Находим скалярное произведение векторов vector{BА} и vector{BC}.
vector{BА}=(3;2;-6)
vector{BА} *vector{BC}=(3)*(-2)+(2)*4-6*4=
=-22
Находим длины .векторов vector{BА} и vector{BC}.
|vector{BА}|^2=3^2+2^2+(-6)^2=49
|vector{BА}|=7
|vector{BC}|^2=(-2)^2+4^2+4^2=36
|vector{BC}|=6
cos(vector{BА},vector{BC})=-22/(7*6)=-11/21
sin(vector{BА},vector{BC})=sqrt(1-(-11/21)^2)=
=sqrt(1-(121/441))=8sqrt(5)/21
AD=7*(8sqrt(5)/21)=8sqrt(5)/3
О т в е т. 8sqrt(5)/3