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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.
Example
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.
Planes in $\mathbb{R}^3$
Examples
1. 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).
2. Find the equation of the plane containing the points $A=(1,2,0)$, $B=(3,0,1)$ and $C=(4,3,-2)$.
Solution: $\vec{AB}=\c2{-2}1$ and $\vec{AC}=\c31{-2}$ are both vectors in this plane. We want to find a normal vector $\nn$ which is be orthogonal to both of these. The cross product of two vectors is orthogonal to both, so we can take the cross product of $\vec{AB}$ and $\vec{AC}$: \[ \nn=\vec{AB}\times\vec{AC}=\def\cp#1#2#3#4#5#6{\begin{vmatrix}\vec\imath&\vec\jmath&\vec k\\#1\\#4\end{vmatrix}}\cp2{-2}131{-2}=\c378\] so the equation of the plane is $3x+7y+8z=d$, and we find $d$ by subbing in a point in the plane, say $A=(1,2,0)$, which gives $d=3(1)+7(2)+8(0)=17$. So the equation is \[ 3x+7y+8z=17.\]
Orthogonal planes and parallel planes
Let $\Pi_1$ be a plane with normal vector $\nn_1$, and let $\Pi_2$ be a plane with normal vector $\nn_2$.
- $\Pi_1$ and $\Pi_2$ are orthogonal or perpendicular planes if they meet at right angles. The following conditions are equivalent:
- $\Pi_1$ and $\Pi_2$ are orthogonal planes;
- $\nn_1\cdot\nn_2=0$;
- $\nn_1$ is a vector in $\Pi_2$;
- $\nn_2$ is a vector in $\Pi_1$.
- $\Pi_1$ and $\Pi_2$ are parallel planes if they have the same normal vectors. In other words, if $\Pi_1$ has equation $ax+by+cz=d_1$ then any parallel plane $\Pi_2$ has an equation with the same left hand side: $ax+by+cz=d_2$.
Examples
1. Find the equation of the plane $\Pi$ passing through $A=(1,3,-3)$ and $B=(4,-2,1)$ which is orthogonal to the plane $x-y+z=5$.
Solution: The plane $x-y+z=5$ has normal vector $\c1{-1}1$, so this is a vector in $\Pi$. Moreover, $\vec{AB}=\c3{-5}4$ is also a vector in $\Pi$, so it has normal vector \[ \nn=\c1{-1}1\times\c3{-5}4=\cp1{-1}13{-5}4=\c1{-1}{-2}.\] So the equation of $\Pi$ is $x-y-2z=d$ and subbing in $A=(1,3,-3)$ gives $d=1-3-2(-3)=4$, so the equation of $\Pi$ is \[ x-y-2z=4.\]
2. The plane parallel to $2x-4y+5z=8$ passing through $(1,2,3)$ is $2x-4y+5z=2(1)-4(2)+5(3) = 10$, or $2x-4y+5z=10$.
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 $x-z=2$.
4. Find the equation of the plane $\Pi$ which contains the line of intersection of the planes \[ \Pi_1: x-y+2z=1\quad\text{and}\quad \Pi_2: 3x+2y-z=4,\] and is perpendicular to the plane $\Pi_3:2x+y+z=3$.
Solution: To find the line of intersection of $\Pi_1$ and $\Pi_2$, we must solve the system of linear equations \begin{gather}x-y+2z=1\\3x+2y-z=4.\end{gather} We can solve this linear system in the usual way, by applying EROs to the matrix $\begin{bmatrix}1&-1&2&1\\3&2&-1&4\end{bmatrix}$: \begin{align*} \def\go#1#2{\begin{bmatrix}#1\\#2\end{bmatrix}} \def\ar#1{\\[6pt]\xrightarrow{#1}&} &\go{1&-1&2&1}{3&2&-1&4} \ar{R2\to R2-3R1} \go{1&-1&2&1}{0&5&-7&1} \ar{R1\to 5R1+R2} \go{5&0&3&6}{0&5&-7&1} \ar{R1\to\tfrac15R1,\,R2\to\tfrac15R2} \go{1&0&3/5&6/5}{0&1&-7/5&1/5} \end{align*} So the line $L$ of intersection is given by \[ L: \c xyz=\c{\tfrac65}{\tfrac15}0+t\c{-\tfrac35}{\tfrac75}1,\quad t\in\mathbb{R}.\] So $\c{-\tfrac35}{\tfrac75}1$ is a direction vector along $L$, and also $5\c{-\tfrac35}{\tfrac75}1=\c{-3}75$ is a vector along $L$. So $\c{-3}75$ is a vector in the plane $\Pi$. Moreover, taking $t=2$ gives the point $(0,3,2)$ in the line $L$, so this is a point in $\Pi$.
Since $\Pi$ is perpendicular to $\Pi_3$, which has normal vector $\nn_3=\c211$, the vector $\c211$ is in $\Pi$.
So a normal vector for $\Pi$ is \[ \nn=\c211\times\c{-3}75 = \cp211{-3}75=\c{-2}{-13}{17}\] hence $\Pi$ has equation $-2x-13y+17z=d$, and subbing in the point $(0,3,2)$ gives $d=0-13(3)+17(2)=-39+34=-5$, so $\Pi$ has equation $-2x-13y+17z=-5$, or \[ 2x+13y-17z=5.\]