
  
%ho123.tex: 
%%a Plain TeX file by Doron Zeilberger for Math 251(1 page)

%begin macros
\def\L{{\cal L}}
\def\l{{\langle}}
\def\r{{\rangle}}
\baselineskip=14pt
\parskip=10pt
\def\Tilde{\char126\relax}
\font\eightrm=cmr8 \font\sixrm=cmr6 
\font\eighttt=cmtt8
\magnification=\magstephalf
\def\P{{\cal P}}
\def\Q{{\cal Q}}
\parindent=0pt
\overfullrule=0in
\def\frac#1#2{{#1 \over #2}}
%end macros
\centerline
{
\bf Dr. Z's Math251 Handout \#12.3 [The Dot Product]
}

By Doron Zeilberger

{\bf Problem Type 12.3a }: 
Use vectors to decide whether the triangle
with $P(p_1,p_2,p_3)$, $Q(q_1,q_2,q_3)$, $R(r_1,r_2,r_3)$,
is right-angled.
%%General Problem

{\bf Example Problem 12.3a}: 
Use vectors to decide whether the triangle
with $P(2,-6,-4)$, $Q(4,0,-8)$, $R(12,-4,-10)$,
is right-angled.
%%Specific Problem
\smallskip
\hrule
\smallskip
%\halign{#\hfill  & \quad #\hfill \cr
\halign{           
\vtop{\hsize=15pc\noindent #\strut} \qquad
&
\vtop{\hsize=15pc\noindent #\strut} \cr
{\bf Steps} & {\bf Example} \cr  &  \cr
{\bf 1.} 
%%%General Step 1
Form the vectors $PQ$, $QR$, and $RP$ by subtraction.
$$
{\bf PQ}=\l q_1-p_1,q_2-p_2,q_3-p_3 \r,
$$
$$
{\bf QR}= \l r_1-q_1,r_2-q_2,r_3-q_3 \r,
$$
$$
{\bf RP}=\l p_1-r_1,p_2-r_2,p_3-r_3 \r \quad .
$$

%%%End General Step 1
&
{\bf 1.} 
%%%Specific Step 1
$$
{\bf PQ} =\l 4-2,0-(-6),-8-(-4) \r = \l 2,6,-4 \r,
$$
$$
{\bf QR}=(12-4,-4-0,-10-(-8))= \l 8,-4,-2 \r,
$$
$$
{\bf RP}=(2-12,(-6)-(-4),(-4)-(-10))= \l -10,-2,6 \r \quad .
$$

%%%End Specific Step 1
\cr

\smallskip & \smallskip \cr
{\bf 2.} 
%%% General Step 2
Take all three dot products, and see whether any of them is
$0$. If this is the case, then it is indeed right-angled, otherwise not.


%%% End General Step 2
&
{\bf 2.} 
%%%Specific Step 2
$$
{\bf PQ}.{\bf QR}= \l 2,6,-4 \r . \l 8,-4,-2 \r =
$$
$$
2 \cdot 8+ 6 \cdot (-4)+
(-4) \cdot (-2)=16-24+8=0 \quad ,
$$
$$
{\bf QR}.{\bf RP}= \l 8,-4,-2 \r. \l -10,-2,6 \r=
$$
$$
=8 \cdot (-10)+ (-4) \cdot (-2)+
(-2) \cdot 6=-80+8-12=-84 \quad ,
$$
$$
{\bf PQ}.{\bf RP}= \l 2,6,-4 \r. \l -10,-2,6 \r =
$$
$$
=2 \cdot (-10)+ 6 \cdot (-2)+
(-4) \cdot 6=-20-12-24=-56 \quad .
$$
Since one of these dot-products is $0$ it is indeed a
{\bf right-angled} triangle. Since the common vertex where
it happens is $Q$ the right-angle is at vertex $Q$.

{\bf Ans.:} It is a right-angled triangle since one of the
dot-products is $0$.
%%%End Specific Step 2
\cr
}

{\bf Problem Type 12.3b}: 
A constant force with vector 
representation ${\bf F}=f_1 {\bf i}+ f_2 {\bf j}+f_3{\bf k}$
moves an object along a straight line from the point
$(p_1,p_2,p_3)$ to the point $(q_1,q_2,q_3)$. Find the work done
if the distance is measured in meters and the magnitude
of the force is measured in newtons.

%%General Problem


{\bf Example Problem 12.3b}: A constant force with vector 
representation ${\bf F}=5 {\bf i}+ 9 {\bf j}-3{\bf k}$
moves an object along a straight line from the point
$(4,6,0)$ to the point $(8,18,30)$. Find the work done
if the distance is measured in meters and the magnitude
of the force is measured in newtons.

%%Specific Problem
\smallskip
\hrule
\smallskip
%\halign{#\hfill  & \quad #\hfill \cr
\halign{           
\vtop{\hsize=15pc\noindent #\strut} \qquad
&
\vtop{\hsize=15pc\noindent #\strut} \cr
{\bf Steps} & {\bf Example} \cr  &  \cr
{\bf 1.} Find the {\bf displacement vector}
${\bf D}$ that connects the starting
point to the endpoint:
$$
{\bf D}=(q_1-p_1) {\bf i}+(q_2-p_2) {\bf j}+(q_3-p_3) {\bf k}
$$
%%%General Step 1

%%%End General Step 1
&
{\bf 1.} 
%%%Specific Step 1
$$
{\bf D}=(8-4) {\bf i}+(18-6) {\bf j}+(30-0) {\bf k}
$$
$$
=4{\bf i}+12{\bf j}+30{\bf k} \quad .
$$

%%%End Specific Step 1
\cr

\smallskip & \smallskip \cr
{\bf 2.} Take the the dot product 
${\bf D} . {\bf F} \quad$.
%%% General Step 2

%%% End General Step 2
&
{\bf 2.} 
%%%Specific Step 2
$$
{\bf D}.{\bf F}=
(4{\bf i}+12{\bf j}+30{\bf k}).
(5 {\bf i}+ 9 {\bf j}-3{\bf k})
$$
$$
=4 \cdot 5+ 12 \cdot 9 + 30 \cdot (-3)=
20+108-90=38 \quad .
$$
{\bf Ans.:} The work done was $38$ joules.
%%%End Specific Step 2

\cr}

\end

