ just add vectors head to tail starting at $A$ until we get to $C$. Thus,
\begin{align}
\vec{AC} &= p + r + \vec{DC} \\
         &= p + r + \vec{EB} \\
         &= p + r + (p - q) \\
         &= 2p + r - q
\end{align}
(c) Notice that $\vec{AE}=q$ and $\vec{EC} = r + \vec{DC}$. From (b) we know that $\vec{DC}=p-q$ and so $\vec{EC} = r + p - q$. From (a) we know that $q\cdot p = \frac{9}{2}$. Thus
\begin{align}
\vec{AE}\cdot\vec{EC} &= q\cdot(r + p - q) \\
         &= q\cdot r + q\cdot p - q\cdot q \\
         &= q\cdot r + \frac{9}{2} - \|q\|^2 \\
         &= q\cdot r - \frac{9}{2}
\end{align}
But we are told this equals $9\sqrt{3} - \frac{9}{2}$ and so we have that $q\cdot r = 9\sqrt{3}$. We know that $q\cdot r=\|q\|\|r\|\cos{\theta}$ and from simple geometry we see that $\theta = \frac{\pi}{6}$ and so
\begin{align}
q\cdot r = 9\sqrt{3} &= \|q\|\|r\|\cos{\frac{\pi}{6}} \\
&= 3\|r\|\frac{\sqrt{3}}{2} \\
&\implies \\
\|r\| &= 6.
\end{align}
</s>