Find the area of the triangle ABC , where A(-3,-2,0)…

Find the area of the triangle $ABC$, where $A(-3,-2,0)$, $B(3,-3,1)$, $C(5,0,2)$.

Step 1: Side vectors

\[ \overrightarrow{AB}=\{6,-1,1\},\qquad \overrightarrow{AC}=\{8,2,2\}. \]

Step 2: Vector product

\[ \overrightarrow{AB}\times\overrightarrow{AC}=\begin{vmatrix}\vec i&\vec j&\vec k\\6&-1&1\\8&2&2\end{vmatrix}=\vec i(-2-2)-\vec j(12-8)+\vec k(12+8)=\{-4,-4,20\}. \]

Check: $\{-4,-4,20\}\cdot\overrightarrow{AB}=-24+4+20=0$ and $\{-4,-4,20\}\cdot\overrightarrow{AC}=-32-8+40=0$ $\checkmark$

Step 3: Area

\[ |\overrightarrow{AB}\times\overrightarrow{AC}|=\sqrt{16+16+400}=\sqrt{432}=12\sqrt3, \] and the triangle is half of the parallelogram: $S_{\triangle ABC}=\tfrac12\cdot12\sqrt3=6\sqrt3\approx10.39$.

Step 4: Independent check (Problem 53)

These are three vertices of the parallelogram $ABCD$ with $D(-1,1,1)$ from Problem 53. There the diagonals are $|AC|=6\sqrt2$ and $|BD|=4\sqrt2$, and they make $60^\circ$, so

\[ S_{ABCD}=\tfrac12\,|AC|\,|BD|\sin60^\circ=\tfrac12\cdot6\sqrt2\cdot4\sqrt2\cdot\tfrac{\sqrt3}{2}=12\sqrt3=2\,S_{\triangle ABC}.\ \checkmark \]

Lagrange check: $\overrightarrow{AB}\cdot\overrightarrow{AC}=48-2+2=48$, so

\[ |\overrightarrow{AB}|^2|\overrightarrow{AC}|^2-(\overrightarrow{AB}\cdot\overrightarrow{AC})^2=38\cdot72-48^2=2736-2304=432=(12\sqrt3)^2.\ \checkmark \]