The volume of a prism and a pyramid

$\displaystyle B$ – the base’s area, $\displaystyle h$ – the height:

The first for a prism, the second for a pyramid: a pyramid’s volume is a third of the volume of the prism with the same base and height. The formulas hold when the solid is not right as well – the height is always the segment perpendicular to the base, not the side edge.

In a solid that is not right, the height is found in one of these ways:

• the coordinates of both ends of the height are given – its length is the vector’s length;
• an edge perpendicular to the base – for example a pyramid one of whose side edges is perpendicular to the base;
• a proof by the scalar product that the vector is perpendicular to two intersecting edges of the base (the last lesson);
• Pythagoras’ theorem, trigonometry or vectors in a right-angled triangle that has the height as one of its sides.

In the pyramid $\displaystyle ABCDE$ of the last lesson the base is a right trapezoid: $\displaystyle {\overrightarrow{AB}}={(0,1,-2)}$, $\displaystyle {\overrightarrow{DC}}={(0,3,-6)}$, $\displaystyle {\overrightarrow{AD}}={(-5,-4,-2)}$.
$\displaystyle AE\perp ABCD$, and $\displaystyle {\overrightarrow{AE}}={(-4,4,2)}$.

• $\displaystyle {AB}={\sqrt{5}}$, $\displaystyle {DC}={3\sqrt{5}}$, $\displaystyle {AD}={\sqrt{45}}={3\sqrt{5}}$ – the trapezoid’s height.
• $\displaystyle {B}={\frac{\sqrt{5}+3\sqrt{5}}{2}\cdot3\sqrt{5}}={2\sqrt{5}\cdot3\sqrt{5}}={30}$.
• $\displaystyle {h}={|\overrightarrow{AE}|}={\sqrt{16+16+4}}={6}$.
• $\displaystyle {V}={\frac{30\cdot6}{3}}={60}$.

In the prism $\displaystyle ABCDA'B'C'D'$ the base $\displaystyle ABCD$ is a parallelogram: $\displaystyle A(3,3,2)$, $\displaystyle B(5,1,0)$, $\displaystyle D(2,3,1)$, $\displaystyle C'(10,13,-7)$.
Find $\displaystyle C$, prove that the prism is a box, and find its volume.

• $\displaystyle {\overrightarrow{DC}}={\overrightarrow{AB}}={(2,-2,-2)}$, so $\displaystyle {C}={(4,1,-1)}$.
• $\displaystyle {\overrightarrow{AD}}={(-1,0,-1)}$: $\displaystyle {\overrightarrow{AB}\cdot\overrightarrow{AD}}={-2+0+2}={0}$ – the base is a rectangle.
• $\displaystyle {\overrightarrow{CC'}}={(6,12,-6)}$: $\displaystyle {\overrightarrow{CC'}\cdot\overrightarrow{AB}}={12-24+12}={0}$ and $\displaystyle {\overrightarrow{CC'}\cdot\overrightarrow{AD}}={-6+0+6}={0}$ – the side edge is perpendicular to the base, and the prism is right. A right prism whose base is a rectangle is a box.
• $\displaystyle {V}={|\overrightarrow{AB}|\cdot|\overrightarrow{AD}|\cdot|\overrightarrow{CC'}|}={\sqrt{12}\cdot\sqrt{2}\cdot\sqrt{216}}={\sqrt{5184}}={72}$.

In the box $\displaystyle ABCDA'B'C'D'$ the base $\displaystyle ABCD$ is a rectangle.
The rectangle’s diagonal is 5 long, and the acute angle between its diagonals is $\displaystyle 60^{\circ}$.
The angle between the box’s diagonal $\displaystyle BD'$ and the base’s diagonal $\displaystyle BD$ is $\displaystyle 30^{\circ}$.

In the parallelepiped of Exercise 3 of the last lesson ($\displaystyle ABCD$ a rhombus of side 1, $\displaystyle {\angle BAD}={\angle A'AB}={\angle A'AD}={60^{\circ}}$, $\displaystyle {AA'}={1.2}$) you proved that $\displaystyle A'E$ is perpendicular to the base, where $\displaystyle {\overrightarrow{A'E}}={\frac{2}{5}\vec{u}+\frac{2}{5}\vec{v}-\vec{w}}$.

In the pyramid $\displaystyle SABCD$ of Exercise 4 of the last lesson: $\displaystyle {\overrightarrow{AB}}={(3,6,-9)}$, $\displaystyle {\overrightarrow{AD}}={(-6,3,-9)}$, and $\displaystyle {\overrightarrow{SK}}={(-6,18,10)}$ is perpendicular to the base.

The base of the right pyramid $\displaystyle OACBS$ is the rectangle $\displaystyle OACB$: $\displaystyle O(0,0,0)$, $\displaystyle A(4,0,0)$, $\displaystyle B(0,8,0)$.
The height $\displaystyle SD$ is perpendicular to the base, and $\displaystyle {SD}={12}$.