Находим скалярное произведение векторов vector{BА} и vector{BC}.
vector{BА}=(2;4;-4)
vector{BС}=(10;10;0)
vector{BА} *vector{BC}=2*10+4*10-4*0=
=60
Находим длины .векторов vector{BА} и vector{BC}.
|vector{BА}|^2=2^2+4^2+(-4)^2=36
|vector{BА}|=6
|vector{BC}|^2=10^2+10^2=200
|vector{BC}|=10sqrt(2)
cos(vector{BА},vector{BC})=60/(6*10sqrt(2))=1/sqrt(2)
sin(vector{BА},vector{BC})=1/sqrt(2)
CD=10sqrt(2)*(1/sqrt(2))=10
О т в е т. CD=10