i'm slowing down. problem 10 of the fourth lab done

labs-set-1/template.tex

@@ -293,4 +293,42 @@
   \end{align*}
 }
 
+\qs{}{
+  \begin{align*}
+    \vec{u} &= \langle -5, -3, 4 \rangle, \quad
+    \vec{v} = \langle 5, -4, 2 \rangle, \quad
+    \vec{w} = \langle 10, -5, -8 \rangle \\[2mm]
+    \vec{u} \times \vec{v} &= 
+    \begin{vmatrix}
+      \hat{i} & \hat{j} & \hat{k} \\
+      -5 & -3 & 4 \\
+      5 & -4 & 2
+    \end{vmatrix} \\[1mm]
+    &= \hat{i}((-3)(2) - (4)(-4)) 
+     - \hat{j}((-5)(2) - (4)(5)) 
+     + \hat{k}((-5)(-4) - (-3)(5)) \\[1mm]
+    &= \hat{i}( -6 + 16 ) - \hat{j}( -10 - 20 ) + \hat{k}( 20 + 15 ) \\[1mm]
+    &= \langle 10, 30, 35 \rangle \\[1mm]
+    \vec{w} \cdot (\vec{u} \times \vec{v}) &= 
+    \langle 10, -5, -8 \rangle \cdot \langle 10, 30, 35 \rangle \\[1mm]
+    &= 10\cdot10 + (-5)\cdot30 + (-8)\cdot35 \\[1mm]
+    &= 100 - 150 - 280 \\[1mm]
+    &= -330 \\
+    \text{This corresponds with answer choice B}
+  \end{align*}
+}
+
+
+\dfn{Area of a Triangle using a Cross Product}{
+  \begin{align*}
+    \text{Area}_{\triangle ABC} = \frac{1}{2} \|\vec{B} - \vec{A} \times \vec{C} - \vec{A}\|
+  \end{align*}
+}
+
+\dfn{Area of a Parallelogram using a Cross Product}{
+  \begin{align*}
+    \text{Area}_{\text{parallelogram}} = \|\vec{u} \times \vec{v}\|
+  \end{align*}
+}
+
 \end{document}