317. Find the lengths of the vectors $\bar{a}\{2 ; 5 ;-1\}, \bar{b}\{1 ;-1 ;-3\}$ and the scalar product of these vectors.
Solution. The lengths of the vectors will be found using formula (14):

$a=\sqrt{2^{2}+5^{2}+(-1)^{2}}=\sqrt{30}, b=\sqrt{1^{2}+(-1)^{2}+(-3)^{2}}=\sqrt{11} ;$ the scalar product

$$
\vec{a} \bar{b}=2 \cdot 1+5(-1)+(-1)(-3)=0
$$

Therefore, vectors $\bar{a}$ and $\bar{b}$ are perpendicular.