Table of Contents
The area of a parallelogram
Consider a parallelogram, two of whose sides are $\def\bR{\mathbb{R}}\def\vv{\vec v}\def\ww{\vec w}\vv$ and $\ww$.
This has double the area of the triangle considered above, so its area is $\|\vv\times\ww\|$.
Example
A triangle with two sides $\def\c#1#2#3{\begin{bmatrix}#1\\#2\\#3\end{bmatrix}}\vv=\c13{-1}$ and $\ww=\c21{-2}$ has area $\tfrac12\|\vv\times\ww\|=\tfrac12\left\|\c13{-1}\times\c21{-2}\right\|=\tfrac12\left\|\c{-5}0{-5}\right\|=\tfrac52\left\|\c{-1}0{-1}\right\|=\tfrac52\sqrt2$, and the parallelogram with sides $\vv$ and $\ww$ has area $\|\vv\times\ww\|=5\sqrt2$.
The volume of a parallelepiped in $\mathbb R^3$
Let $\def\uu{\vec u}\uu$, $\vv$ and $\ww$ be vectors in $\bR^3$.
Consider a parallelepiped, with three sides given by $\uu$, $\vv$ and $\ww$.
Call the face with sides $\vv$ and $\ww$ the base of the parallelpiped. The area of the base is $A=\|\vv\times\ww\|$, and the volume of the parallelpiped is $Ah$ where $h$ is the height, measured at right-angles to the base.
One vector which is at right-angles to the base is $\vv\times\ww$. It follows that $h$ is the length of $\vec p=\text{proj}_{\vv\times\ww}\uu$, so \[ h=\|\text{proj}_{\vv\times\ww}\uu\|=\left\|\frac{\uu\cdot(\vv\times\ww)}{\|\vv\times\ww\|^2}\vv\times\ww\right\| = \frac{\uu\cdot(\vv\times\ww)}{\|\vv\times\ww\|^2}\|\vv\times\ww\| = \frac{|\uu\cdot(\vv\times\ww)|}{\|\vv\times\ww\|}\] so the volume is \[ V=Ah=\|\vv\times\ww\|\frac{|\uu\cdot(\vv\times\ww)|}{\|\vv\times\ww\|}\] or \[ V=|\uu\cdot(\vv\times\ww)|.\] Now $\uu\cdot(\vv\times \ww)=\det\left(\begin{bmatrix}u_1&u_2&u_3\\v_1&v_2&v_3\\w_1& w_2&w_3\end{bmatrix}\right)$, so $V$ is the absolute value of this determinant: \[ V=\left|\quad\det\left( \begin{bmatrix}u_1&u_2&u_3\\v_1&v_2&v_3\\w_1& w_2&w_3\end{bmatrix}\right)\quad \right|.\]
Example
Find volume of a parallelepiped whose vertices include $A=(1,1,1)$, $B=(2,1,3)$, $C=(0,2,2)$ and $D=(3,4,1)$, where $A$ is an adjacent vertex to $B$, $C$ and $D$.
Solution
The vectors $\vec{AB}=\c102$, $\vec{AC}=\c{-1}11$ and $\vec{AD}=\c230$ are all edges of this parallepiped, so the volume is \[ V=\left|\quad \begin{vmatrix}1&0&2\\-1&1&1\\2&3&0\end{vmatrix}\quad \right| = | 1(0-3)-0+2(-3-2)| = |-13| = 13.\]
Planes and lines in $\mathbb{R}^3$
Recall that a typical plane in $\bR^3$ has equation
\[ ax+by+cz=d\]
where $a,b,c,d$ are constants. If we write
\[ \def\nn{\vec n}\nn=\c abc\]
then we can rewrite the equation of this plane in the form
\[ \nn\cdot \c xyz=d.\]
If $A=\def\cc#1{(x_{#1},y_{#1},z_{#1})}\cc1$ and $B=\cc2$ are both points in this plane, then the vector $\vec{AB}$ is said to be in the plane, or to be parallel to the plane. Observe that
\[ \vec n\cdot \vec{AB}=\nn\cdot\def\cp#1{\c{x_{#1}}{y_{#1}}{z_{#1}}}\left(\cp2-\cp1\right) = \nn\cdot\cp2-\nn\cdot\cp1=d-d=0,\]
so \[\nn\cdot\vv=0\]
for every vector $\vv$ in the plane. In other words: the vector $\nn$ is orthogonal to every vector in the plane.
We call a vector with this property a normal vector to the plane.
Examples
1. Find a unit normal vector to the plane $x+y-3z=4$.
Solution: The vector $\nn=\c11{-3}$ is a normal vector to this plane, so $\vv=\frac1{\|\nn\|}\nn=\frac1{\sqrt{11}}\c11{-3}$ is a unit normal vector to this plane. Indeed, $\vv$ is a unit vector and it's in the same direction as the normal vector $\nn$, so $\vv$ is also a normal vector.
2. Find the equation of the plane with normal vector $\def\nn{\vec n}\def\c#1#2#3{\begin{bmatrix}#1\\#2\\#3\end{bmatrix}}\c1{-3}2$ which contains the point $(1,-2,1)$. Then find three other points in this plane.
Solution: the equation is $x-3y+2z=d$, and we can find $d$ by subbing in $(x,y,z)=(1,-2,1)$: $1-3(-2)+2(1)=d$, so $d=9$ and the equation of the plane is \[ x-3y+2z=9.\] Some other points in this plane are $(9,0,0)$, $(0,1,6)$, $(1,1,\tfrac{11}2)$. (We can find these by inspection).
3. Find the equation of the plane parallel to the vectors $\c111$ and $\c1{-1}1$ containing the point $(3,0,1)$.
Solution: a normal vector is $\nn=\c111\times\c1{-1}1=\cp1111{-1}1=\c{2}0{-2}$, so the equation is $2x+0y-2z=2(3)-2(1)=4$, or $2x-2z=4$, or $x-z=2$.