diff --git a/.gitignore b/.gitignore
index 0a2d00d..5251fd9 100644
--- a/.gitignore
+++ b/.gitignore
@@ -7,3 +7,5 @@
 **/*.bcf
 **/*.run.xml
 **/*.bbl
+**/**.dat 
+**/**.script
diff --git a/general-template/preamble.tex b/general-template/preamble.tex
index bc65870..9f070cd 100644
--- a/general-template/preamble.tex
+++ b/general-template/preamble.tex
@@ -33,6 +33,12 @@
 \usepackage{tikz}
 \usepackage{titletoc}
 
+% add this to the preamble (near your other \usepackage lines)
+\usepackage{pgfplots}
+\pgfplotsset{compat=1.18} % or compat=newest if you prefer
+\usepgfplotslibrary{colormaps} % enables 'viridis' etc.
+
+
 \newcommand\mycommfont[1]{\footnotesize\ttfamily\textcolor{blue}{#1}}
 \SetCommentSty{mycommfont}
 \newcommand{\incfig}[1]{%
@@ -40,6 +46,8 @@
     \import{./figures/}{#1.pdf_tex}
 }
 
+
+
 \usepackage{tikzsymbols}
 \tikzset{
 	symbol/.style={
diff --git a/labs-set-1/preamble.tex b/labs-set-1/preamble.tex
index 5ba904f..9c37981 100644
--- a/labs-set-1/preamble.tex
+++ b/labs-set-1/preamble.tex
@@ -32,6 +32,12 @@
 \usepackage{tikz}
 \usepackage{titletoc}
 
+% add this to the preamble (near your other \usepackage lines)
+\usepackage{pgfplots}
+\pgfplotsset{compat=1.18} % or compat=newest if you prefer
+\usepgfplotslibrary{colormaps} % enables 'viridis' etc.
+
+
 \newcommand\mycommfont[1]{\footnotesize\ttfamily\textcolor{blue}{#1}}
 \SetCommentSty{mycommfont}
 \newcommand{\incfig}[1]{%
diff --git a/labs-set-1/template.pdf b/labs-set-1/template.pdf
index 5b11906..c04ee06 100644
Binary files a/labs-set-1/template.pdf and b/labs-set-1/template.pdf differ
diff --git a/labs-set-1/template.tex b/labs-set-1/template.tex
index 4052c60..7fdb26e 100644
--- a/labs-set-1/template.tex
+++ b/labs-set-1/template.tex
@@ -331,4 +331,216 @@
   \end{align*}
 }
 
+\qs{}{
+  In order to aptly encode the direction and magnitude of the side of the triangle let $\vec{s}_1, \vec{s}_2$ originate from $P$.
+  \begin{align*}
+    \vec{s}_1 = \langle 6-1,6-1,-3-1 \rangle = \langle 5,5,-4 \rangle \quad \vec{s}_2 = \langle 10-1,4-1,2-1 \rangle =\langle 9,3,1 \rangle \\
+    A=\tfrac{1}{2}\|\vec{s}_1\times\vec{s}_2\| =  \begin{vmatrix}
+      \hat{i} & \hat{j} & \hat{k} \\
+      5 & 5 & -4 \\
+      9 & 3 & 1
+    \end{vmatrix} \\
+    \implies \begin{vmatrix}
+      \hat{i} & \hat{j} & \hat{k} \\
+      5 & 5 & -4 \\
+      9 & 3 & 1
+    \end{vmatrix} = \hat{i}(5+12)-\hat{j}(5+36)+\hat{k}(15-45)=\langle 17,-41,-30\rangle \\
+    \implies \tfrac{1}{2}\|\vec{s}_1\times\vec{s}_2\| = \tfrac{1}{2}\cdot\sqrt{17^2+41^2+30^2} = \tfrac{1}{2}\sqrt{2870} \\
+    \text{This corresponds with answer choice D}
+  \end{align*}
+}
+
+\qs{}{
+  \begin{align*}
+    \|\vec{F}\times\vec{PQ}\| =\|\vec{\tau}\| = \|\vec{PQ}\|\cdot\|\vec{F}\|\sin{45\degree}=\tfrac{9}{12}\cdot5\frac{\sqrt{2}}{2}=\frac{45\sqrt{2}}{24} = \tfrac{15}{8}\sqrt{2}\;\mathrm{ft}\cdot\mathrm{lb} \\
+    \text{This corresponds with answer choice D}
+  \end{align*}
+}
+
+\chapter{Lab 5 - 13.5}
+\section{Work}
+
+\qs{}{
+  \begin{align*}
+    \vec{r}=\langle 0,1,0 \rangle + \langle 3,0,-1 \rangle t \\
+    \text{breaking it down into component form:}\\
+    x=3t,\quad y=1,\quad z=-t \\
+    \text{This corresponds with answer choice C}
+  \end{align*}
+}
+
+\qs{}{
+  \begin{align*}
+    \vec{r}=\langle -6,5,-5 \rangle t + \langle 5,-1,-5 \rangle \\
+    \text{This corresponds with answer choice C}
+  \end{align*}
+}
+
+\qs{}{
+  \begin{align*}
+    \vec{v}=P_2-P_1=\langle 3,0,4 \rangle \quad \vec{r}=\langle 3,0,4 \rangle t + \langle -3,7,3 \rangle \lor\vec{r} = \langle 3,0,4 \rangle t + \langle 0,7,7 \rangle  \\
+    \text{This corresponds with answer choice B}
+  \end{align*}
+}
+
+\qs{}{
+  \begin{align*}
+      \vec{u}=\langle 6,4,4 \rangle \quad \vec{v}= \langle -7,-6,-4 \rangle \\
+  \vec{v}_1\coloneqq \begin{vmatrix}
+  \hat{i} & \hat{j} & \hat{k} \\
+  6 &4 & 4 \\
+  -7 & -6 & -4 
+\end{vmatrix} = \hat{i}(-16+24)-\hat{j}(-24+28)+\hat{k}(-36+28)=\langle 8,-4,-8\rangle \\
+\vec{r}=\vec{v}_1 t+\langle -4,-7,4 \rangle \\
+\text{This corresponds with answer choice D}
+  \end{align*}
+}
+
+\qs{}{
+  \begin{align*}
+    \vec{v}_1=\langle 3,-3,-1 \rangle \quad \vec{v}_2 = \langle 4,-2,-5 \rangle \\ \nexists k | k\in \mathbb{R} \land k\cdot\vec{v}_1=\vec{v}_2\\ \therefore l_1 \nparallel l_2 \\
+1+3t &= 3+4s \\
+3-3t &= -1-2s \\
+z=-t &= 3-5s \\[6pt]
+s &= \tfrac{1}{4}(3t-2) \\
+3-3t &= -1-2\cdot\tfrac{1}{4}(3t-2) \\
+3-3t &= -1-\tfrac{1}{2}(3t-2) \\
+3-3t &= -\tfrac{3}{2}t \\
+3 &= \tfrac{3}{2}t \\
+t &= 2 \\
+s &= \tfrac{1}{4}(3\cdot 2-2)=1 \\[6pt]
+x &= 1+3(2)=7 \\
+y &= 3-3(2)=-3 \\
+z &= -2 \\[6pt]
+x &= 3+4(1)=7 \\
+y &= -1-2(1)=-3 \\
+z &= 3-5(1)=-2 \\[6pt]
+\therefore (x,y,z) &= (7,-3,-2) \\
+\text{This corresponds with answer choice D}
+\end{align*}
+
+}
+\qs{}{
+  \begin{align*}
+    \exists k | k\in \mathbb{R} \land k\cdot\vec{v}_1=\vec{v}_2\\ \therefore l_1 \parallel l_2  \\
+    3-t&=-6+5s \\
+    9-t&=5s \\
+    9-5s&=t \\
+    1+6(9-5s)&=55-30s=1+54-30s \\
+    55-30s&=55-30s \forall s\in\mathbb{R} \\
+    4-2(9-5s)&=-14+10s=4-18+10s \\
+    -14+10s&=-14+10s \forall s\in\mathbb{R} \\
+    \therefore \text{$l_1$ and $l_2$ are identical. This corresponds to answer choice B}
+  \end{align*}
+}
+
+\qs{}{
+  \begin{align*}
+    \vec{n}\cdot(\vec{r}-\vec{r}_0)=0 \quad \langle 2,7,6 \rangle \cdot(\vec{r}-\langle 4,-3,2 \rangle) = 0 \\
+    \vec{r}|\vec{r}=\langle x,y,z \rangle \implies \langle 2,7,6 \rangle \cdot \langle x-4,y+3,z-2 \rangle  \\
+    2(x-4)+7(y+3)+6(z-2)=2x-8+7y+21+6z-12=0 \\
+    2x+7y+6z-8+21-12=0 \quad 2x+7y+6z=-1 \\
+    \text{This corresponds with answer choice C}
+  \end{align*}
+}
+
+\qs{}{
+  \begin{align*}
+    \text{Considering answer choice A: } \\
+    5(-1)-7(8)+58=-5-56+58=-3\ne 3 \therefore \cancel{\mathrm{A}} \\
+    5(-1)+7(8)-58=-5+56-58=51-58=-7\ne3\therefore\cancel{\mathrm{B}} \\
+    5(-1)+y(8)-58\ne-3 \therefore \cancel{\mathrm{C}} \\
+    5(-1)-7(8)+58=-5-56+58=-3 \; \therefore \; \text{The answer is D} 
+  \end{align*}
+}
+
+\qs{}{
+  \begin{align*}
+    -2x+2&=-2y \\
+    -2x=2+5z&=2 \\
+    z_0\coloneqq0 \quad -2y+5(0)=2 \; \implies y=-2 \\
+    -2x+2(-2)&=-2=-2x-4 \\
+    -2x&=2 \\
+    \implies x&=-1 \\
+    r_0=(-1,-2,0) \\
+    \text{By parameterizing $r_0$, only answer choice D has the necessary constant terms.}
+  \end{align*}
+}
+
+\qs{}{
+  \begin{align*}
+    x+y&=7-z \\
+    7-z&=12 \\
+    z&=-5 \\
+    x&=t \\
+    t+y&=12 \\
+    y&=12-t \\
+    \therefore x=t \quad y=12-t \quad z=-5     
+  \end{align*}
+  This corresponds with answer choice B. Answer Choice A is the same line with different parametriazation to $x=-t$ instead of $x=t$
+}
+
+\chapter{Lab 6 - 13.6}
+\section{Work}
+
+\qs{}{
+  \begin{tikzpicture}
+  \begin{axis}[
+  legend pos=outer north east,
+  title=Q1,
+  axis lines = box,
+  xlabel = $x$,
+  ylabel = $y$,
+  zlabel = $z$,
+  view={45}{30},
+  ]
+  \addplot3[
+    mesh,
+    samples=50,
+    color=blue,
+  ]
+  {sqrt(10*x - 2*y^2/5)};
+\addplot3[
+  mesh,
+  samples=50,
+  color=red,
+]
+{-sqrt(10*x - 2*y^2/5)};
+  \addlegendentry{$z = \pm\sqrt{10x - \frac{2y^2}{5}}$}
+  \end{axis}
+  \end{tikzpicture}
+
+  This image corresponds to that of an elliptical paraboloid, corresponding with answer choice A.
+}
+
+\qs{}{
+  Since all terms are raised to the second degree, all behave linearly with respect to each other. Therefore, this is not a curved surface. Furthermore, the coefficient of the term containing $x$ is negative. Therefore, the equation represents an elliptical cone along the $x$-axis. This corresponds with answer choice D.  
+}
+
+\qs{}{
+  This surface is linear along the $y$-axis, and has a negative coefficient on the $x$ term. Therefore, the surface is a Hyperbolic paraboloid, corresponding with answer choice A. 
+}
+\newpage
+\qs{}{
+  The surface is linear along the $z$-axis, guaranteeing it to be a paraboloid. The $xz$-trace occurs when  $y=0$. The $yz$-trace occurs when $x=0$. 
+  \begin{align*}
+    \text{$xz$-trace: }\quad 16x^2-16(0)^2=z=16x^2 \\
+    \text{$yz$-trace: }\quad 16(0)^2-16y^2-z=0=16y^2+z
+  \end{align*}
+  This corresponds with answer choice D.
+}
+
+\qs{}{
+  Since two terms have a negative coefficient, and all terms are quadratic, the surface must be a hyperboloid of two sheets, leaving only answer choices B and D. A hyperboloid of two sheets only has hyperbolic cross-sections, so B must be the answer.
+}
+
+\qs{}{
+  Since the equation only yields circular cross sections, and varies quadratically over $\hat{i},\hat{j},\hat{k}$, The answer is Figure 1, A.
+}
+
+\qs{}{
+  The equation yield a circular cross section for any plane parallel to the $yz$-plane, and only discontinuous sets for the $xy$-plane and the $xz$-plane, the surface must be elliptical $\forall (y,z)$. Only Figure 3 satisfies the conditions, validating answer choice C.
+}
+
+
 \end{document}
diff --git a/labs-set-1/template.toc b/labs-set-1/template.toc
index d42b737..10736c8 100644
--- a/labs-set-1/template.toc
+++ b/labs-set-1/template.toc
@@ -4,4 +4,8 @@
 \contentsline {section}{\numberline {2.1}Work}{5}{section.2.1}%
 \contentsline {chapter}{\numberline {3}Lab 4 - 13.3, 13.4}{7}{chapter.3}%
 \contentsline {section}{\numberline {3.1}Work}{7}{section.3.1}%
+\contentsline {chapter}{\numberline {4}Lab 5 - 13.5}{11}{chapter.4}%
+\contentsline {section}{\numberline {4.1}Work}{11}{section.4.1}%
+\contentsline {chapter}{\numberline {5}Lab 6 - 13.6}{14}{chapter.5}%
+\contentsline {section}{\numberline {5.1}Work}{14}{section.5.1}%
 \contentsfinish