Find the area and the height BD of the triangle ABC , where…

Find the area and the height $BD$ of the triangle $ABC$, where $A(1,-2,8)$, $B(0,0,4)$, $C(6,2,0)$.

Step 1: Side vectors and vector product

\[ \overrightarrow{AB}=\{-1,2,-4\},\qquad \overrightarrow{AC}=\{5,4,-8\}, \]

\[ \overrightarrow{AB}\times\overrightarrow{AC}=\begin{vmatrix}\vec i&\vec j&\vec k\\-1&2&-4\\5&4&-8\end{vmatrix}=\vec i(-16+16)-\vec j(8+20)+\vec k(-4-10)=\{0,-28,-14\}. \]

Check: $\{0,-28,-14\}\cdot\overrightarrow{AB}=-56+56=0$ and $\{0,-28,-14\}\cdot\overrightarrow{AC}=-112+112=0$ $\checkmark$

Step 2: Area

\[ |\overrightarrow{AB}\times\overrightarrow{AC}|=\sqrt{784+196}=\sqrt{980}=14\sqrt5,\qquad S=\tfrac12\cdot14\sqrt5. \]

\[S_{\triangle ABC}=7\sqrt5\approx15.65\]

Step 3: Height $BD$ onto $AC$

\[ |AC|=\sqrt{25+16+64}=\sqrt{105},\qquad BD=\frac{2S}{|AC|}=\frac{14\sqrt5}{\sqrt{105}}=\frac{14}{\sqrt{21}}. \]

\[BD=\frac{2\sqrt{21}}{3}\approx3.06\]

Step 4: The foot $D$ (added check)

\[ t=\frac{\overrightarrow{AB}\cdot\overrightarrow{AC}}{|AC|^2}=\frac{-5+8+32}{105}=\frac13,\qquad D=A+\tfrac13\overrightarrow{AC}=\left(\tfrac83,-\tfrac23,\tfrac{16}{3}\right), \]

\[ \overrightarrow{BD}=\left\{\tfrac83,-\tfrac23,\tfrac43\right\},\qquad |BD|=\tfrac13\sqrt{64+4+16}=\tfrac{\sqrt{84}}{3}=\tfrac{2\sqrt{21}}{3}.\ \checkmark \]