6
$\begingroup$

I have a list of problems

lis = {{{0, 1, 5}, {7, 4, 2}, {9, 3, 1}}, {{6, 8, 5}, {7, 4, 3}, {9, 
     1, 2}}, {{6, 8, 5}, {7, 4, 3}, {9, 2, 1}}, {{6, 9, 1}, {7, 2, 
     3}, {8, 4, 5}}, {{6, 9, 1}, {7, 2, 3}, {8, 5, 4}}, {{6, 9, 
     1}, {7, 2, 4}, {8, 3, 5}}};
data = Table[{pA, pB, pC} = points;
   mylist = {pA, pB, pC, 
     TraditionalForm[
       Subtract @@ Expand[CoplanarPoints[{pA, pB, pC, {x, y, z}}]]] ==
       0}, {points, lis}];

enter image description here

and I want to write solutions of all problems.

SetDirectory[NotebookDirectory[]]
lis = {{{0, 1, 5}, {7, 4, 2}, {9, 3, 1}}, {{6, 8, 5}, {7, 4, 3}, {9, 
     1, 2}}, {{6, 8, 5}, {7, 4, 3}, {9, 2, 1}}, {{6, 9, 1}, {7, 2, 
     3}, {8, 4, 5}}, {{6, 9, 1}, {7, 2, 3}, {8, 5, 4}}, {{6, 9, 
     1}, {7, 2, 4}, {8, 3, 5}}};
data = Table[{pA, pB, pC} = points;
   mylist = {pA, pB, pC, 
     TraditionalForm[
       Subtract @@ Expand[CoplanarPoints[{pA, pB, pC, {x, y, z}}]]] ==
       0}, {points, lis}];
toX[e_] := 
 StringReplace[ToString[TeXForm[e]], {"\\}" -> ")", "\\{" -> "("}]
fileName = FileNameJoin[{Directory[], "my_HW.tex"}]
If[FileExistsQ[fileName], DeleteFile[fileName]];
file = OpenWrite[fileName, PageWidth -> Infinity];
WriteString[file, 
  "\\documentclass[12pt,a4paper]{article}\n" <> 
   "\\usepackage[left=2cm, right=2cm, top=2cm, bottom=2cm]{geometry}\n\
" <> "\\usepackage{amsmath}\n" <> "\\usepackage{amsthm}\n" <> 
   "\\usepackage{esvect}\n" <> "\\theoremstyle{definition}\n" <> 
   "\\newtheorem{ex}{Exercise}\n" <> "\\newtheorem{sol}{Solution}\n" <>
    "\\begin{document}\n"];
Do[WriteString[file, 
   "\\begin{ex}\n" <> 
    "The equation of the plane passing through three points $A" <> 
    toX[data[[n, 1]]] <> "$, $B" <> toX[data[[n, 2]]] <> "$, $C" <> 
    toX[data[[n, 3]]] <> "$ is $" <> toX[data[[n, 4]]] <> 
    ".$\n\\end{ex}\n
\\begin{sol}
We have
$\vv{AB} == data[[n,2]] - data[[n,1]]$;
$\vv{AC}==data[[n,3]] - data[[n,2]]$;
Noralvector of the plane $(ABC)$ is $\vv{n} = Cross[AB,AC]$.
The equation of the plane $(ABC)$ is 
\\end{sol}
"], {n, 1, Length@data}];
WriteString[file, "\\end{document}\n"];
Close[file]

My LaTeX file

    \documentclass[12pt,a4paper]{article}
    \usepackage[left=2cm, right=2cm, top=2cm, bottom=2cm]{geometry}
    \usepackage{amsmath}
    \usepackage{amsthm}
    \usepackage{esvect}
    \theoremstyle{definition}
    \newtheorem{ex}{Exercise}
    \newtheorem{sol}{Solution}
    \begin{document}
    \begin{ex}
    The equation of the plane passing through three points $A(0,1,5)$, $B(7,4,2)$, $C(9,3,1)$ is $6 x-y+13 z-64=0.$
    \end{ex}
    
    \begin{sol}
We have
$\vv{AB} == data[[n,2]] - data[[n,1]]$;
$\vv{AC}==data[[n,3]] - data[[n,2]]$;
Noralvector of the plane $(ABC)$ is $\vv{n} = Cross[AB,AC]$.
The equation of the plane $(ABC)$ is 
\end{sol}
    \begin{ex}
The equation of the plane passing through three points $A(6,8,5)$, $B(7,4,3)$, $C(9,1,2)$ is $2 x+3 y-5 z-11=0.$
\end{ex}
    
    \begin{sol}
We have
$\vv{AB} == data[[n,2]] - data[[n,1]]$;
$\vv{AC}==data[[n,3]] - data[[n,2]]$;
Noralvector of the plane $(ABC)$ is $\vv{n} = Cross[AB,AC]$.
The equation of the plane $(ABC)$ is 
\end{sol}
    \begin{ex}
The equation of the plane passing through three points $A(6,8,5)$, $B(7,4,3)$, $C(9,2,1)$ is $2 x-y+3 z-19=0.$
\end{ex}
    
    \begin{sol}
We have
$\vv{AB} == data[[n,2]] - data[[n,1]]$;
$\vv{AC}==data[[n,3]] - data[[n,2]]$;
Noralvector of the plane $(ABC)$ is $\vv{n} = Cross[AB,AC]$.
The equation of the plane $(ABC)$ is 
\end{sol}
    \begin{ex}
The equation of the plane passing through three points $A(6,9,1)$, $B(7,2,3)$, $C(8,4,5)$ is $2 x-z-11=0.$
\end{ex}
    
    \begin{sol}
We have
$\vv{AB} == data[[n,2]] - data[[n,1]]$;
$\vv{AC}==data[[n,3]] - data[[n,2]]$;
Noralvector of the plane $(ABC)$ is $\vv{n} = Cross[AB,AC]$.
The equation of the plane $(ABC)$ is 
\end{sol}
    \begin{ex}
The equation of the plane passing through three points $A(6,9,1)$, $B(7,2,3)$, $C(8,5,4)$ is $13 x-y-10 z-59=0.$
\end{ex}
    
    \begin{sol}
We have
$\vv{AB} == data[[n,2]] - data[[n,1]]$;
$\vv{AC}==data[[n,3]] - data[[n,2]]$;
Noralvector of the plane $(ABC)$ is $\vv{n} = Cross[AB,AC]$.
The equation of the plane $(ABC)$ is 
\end{sol}
    \begin{ex}
The equation of the plane passing through three points $A(6,9,1)$, $B(7,2,4)$, $C(8,3,5)$ is $5 x-y-4 z-17=0.$
\end{ex}
    
    \begin{sol}
We have
$\vv{AB} == data[[n,2]] - data[[n,1]]$;
$\vv{AC}==data[[n,3]] - data[[n,2]]$;
Noralvector of the plane $(ABC)$ is $\vv{n} = Cross[AB,AC]$.
The equation of the plane $(ABC)$ is 
\end{sol}
    \end{document}

I got

enter image description here

My LaTeX based on

Table[{ab = data[[n, 2]] - data[[n, 1]], 
  ac = data[[n, 3]] - data[[n, 1]], n = Cross[ab, ac]}, {n, 1, Length@data}]

enter image description here

$\endgroup$

2 Answers 2

7
$\begingroup$

It looks like some of your solutions are wrong. At least this is what my code shows. For example problem 3 gives different solution that what you have. You might want double checks how to you obtained the equations of the plane. I used standard method in text book.

Since I wrote this in code cells, it will be messed up when posting. So I include also links to notebook,PDF and latex

PDF

LATEX

NOTEBOOK

Here is screen shot of first page of the PDF

enter image description here

Mathematica code

SetDirectory[NotebookDirectory[]]
lis={{{0,1,5},{7,4,2},{9,3,1}},{{6,8,5},{7,4,3},{9,1,2}},{{6,8,5},{7,4,3},{9,2,1}},{{6,9,1},{7,2,3},{8,4,5}},{{6,9,1},{7,2,3},{8,5,4}},{{6,9,1},{7,2,4},{8,3,5}}};
data = Table[{pA, pB, pC} = points;
   mylist = {pA, pB, pC, 
     TraditionalForm[
       Subtract @@ Expand[CoplanarPoints[{pA, pB, pC, {x, y, z}}]]] ==
       0}, {points, lis}];
fileName=FileNameJoin[{Directory[],"my_HW.tex"}];

toX[e_] :=ToString[TeXForm[e]]
fix[s_String]:=StringReplace[s,{"\\}" -> ")", "\\{" -> "("}]


solvePlaneEquation[p1_List,p2_List,p3_List,sol_,x_Symbol,y_Symbol,z_Symbol]:=Module[{n,v1,v2,a,b,c,x0,y0,z0,eq,s},
  s="\\begin{ex}\n" <> 
  "The equation of the plane passing through three points $A"<> 
    fix[toX[p1]] <> "$, $B" <> fix[toX[p2]] <> "$, $C" <> 
    fix[toX[p3]] <> "$ is $" <> toX[sol] <> "$\n\\end{ex}\n"<>
  "\\begin{sol}\n";
  v1=p2-p1;
  v2=p3-p1;
  n=makeCrossProduct[v1,v2];
  {a,b,c}=n;
  {x0,y0,z0}=p1;
   eq=Expand[a*(x-x0)+b*(y-y0)+c*(z-z0)]==0;
  (*below ref cvgmt from https://mathematica.stackexchange.com/questions/290284/move-variables-to-one-side-of-equation*)
  (*move constant to rhs to make it better looking*)
  Print["eq=",eq];
  eq=SubtractSides[SubtractSides[eq],First@CoefficientArrays[eq]];
  
  
  s=s<>"We have\n\\begin{align*}\n\\overrightarrow{AB}&="<>fix@toX[p2]<>"-"<>fix@toX[p1]<>"\\\\ \n"<>
  "&="<>fix@toX[v1]<>"\\\\ \n"<>
  "\\overrightarrow{AC}&="<>fix@toX[p3]<>"-"<>fix@toX[p1]<>"\\\\ \n"<>
  "&="<>fix@toX[v2]<>"\n"<>
  "\\end{align*}\n"<>
  "Therefore the normal vector $\\vec{n}$ to the plane is given by\n"<>
  "\\begin{align*}\n"<>
  "\\vec{n} &= \\overrightarrow{AB} \\times \\overrightarrow{AC}\\\\ \n"<>
  "&="<>fix@toX[v1]<>"\\times"<>fix@toX[v2]<>"\\\\ \n"<>
  "&="<>fix@toX[n]<>"\n"<>
  "\\end{align*}\n"<>
  "Assigning $a,b,c$ to the coordinates of the normal vector $\\vec{n}$ and "<>
  "Assigning $x_0,y_0,z_0$ to the coordinates of the vector $A$, then the "<>
  "equation of the plane is given by\n"<>
  "\\begin{align*}\n"<>
  "a(x-x_0)+b (y-y_0)+c (z-z_0) &=0\n"<>
  "\\end{align*}\n"<>
  "Which results in\n"<>
  "\\begin{align*}\n"<>
    toX[a]<>"\\left(x"<>If[x0>=0,"-","+"]<>toX[x0]<>"\\right)"<>If[b>=0,"+",""]<>
    toX[b]<>"\\left(y"<>If[y0>=0,"-","+"]<>toX[y0]<>"\\right)"<>If[c>=0,"+",""]<>
    toX[c]<>"\\left(z"<>If[z0>=0,"-","+"]<>toX[z0]<>"\\right)&=0\\\\ \n"<>
    toX[eq[[1]]]<>"&="<>toX[eq[[2]]]<>"\n"<>
  "\\end{align*}\n"<>
  "\\end{sol}\n";
  
  s
]


makeCrossProduct[v1_List,v2_List]:=Module[{},
   Cross[v1,v2]
]

processList[L_List,fileName_String,x_Symbol,y_Symbol,z_Symbol]:=Module[{s,file,p1,p2,p3,eq,n},
   If[FileExistsQ[fileName],DeleteFile[fileName]];
   file=OpenWrite[fileName,PageWidth->Infinity];
   WriteString[file, 
   "\\documentclass[12pt,a4paper]{article}\n" <> 
   "\\usepackage[left=2cm, right=2cm, top=2cm, bottom=2cm]{geometry}\n"<>
   "\\usepackage{amsmath}\n" <>
   "\\usepackage{amsthm}\n" <>
   "\\usepackage{esvect}\n" <> 
   "\\theoremstyle{definition}\n" <> 
   "\\newtheorem{ex}{Exercise}\n" <>
   "\\newtheorem{sol}{Solution}\n" <>
   "\\begin{document}\n"];
   Do[p1=L[[n,1]];p2=L[[n,2]];p3=L[[n,3]];eq=L[[n,4]];
       s=solvePlaneEquation[p1,p2,p3,eq,x,y,z];
       WriteString[file,s]
     ,{n,1,Length[L]} 
  ];
  WriteString[file, "\\end{document}\n"];
  Close[file]
]

To run and generate the Latex file

    processList[data, fileName, x, y, z]

Latex generated by Mathematica

\documentclass[12pt,a4paper]{article}
\usepackage[left=2cm, right=2cm, top=2cm, bottom=2cm]{geometry}
\usepackage{amsmath}
\usepackage{amsthm}
\usepackage{esvect}
\theoremstyle{definition}
\newtheorem{ex}{Exercise}
\newtheorem{sol}{Solution}
\begin{document}
\begin{ex}
The equation of the plane passing through three points $A(0,1,5)$, $B(7,4,2)$, $C(9,3,1)$ is $6 x-y+13 z-64=0$
\end{ex}
\begin{sol}
We have
\begin{align*}
\overrightarrow{AB}&=(7,4,2)-(0,1,5)\\ 
&=(7,3,-3)\\ 
\overrightarrow{AC}&=(9,3,1)-(0,1,5)\\ 
&=(9,2,-4)
\end{align*}
Therefore the normal vector $\vec{n}$ to the plane is given by
\begin{align*}
\vec{n} &= \overrightarrow{AB} \times \overrightarrow{AC}\\ 
&=(7,3,-3)\times(9,2,-4)\\ 
&=(-6,1,-13)
\end{align*}
Assigning $a,b,c$ to the coordinates of the normal vector $\vec{n}$ and Assigning $x_0,y_0,z_0$ to the coordinates of the vector $A$, then the equation of the plane is given by
\begin{align*}
a(x-x_0)+b (y-y_0)+c (z-z_0) &=0
\end{align*}
Which results in
\begin{align*}
-6\left(x-0\right)+1\left(y-1\right)-13\left(z-5\right)&=0\\ 
-6 x+y-13 z&=-64
\end{align*}
\end{sol}
\begin{ex}
The equation of the plane passing through three points $A(6,8,5)$, $B(7,4,3)$, $C(9,1,2)$ is $2 x+3 y-5 z-11=0$
\end{ex}
\begin{sol}
We have
\begin{align*}
\overrightarrow{AB}&=(7,4,3)-(6,8,5)\\ 
&=(1,-4,-2)\\ 
\overrightarrow{AC}&=(9,1,2)-(6,8,5)\\ 
&=(3,-7,-3)
\end{align*}
Therefore the normal vector $\vec{n}$ to the plane is given by
\begin{align*}
\vec{n} &= \overrightarrow{AB} \times \overrightarrow{AC}\\ 
&=(1,-4,-2)\times(3,-7,-3)\\ 
&=(-2,-3,5)
\end{align*}
Assigning $a,b,c$ to the coordinates of the normal vector $\vec{n}$ and Assigning $x_0,y_0,z_0$ to the coordinates of the vector $A$, then the equation of the plane is given by
\begin{align*}
a(x-x_0)+b (y-y_0)+c (z-z_0) &=0
\end{align*}
Which results in
\begin{align*}
-2\left(x-6\right)-3\left(y-8\right)+5\left(z-5\right)&=0\\ 
-2 x-3 y+5 z&=-11
\end{align*}
\end{sol}
\begin{ex}
The equation of the plane passing through three points $A(6,8,5)$, $B(7,4,3)$, $C(9,2,1)$ is $2 x-y+3 z-19=0$
\end{ex}
\begin{sol}
We have
\begin{align*}
\overrightarrow{AB}&=(7,4,3)-(6,8,5)\\ 
&=(1,-4,-2)\\ 
\overrightarrow{AC}&=(9,2,1)-(6,8,5)\\ 
&=(3,-6,-4)
\end{align*}
Therefore the normal vector $\vec{n}$ to the plane is given by
\begin{align*}
\vec{n} &= \overrightarrow{AB} \times \overrightarrow{AC}\\ 
&=(1,-4,-2)\times(3,-6,-4)\\ 
&=(4,-2,6)
\end{align*}
Assigning $a,b,c$ to the coordinates of the normal vector $\vec{n}$ and Assigning $x_0,y_0,z_0$ to the coordinates of the vector $A$, then the equation of the plane is given by
\begin{align*}
a(x-x_0)+b (y-y_0)+c (z-z_0) &=0
\end{align*}
Which results in
\begin{align*}
4\left(x-6\right)-2\left(y-8\right)+6\left(z-5\right)&=0\\ 
4 x-2 y+6 z&=38
\end{align*}
\end{sol}
\begin{ex}
The equation of the plane passing through three points $A(6,9,1)$, $B(7,2,3)$, $C(8,4,5)$ is $2 x-z-11=0$
\end{ex}
\begin{sol}
We have
\begin{align*}
\overrightarrow{AB}&=(7,2,3)-(6,9,1)\\ 
&=(1,-7,2)\\ 
\overrightarrow{AC}&=(8,4,5)-(6,9,1)\\ 
&=(2,-5,4)
\end{align*}
Therefore the normal vector $\vec{n}$ to the plane is given by
\begin{align*}
\vec{n} &= \overrightarrow{AB} \times \overrightarrow{AC}\\ 
&=(1,-7,2)\times(2,-5,4)\\ 
&=(-18,0,9)
\end{align*}
Assigning $a,b,c$ to the coordinates of the normal vector $\vec{n}$ and Assigning $x_0,y_0,z_0$ to the coordinates of the vector $A$, then the equation of the plane is given by
\begin{align*}
a(x-x_0)+b (y-y_0)+c (z-z_0) &=0
\end{align*}
Which results in
\begin{align*}
-18\left(x-6\right)+0\left(y-9\right)+9\left(z-1\right)&=0\\ 
9 z-18 x&=-99
\end{align*}
\end{sol}
\begin{ex}
The equation of the plane passing through three points $A(6,9,1)$, $B(7,2,3)$, $C(8,5,4)$ is $13 x-y-10 z-59=0$
\end{ex}
\begin{sol}
We have
\begin{align*}
\overrightarrow{AB}&=(7,2,3)-(6,9,1)\\ 
&=(1,-7,2)\\ 
\overrightarrow{AC}&=(8,5,4)-(6,9,1)\\ 
&=(2,-4,3)
\end{align*}
Therefore the normal vector $\vec{n}$ to the plane is given by
\begin{align*}
\vec{n} &= \overrightarrow{AB} \times \overrightarrow{AC}\\ 
&=(1,-7,2)\times(2,-4,3)\\ 
&=(-13,1,10)
\end{align*}
Assigning $a,b,c$ to the coordinates of the normal vector $\vec{n}$ and Assigning $x_0,y_0,z_0$ to the coordinates of the vector $A$, then the equation of the plane is given by
\begin{align*}
a(x-x_0)+b (y-y_0)+c (z-z_0) &=0
\end{align*}
Which results in
\begin{align*}
-13\left(x-6\right)+1\left(y-9\right)+10\left(z-1\right)&=0\\ 
-13 x+y+10 z&=-59
\end{align*}
\end{sol}
\begin{ex}
The equation of the plane passing through three points $A(6,9,1)$, $B(7,2,4)$, $C(8,3,5)$ is $5 x-y-4 z-17=0$
\end{ex}
\begin{sol}
We have
\begin{align*}
\overrightarrow{AB}&=(7,2,4)-(6,9,1)\\ 
&=(1,-7,3)\\ 
\overrightarrow{AC}&=(8,3,5)-(6,9,1)\\ 
&=(2,-6,4)
\end{align*}
Therefore the normal vector $\vec{n}$ to the plane is given by
\begin{align*}
\vec{n} &= \overrightarrow{AB} \times \overrightarrow{AC}\\ 
&=(1,-7,3)\times(2,-6,4)\\ 
&=(-10,2,8)
\end{align*}
Assigning $a,b,c$ to the coordinates of the normal vector $\vec{n}$ and Assigning $x_0,y_0,z_0$ to the coordinates of the vector $A$, then the equation of the plane is given by
\begin{align*}
a(x-x_0)+b (y-y_0)+c (z-z_0) &=0
\end{align*}
Which results in
\begin{align*}
-10\left(x-6\right)+2\left(y-9\right)+8\left(z-1\right)&=0\\ 
-10 x+2 y+8 z&=-34
\end{align*}
\end{sol}
\end{document}
$\endgroup$
5
  • $\begingroup$ The result of your code incorrect at "Which results in". Can I use the result of mylist[[4]]. It is correct and was reduced. $\endgroup$ Commented May 26 at 9:48
  • $\begingroup$ @JohnPaulPeter I think there might be a bug in my code give me time to look more will update if needed. $\endgroup$
    – Nasser
    Commented May 26 at 10:10
  • $\begingroup$ I surprise, the results in your PDF file are correct. Only not similart to mylist. $\endgroup$ Commented May 26 at 10:30
  • $\begingroup$ @JohnPaulPeter I doubled checked and see nothing wrong. Added extra step to make it more clear. So I do not know why your solution for few of them are different. I also updated the links. I thought I copied your list OK. Anyway, if you find bug just let me know and will check again. $\endgroup$
    – Nasser
    Commented May 26 at 10:34
  • $\begingroup$ @Nasser Have you ever done a multiple choice in Mathematica and make a latex file? I do know make it. $\endgroup$ Commented Jun 9 at 3:55
1
$\begingroup$

You can try this code

SetDirectory[NotebookDirectory[]]
lis = {{{0, 1, 5}, {7, 4, 2}, {9, 3, 1}}, {{6, 8, 5}, {7, 4, 3}, {9, 
     1, 2}}, {{6, 8, 5}, {7, 4, 3}, {9, 2, 1}}, {{6, 9, 1}, {7, 2, 
     3}, {8, 4, 5}}, {{6, 9, 1}, {7, 2, 3}, {8, 5, 4}}, {{6, 9, 
     1}, {7, 2, 4}, {8, 3, 5}}};
data = Table[{pA, pB, pC} = points;
   mylist = {pA, pB, pC, pB - pA, pC - pA, 
     cr = Cross[pB - pA, pC - pA], 
     TraditionalForm[
       cr[[1]] (x - pA[[1]]) + cr[[2]] (y - pA[[2]]) + 
        cr[[3]] (z - pA[[3]])] == 0, 
     TraditionalForm[
       Subtract @@ Expand[CoplanarPoints[{pA, pB, pC, {x, y, z}}]]] ==
       0}, {points, lis}];
toX[e_] := 
 StringReplace[ToString[TeXForm[e]], {"\\}" -> ")", "\\{" -> "("}]
fileName = FileNameJoin[{Directory[], "answer.tex"}]
If[FileExistsQ[fileName], DeleteFile[fileName]];
file = OpenWrite[fileName, PageWidth -> Infinity];
WriteString[file, 
  "\\documentclass[12pt,a4paper]{article}\n" <> 
   "\\usepackage[left=2cm, right=2cm, top=2cm, bottom=2cm]{geometry}\n\
" <> "\\usepackage{amsmath}\n" <> "\\usepackage{amsthm}\n" <> 
   "\\usepackage{esvect}\n" <> "\\usepackage{fouriernc}\n" <> 
   "\\theoremstyle{definition}\n" <> "\\newtheorem{ex}{Exercise}\n" <>
    "\\newtheorem{sol}{Solution}\n" <> "\\begin{document}\n"];
Do[WriteString[file, 
   "\\begin{ex}\n" <> 
    "The equation of the plane passing through three points $A" <> 
    toX[data[[n, 1]]] <> "$, $B" <> toX[data[[n, 2]]] <> "$, $C" <> 
    toX[data[[n, 3]]] <> "$ is $" <> toX[data[[n, 8]]] <> 
    ".$\n\\end{ex}\n
\\begin{sol}
\mbox{}
\\begin{itemize}
 \item We have 
\[\vv{AB} = " <> toX[data[[n, 2]]] <> " - " <> toX[data[[n, 1]]] <> 
    "=" <> toX[data[[n, 4]]] <> "\]
and
\[\vv{AC} = " <> toX[data[[n, 3]]] <> " - " <> toX[data[[n, 1]]] <> 
    "=" <> toX[data[[n, 5]]] <> ".\]
\item Therefore the normal vector  to the plane is given by  $\vv{n}= \
\vv{AB}\\times \vv{AC} = " <> toX[data[[n, 6]]] <> ".$
\item Equation of the plane is given by   \[" <> toX[data[[n, 7]]] <> 
    ".\]
Expand and simplify, we get
  $" <> toX[data[[n, 8]]] <> "$.
\\end{itemize} \qed
\\end{sol}
"], {n, 1, Length@data}];
WriteString[file, "\\end{document}\n"];
Close[file]

The LaTeX file

\documentclass[12pt,a4paper]{article}
\usepackage[left=2cm, right=2cm, top=2cm, bottom=2cm]{geometry}
\usepackage{amsmath}
\usepackage{amsthm}
\usepackage{esvect}
\usepackage{fouriernc}
\theoremstyle{definition}
\newtheorem{ex}{Exercise}
\newtheorem{sol}{Solution}
\begin{document}
\begin{ex}
The equation of the plane passing through three points $A(0,1,5)$, $B(7,4,2)$, $C(9,3,1)$ is $6 x-y+13 z-64=0.$
\end{ex}

\begin{sol}
\mbox{}
\begin{itemize}
 \item We have 
\[\vv{AB} = (7,4,2) - (0,1,5)=(7,3,-3)\]
and
\[\vv{AC} = (9,3,1) - (0,1,5)=(9,2,-4).\]
\item Therefore the normal vector  to the plane is given by  $\vv{n}= \vv{AB}\times \vv{AC} = (-6,1,-13).$
\item Equation of the plane is given by   \[-6 x+y-13 (z-5)-1=0.\]
Expand and simplify, we get
  $6 x-y+13 z-64=0$.
\end{itemize} \qed
\end{sol}
\begin{ex}
The equation of the plane passing through three points $A(6,8,5)$, $B(7,4,3)$, $C(9,1,2)$ is $2 x+3 y-5 z-11=0.$
\end{ex}

\begin{sol}
\mbox{}
\begin{itemize}
 \item We have 
\[\vv{AB} = (7,4,3) - (6,8,5)=(1,-4,-2)\]
and
\[\vv{AC} = (9,1,2) - (6,8,5)=(3,-7,-3).\]
\item Therefore the normal vector  to the plane is given by  $\vv{n}= \vv{AB}\times \vv{AC} = (-2,-3,5).$
\item Equation of the plane is given by   \[-2 (x-6)-3 (y-8)+5 (z-5)=0.\]
Expand and simplify, we get
  $2 x+3 y-5 z-11=0$.
\end{itemize} \qed
\end{sol}
\begin{ex}
The equation of the plane passing through three points $A(6,8,5)$, $B(7,4,3)$, $C(9,2,1)$ is $2 x-y+3 z-19=0.$
\end{ex}

\begin{sol}
\mbox{}
\begin{itemize}
 \item We have 
\[\vv{AB} = (7,4,3) - (6,8,5)=(1,-4,-2)\]
and
\[\vv{AC} = (9,2,1) - (6,8,5)=(3,-6,-4).\]
\item Therefore the normal vector  to the plane is given by  $\vv{n}= \vv{AB}\times \vv{AC} = (4,-2,6).$
\item Equation of the plane is given by   \[4 (x-6)-2 (y-8)+6 (z-5)=0.\]
Expand and simplify, we get
  $2 x-y+3 z-19=0$.
\end{itemize} \qed
\end{sol}
\begin{ex}
The equation of the plane passing through three points $A(6,9,1)$, $B(7,2,3)$, $C(8,4,5)$ is $2 x-z-11=0.$
\end{ex}

\begin{sol}
\mbox{}
\begin{itemize}
 \item We have 
\[\vv{AB} = (7,2,3) - (6,9,1)=(1,-7,2)\]
and
\[\vv{AC} = (8,4,5) - (6,9,1)=(2,-5,4).\]
\item Therefore the normal vector  to the plane is given by  $\vv{n}= \vv{AB}\times \vv{AC} = (-18,0,9).$
\item Equation of the plane is given by   \[9 (z-1)-18 (x-6)=0.\]
Expand and simplify, we get
  $2 x-z-11=0$.
\end{itemize} \qed
\end{sol}
\begin{ex}
The equation of the plane passing through three points $A(6,9,1)$, $B(7,2,3)$, $C(8,5,4)$ is $13 x-y-10 z-59=0.$
\end{ex}

\begin{sol}
\mbox{}
\begin{itemize}
 \item We have 
\[\vv{AB} = (7,2,3) - (6,9,1)=(1,-7,2)\]
and
\[\vv{AC} = (8,5,4) - (6,9,1)=(2,-4,3).\]
\item Therefore the normal vector  to the plane is given by  $\vv{n}= \vv{AB}\times \vv{AC} = (-13,1,10).$
\item Equation of the plane is given by   \[-13 (x-6)+y+10 (z-1)-9=0.\]
Expand and simplify, we get
  $13 x-y-10 z-59=0$.
\end{itemize} \qed
\end{sol}
\begin{ex}
The equation of the plane passing through three points $A(6,9,1)$, $B(7,2,4)$, $C(8,3,5)$ is $5 x-y-4 z-17=0.$
\end{ex}

\begin{sol}
\mbox{}
\begin{itemize}
 \item We have 
\[\vv{AB} = (7,2,4) - (6,9,1)=(1,-7,3)\]
and
\[\vv{AC} = (8,3,5) - (6,9,1)=(2,-6,4).\]
\item Therefore the normal vector  to the plane is given by  $\vv{n}= \vv{AB}\times \vv{AC} = (-10,2,8).$
\item Equation of the plane is given by   \[-10 (x-6)+2 (y-9)+8 (z-1)=0.\]
Expand and simplify, we get
  $5 x-y-4 z-17=0$.
\end{itemize} \qed
\end{sol}
\end{document}

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