diff --git a/labs-set-1/template.pdf b/labs-set-1/template.pdf
index 78d5b1f..fc4ca0d 100644
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diff --git a/labs-set-1/template.tex b/labs-set-1/template.tex
index 918b904..5ab4680 100644
--- a/labs-set-1/template.tex
+++ b/labs-set-1/template.tex
@@ -197,4 +197,100 @@
   \end{align*}
 }
 
+\chapter{Lab 4 - 13.3, 13.4}
+\section{Work}
+
+\qs{}{
+  \begin{align*}
+    \vec{u}\coloneqq \langle -5,7 \rangle \quad \vec{v} \coloneqq \langle 1,6 \rangle \quad \vec{w} \coloneqq
+    \langle -11,2 \rangle \\
+    \vec{u}\cdot(\vec{v}+\vec{w}) = \vec{u}\cdot(\langle1-11,6+2\rangle)=\vec{u}\cdot\langle-10,8\rangle \\
+    \implies \vec{u}\cdot(\vec{v}+\vec{w}) = -5(-10)+7(8)=50+56=106
+  \end{align*}
+}
+
+\qs{}{
+  \begin{align*}
+       \vec{r}\coloneqq \langle 7,-1,-3 \rangle \quad \vec{v} \coloneqq \langle 2,6,6 \rangle \quad \vec{w} \coloneqq
+    \langle 7,4,7 \rangle \\ 
+    \vec{w}\cdot(\vec{v}+\vec{r}) = \vec{w}\cdot\langle9,5,3\rangle=63+20+21=104
+  \end{align*}
+}
+
+\qs{}{
+  Let $o$ be the work done by the system. 
+  \begin{align*}
+    o=\langle 158;180\degree-15\degree \rangle \cdot \langle -6,0 \rangle = \langle 158\cos{180\degree-15\degree},158\sin{180\degree-15\degree} \rangle \cdot \langle -6,0 \rangle \\
+    o=\langle -152.616280554, 40.8934091262 \rangle \cdot \langle -6,0 \rangle = 915.697683324 \\
+    \text{This corresponds with answer choice B}
+  \end{align*}
+}
+
+\qs{}{
+  \begin{align*}
+    \vec{u}\cdot\vec{v}=\|\vec{u}\|\cdot\|\vec{v}\|\cos{\theta} \\
+    \implies \langle 1,-1 \rangle \cdot \langle 4,5 \rangle = \sqrt{2}\sqrt{4^2+5^2}\cos{\theta} = 4-5=-1=\sqrt{2}\sqrt{25+16}\cos{\theta} \\
+    \implies \theta = \arccos{\frac{-1}{\sqrt{2}\sqrt{41}}}= 96.3401917459\degree \\ 
+    \text{This corresponds with answer choice D}
+  \end{align*}
+}
+
+\qs{}{
+  \begin{align*}
+    \langle 4,3 \rangle \cdot \langle 20,-8 \rangle = 80-24 \ne 0 \\
+    \langle 15,-6 \rangle \cdot \langle 20,-8 \rangle = 300+48 \ne 0 \\
+    \langle 20,4 \rangle \cdot \langle 20,-8 \rangle = 400-32  \ne 0 \\
+    \because \langle -10,-25 \rangle \cdot \langle 20,-8 \rangle = -200+200 = 0 \\
+    \implies \text{Answer choice D is correct.}
+  \end{align*}
+}
+
+\qs{}{
+  If $\vec{v}\cdot\vec{w}=0$:
+  
+  \begin{align*}
+    \langle 1,-x \rangle \cdot \langle -4,-3 \rangle = -4+3x=0 \\
+    \implies 3x=4 \quad \therefore x=\tfrac{4}{3}
+  \end{align*}
+}
+
+\dfn{Scalar component of a projection}{
+  \begin{align*}
+    \mathrm{scal}_{\vec{b}}(\vec{a}) = \frac{\vec{a} \cdot \vec{b}}{\|\vec{b}\|}
+  \end{align*}
+}
+
+\dfn{Vector projection}{
+  \begin{align*}
+    \mathrm{proj}_{\vec{b}}(\vec{a}) = \left( \frac{\vec{a} \cdot \vec{b}}{\|\vec{b}\|} \right)\frac{\vec{b}}{\|\vec{b}\|}
+  \end{align*}
+}
+
+\qs{}{
+  \begin{align*}
+    \vec{v}=\langle 2,3 \rangle \quad \vec{w}=\langle 8,-6 \rangle \\
+    \mathrm{proj}_{\vec{w}}{\vec{v}} = \frac{\vec{v}\cdot\vec{w}}{\|\vec{w}\|} \frac{\vec{w}}{\|\vec{w}\|} = \frac{16-18}{\sqrt{8^2+6^2}}\frac{\langle 8,-6 \rangle}{\sqrt{6^2+8^2}} \\
+    \implies \mathrm{proj}_{\vec{w}}{\vec{v}} = \frac{-2}{10}\cdot\langle \tfrac{8}{10}, \tfrac{-6}{10} \rangle = \tfrac{-2}{10}\cdot\tfrac{2}{10}\cdot\langle 4,-3 \rangle = \tfrac{-4}{100}\langle 4,-3 \rangle=\tfrac{-1}{25}\langle 4,-3 \rangle \\
+    \text{This corresponds with answer choice C}
+  \end{align*}
+}
+
+\qs{}{
+  \begin{align*}
+    \vec{u}=\langle 4,5 \rangle \quad \vec{v} = \langle 1,1 \rangle \\
+    \mathrm{scal}_{\vec{v}}{\vec{u}}=\frac{\langle 4,5 \rangle \cdot \langle 1,1 \rangle }{\sqrt{2}} \\
+    \implies \mathrm{scal}_{\vec{v}}{\vec{u}} = \frac{4+5}{\sqrt{2}} = \tfrac{9}{\sqrt{2}} \\
+    \text{This corresponds with answer choice A}
+  \end{align*}
+}
+
+\qs{}{
+  \begin{align*}
+    \vec{u}=\langle 0,7,-2 \rangle \quad \vec{v}=\langle 4,-6,-7 \rangle \\
+    \vec{u}\cdot\vec{v} = \|\vec{u}\|\cdot\|\vec{v}\|\cos{\theta} = 0-42+14 =-28=\sqrt{7^2+2^2}\sqrt{4^2+6^2+7^2}\cos{\theta} \\
+    \theta = \arccos{\frac{-28}{\sqrt{53}\sqrt{101}}}= 1.9635142449\;\mathrm{rad} \\
+    \text{This corresponds with answer choice B}
+  \end{align*}
+}
+
 \end{document}
diff --git a/labs-set-1/template.toc b/labs-set-1/template.toc
index a5b793d..d42b737 100644
--- a/labs-set-1/template.toc
+++ b/labs-set-1/template.toc
@@ -2,4 +2,6 @@
 \contentsline {section}{\numberline {1.1}Work}{2}{section.1.1}%
 \contentsline {chapter}{\numberline {2}Lab 3 - 13.2}{5}{chapter.2}%
 \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}%
 \contentsfinish