向量P和Q的模分别为2倍根号2和3,向量P,Q夹角为45度,向量AB=5P+2Q,AC=P-3Q,D为BC中点,求AD长

向量P*向量Q=|P|*|Q|*cos45=2√2*3*√2/2=6.
向量AB=5P+2Q,
向量AC=P-3Q,
则向量BC=向量AC-向量AB=(P-3Q)-(5P+2Q)=(-4P-5Q),
向量DC=1/2*向量BC=(-2P-2.5Q),
向量AD=向量AC-向量DC=(P-3Q)-(-2P-2.5Q)
=(3P-0.5Q),
|AD|^2=(3P-0.5Q)^2=9*P^2+0.25*Q^2-3*PQ,
而,PQ=6,
P^2=(2√2)^2=8,
Q^2=9.
则有,
|AD|^2=(3P-0.5Q)^2=9*P^2+0.25*Q^2-3*PQ
=9*8+0.25*9-3*6
=54+0.25*9
=(54*4+9)/4
=225/4
|AD|=15/2.