finished the 3rd lab

labs-set-1/template.tex

@@ -111,4 +111,90 @@
   \end{align*}
 }
 
+\pagebreak
+
+\chapter{Lab 3 - 13.2}
+\section{Work}
+\qs{}{
+  Let $P$ be the plane in question. 
+  \begin{align*}
+    P = \{ (x,y,z) \in \mathbb{R}^3 : x = 1 \}
+  \end{align*}
+}
+
+\qs{}{
+  \begin{align*}
+    \mathrm{dist}(P_1,P_2) = \sqrt{(5-1)^2 + (-6+1)^2 + (-5+2)^2} = \sqrt{4^2 + 5^2 +3^2} = \sqrt{16+25+9}=\sqrt{50} = 5\sqrt{2}
+  \end{align*}
+}
+
+\qs{}{
+  \begin{align*}
+    5^2 = (x+8)^2 + (y-10)^2+z^2 = x^2 + 8^2 + 16x +y^2 + 100 -20y +z^2 \\ 
+    \text{This corresponds with answer choice A}
+  \end{align*}
+}
+\qs{}{
+  Let $C$ be the center point of the sphere, and $r$ be the radius.
+  \begin{align*}
+    x^2+y^2+z^2-18x-10y-6z=-15 \\ 
+    \implies x^2 - 18x + (\tfrac{18}{2})^2 + y^2 - 10y + (\tfrac{10}{2})^2 + z^2 - 6z + (\tfrac{6}{2})^2 = -15 + (\tfrac{18}{2})^2 + (\tfrac{10}{2})^2 + (\tfrac{6}{2})^2 \\
+    \implies (x-9)^2 + (y-5)^2 + (z-3)^2 = -15+81+25+9=10+81+9=100=10^2 \\ 
+    \therefore C = (9,5,3) \quad r=10
+  \end{align*}
+}
+
+\qs{}{
+  The set $\{ (x,y,z) \in \mathbb{R}^3 : x^2 +y^2 +z^2 >1 \}$ can be described as the set of all real points outside of a sphere with radius one centered at the origin, non-inclusive of the boundary where $x^2+y^2+z^2=1$. This corresponds with answer choice C.
+}
+
+\qs{}{
+  \begin{align*}
+    \vec{v}=\vec{PQ}=Q-P \quad Q = (4,3,-3) \quad P = (-1,-3,0) \\
+    \vec{v}=\langle 4+1,3+3,-3 \rangle = \langle 5,6,-3 \rangle = 5\hat{i} + 6\hat{j} -3\hat{k} \\
+    \text{This corresponds with answer choice A}
+  \end{align*} 
+}
+
+\qs{}{
+  \begin{align*}
+    \vec{v}=\vec{AB}=B-A \quad B = (-2,-13,-2) \quad A = (-7,-6,-5) \\
+    \vec{v}=\langle -2+7,-13+6,-2+5 \rangle = \langle 5,-7,3 \rangle = 5\hat{i} - 7\hat{j} +3\hat{k} \\
+    \text{This corresponds with answer choice B}
+  \end{align*} 
+}
+
+\qs{}{
+  \begin{align*}
+    M \coloneqq \mathrm{midpoint}(A,B)=(\frac{3+5}{2},\frac{5+2}{2},\frac{5+4}{2})=(4,3.5,4.5) \\
+    \vec{v} = M-C = \langle 4-1,3.5-1,4.5-1 \rangle = \langle 3,2.5,3.5 \rangle \\
+    \text{This corresponds with answer choice B}
+  \end{align*}
+}
+
+\qs{}{
+  \begin{align*}
+    \vec{v} \coloneqq 2\vec{u}-6\vec{v} \quad \vec{u}= \langle 1,1,0 \rangle \quad \vec{v}=\langle 3,0,1 \rangle \\
+    \vec{v} = \langle 2,2,0 \rangle - \langle 18,0,6 \rangle = \langle -16,2,-6 \rangle = -16\hat{i}+2\hat{j}-6\hat{k} \\ 
+    \text{This corresponds with answer choice B}
+  \end{align*}
+}
+
+\qs{}{
+  The provided vector corresponds with answer choice D since:
+  \begin{align*}
+    5\hat{i}+10\hat{j}+10\hat{k}=15( \tfrac{1}{3}\hat{i}+\tfrac{2}{3}\hat{j}+\tfrac{2}{3}\hat{k}) \\
+    \implies  5\hat{i}+10\hat{j}+10\hat{k}= \tfrac{15}{3}\hat{i}+\tfrac{30}{3}\hat{j}+\tfrac{30}{3}\hat{k} \\
+    \implies \cancel{ 5\hat{i}+10\hat{j}+10\hat{k}}=  \cancel{5\hat{i}+10\hat{j}+10\hat{k}} \\
+    \implies 0=0
+  \end{align*}
+}
+
+\qs{}{
+  Let $\vec{v}=\langle -1,6,0 \rangle$ 
+  \begin{align*}
+    \|\vec{v}\|=\sqrt{1^2+6^2+0^2}=\sqrt{1+36}=\sqrt{37}
+  \end{align*}
+}
+
 \end{document}

labs-set-1/template.toc

@@ -1,3 +1,5 @@
 \contentsline {chapter}{\numberline {1}Lab 2 - 13.1 Apps}{2}{chapter.1}%
 \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}%
 \contentsfinish