По условию:
x_(a)=4
y_(a)=-2
z_(a)=3
|vector{a}|=sqrt{4^2+(-2)^2+3^2}=sqrt(29)
y_(b)=y_(a)=-2
x_(b)=0
vector{b}= 0*vector{i} – 2*vector{j} + z_(b)*vector{k}
|vector{b}|=sqrt{0^2+(-2)^2+(z_(b))^2}
По условию:
|vector{b}|=|vector{a}|
sqrt{0^2+(-2)^2+(z_(b))^2}=sqrt(29)
Возводим в квадрат:
0^2+(-2)^2+(z_(b))^2=29
(z_(b))^2=25
z_(b)= ± 5
О т в е т. vector{b}= 0*vector{i} – 2*vector{j} + 5*vector{k} или vector{b}= 0*vector{i} – 2*vector{j} - 5*vector{k}