Vectors:
a)
Find CB:
−DC+DA+AB=CB−p+2q−p+5p=CB2q+3p=CB→CB=2q+3p
b)
AQ=25AB|AB=5p=25⋅5pAQ=2pQB=35AB|AB=5p=35⋅5pQB=3p
DA2+PQ+QB−CB−DC=0|DA=2q−pQB=3pDC=p2q−p2+PQ+3p−CB−p=02q−p2+PQ+2p−CB=02q−p2+PQ+2p=CB→CB=2q+3p2q=CB−3pq=CB−3p22(CB−3p2)−p2+PQ+2p=CBCB−3p−p2+PQ+2p=CBCB−4p2+PQ+2p=CBCB2−2p+PQ+2p=CBCB2+PQ=CBPQ=CB−CB2PQ=12⋅CB