4)

BD =AD-AB =b-a

EF =1/2 BD =1/2 (b-a)

6)

Найти косинус угла между векторами a=3k+p и b=k+2p

если |k|=|p|=1 и k⊥p

k⊥p => kp=0

(a,b) =(3k+p)(k+2p) =3k^2 +pk +6pk +2p^2 =3*1 +2*1 =5

|a|^2 =(3k+p)^2 =9k^2 +6kp +p^2 =9*1 +1 =10 => |a|=√10

|b|^2 =(k+2p)^2 =k^2 +4kp +4p^2 =1 +4*1 =5 => |b|=√5

cosф = (a,b) / |a|*|b| = 5/√10√5 =1/√2