Находим скалярное произведение
[b]vector{d}*vector{b}[/b]= (vector{c}*(vector{b}*vector{a})-vector{a}*(vector{b}*vector{c}))*vector{b}=
=(vector{c}*vector{b}*vector{a}-vector{a}*vector{b}*vector{c})*vector{b}=0*vector{b}=[b]0[/b]
Доказано
vector{d}=vector{c}*(vector{b}*vector{a})-vector{a}*(vector{b}*vector{c})=vector{c}*vector{b}*vector{a}-vector{a}*vector{b}*vector{c}=vector{0}
vector{0} перпендикулярен любому вектору