Answer
(a) $-1$
(b) $92.8^{\circ}$
Work Step by Step
(a) Given $\vec u=\langle 3,-2,4 \rangle$ and $\vec u=\langle 3,1,-2 \rangle$, by definition the dot product is given by $\vec u\cdot\vec v=3\times3-2\times1+4\times(-2)=-1$
(b) Since $\vec u\cdot\vec v\ne0$, the two vectors are not perpendicular. and the angle between them can be found as $cos\theta=\frac{\vec u\cdot\vec v}{|\vec u||\vec v|}=\frac{-1}{\sqrt {(3)^2+(-2)^2+(4)^2}\sqrt {(3)^2+(1)^2+(-2)^2}}=\frac{-1}{\sqrt {406}}$, thus $\theta\approx92.8^{\circ}$
