5.15. Given the vector $\bar{a}(1; -2; 5)$. Find the coordinates of the vector $\bar{b}$, lying in the $x Q y$ plane and perpendicular to the vector $\bar{a}$, if $|\vec{b}|=2 \sqrt{5}$
5.15. Since the vector $\bar{b} \epsilon=x O y$, it has coordinates $(x ; y ; 0)$. Using the conditions $|\vec{b}|=2 \sqrt{5}, \bar{b} \perp \bar{a}$, we can form the system of equations

$$
\left\{\begin{array}{l}
x^{2}+y^{2}=20 \\
x-2 y=0
\end{array}\right.
$$

Solving it, we get two vectors $(4 ; 2 ; 0)$ and $(-4 ;-2 ; 0)$.

Answer: $(4 ; 2 ; 0)$ or $(-4 ;-2 ; 0)$.