Question

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

Answer 1

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.