## Calculus 8th Edition

$-pq$
Given: $a = (p,-p, 2p)$ and $b=(2q,q,-q)$ $a \cdot b=p \cdot 2q+(-p) \cdot q+ 2p \cdot q$ $a \cdot b=p \cdot 2q+(-p) \cdot q+ 2p \cdot q=2pq-pq-2pq$ $=-pq$