The question specifies a three-dimensional region via algebraic inequalities and asks for its volume without involving revolution or standard geometric solid formulas.
8. Two unit vectors $e _ { 1 } , e _ { 2 }$ have an angle of $60 ^ { \circ }$ between them. Vector $m = t e _ { 1 } + 2 e _ { 2 } ( t < 0 )$. Then A. The maximum value of $\frac { | m | } { t }$ is $\frac { \sqrt { 3 } } { 2 }$ B. The minimum value of $\frac { | m | } { t }$ is $- 2$ C. The minimum value of $\frac { | m | } { t }$ is $\frac { \sqrt { 3 } } { 2 }$ D. The maximum value of $\frac { | m | } { t }$ is $- 2$
In coordinate space, a parallelepiped has three vertices of one base at $( - 1,2,1 ) , ( - 4,1,3 ) , ( 2,0 , - 3 )$ , and one vertex of another face lies on the $xy$-plane at distance 1 from the origin. Among parallelepipeds satisfying the above conditions, the maximum volume is (17-1)(17-2).
In coordinate space, let $S$ be the surface obtained by rotating the line segment $AB$ connecting the point $A(0,\ 0,\ 2)$ and the point $B(1,\ 0,\ 1)$ once around the $z$-axis. Let $P$ be a point on $S$ and $Q$ be a point on the $xy$-plane such that $PQ = 2$. As $P$ and $Q$ move subject to this condition, let $K$ be the region that the midpoint $M$ of the line segment $PQ$ can pass through. Find the volume of $K$. %% Page 6
In the three-dimensional orthogonal $x y z$ coordinate system, consider the region $V$ that satisfies Equations (1) and (2). $$\begin{aligned}
& x ^ { 2 } + y ^ { 2 } - z ^ { 2 } \geq 0 \\
& x ^ { 2 } + y ^ { 2 } + 2 x \leq 0
\end{aligned}$$ I. Sketch the cross-sectional shape of the region $V$ at $z = 1$. II. Obtain the surface area of the region $V$.