Answer
(a) $1$
(b) No. $60^{\circ}$
Work Step by Step
(a) Given $\vec u=0i+j-k$ and $\vec v=i+j+0k$, by definition the dot product is given by $\vec u\cdot\vec v=0\times1+1\times1-1\times0=1$
(b) Since $\vec u\cdot\vec v\ne0$, the two vectors are not perpendicular to each other. and the angle between them can be found as $cos\theta=\frac{\vec u\cdot\vec v}{|\vec u||\vec v|}=\frac{1}{\sqrt {(0)^2+(1)^2+(-1)^2}\sqrt {(1)^2+(1)^2+(0)^2}}=\frac{1}{2}$, thus $\theta\approx60^{\circ}$
