## Thomas' Calculus 13th Edition

$\sqrt 2$
The distance formula for two vectors $p$ and $q$ is defined as: $d=\dfrac{|p \cdot q|}{|p|}$ Thus, we have $p \cdot q=-2(1)+(0)(-1)+6(0)=-2$ and $|p \cdot q|=|-2|=2$ Now, $d=\dfrac{2}{ \sqrt {(1)^2+(-1)^2+(0)^2}}=\dfrac{2}{\sqrt {2}}=\sqrt 2$