2021-M2-12

1. Since $\overrightarrow{AB}$ is parallel to $5\mathbf{i}-4\mathbf{j}-2\mathbf{k}$, then for some $\mu \in \mathbb{R}$, we have

   $\begin{array}{rcl}\overrightarrow{AB}& =& \mu (5\mathbf{i}-4\mathbf{j}-2\mathbf{k})\\ \overrightarrow{OB}-\overrightarrow{OA}& =& 5\mu \mathbf{i}-4\mu \mathbf{j}-2\mu \mathbf{k}\\ 12\mathbf{i}-s\mathbf{j}-2\mathbf{k}-t\mathbf{i}-14\mathbf{j}-s\mathbf{k}& =& 5\mu \mathbf{i}-4\mu \mathbf{j}-2\mu \mathbf{k}\\ (12-t)\mathbf{i}+(-s-14)\mathbf{j}+(-2-s)\mathbf{k}& =& 5\mu \mathbf{i}-4\mu \mathbf{j}-2\mu \mathbf{k}\end{array}$

   By comparing the components of both sides, we have

   $\{\begin{array}{ll}12-t=5\mu & \dots \text{①}\\ -s-14=-4\mu & \dots \text{②}\\ -2-s=-2\mu & \dots \text{③}\end{array}$

   $\text{②}-2\times \text{③}$, we have

   $\begin{array}{rcl}-s-14-2(-2-s)& =& 0\\ s-10& =& 0\\ s& =& 10\end{array}$

   Sub. $s=10$ into $\text{③}$, we have

   $\begin{array}{rcl}-2-10& =& -2\mu \\ \mu & =& 6\end{array}$

   Sub. $\mu =6$ into $\text{①}$, we have

   $\begin{array}{rcl}12-t& =& 5(6)\\ 12-t& =& 30\\ t& =& -18\end{array}$

2. The required area
   

   $\begin{array}{cl}=& {\displaystyle \frac{1}{2}}|\overrightarrow{AB}\times \overrightarrow{AC}|\\ =& {\displaystyle \frac{1}{2}}|(5\mu \mathbf{i}-4\mu \mathbf{j}-2\mu \mathbf{k})\times (\overrightarrow{OC}-\overrightarrow{OA})|\\ =& {\displaystyle \frac{1}{2}}|(30\mathbf{i}-24\mathbf{j}-12\mathbf{k})\times (12\mathbf{i}-16\mathbf{j}+10\mathbf{k}+18\mathbf{i}-14\mathbf{j}-10\mathbf{k})|\\ =& {\displaystyle \frac{1}{2}}|(30\mathbf{i}-24\mathbf{j}-12\mathbf{k})\times (30\mathbf{i}-30\mathbf{j})|\\ =& {\displaystyle \frac{1}{2}}\left|\left|\begin{array}{ccc}\mathbf{i}& \mathbf{j}& \mathbf{k}\\ 30& -24& -12\\ 30& -30& 0\end{array}\right|\right|\\ =& {\displaystyle \frac{1}{2}}|\left|\begin{array}{cc}-24& -12\\ -30& 0\end{array}\right|\mathbf{i}-\left|\begin{array}{cc}30& -12\\ 30& 0\end{array}\right|\mathbf{j}+\left|\begin{array}{cc}30& -24\\ 30& -30\end{array}\right|\mathbf{k}|\\ =& {\displaystyle \frac{1}{2}}|-360\mathbf{i}-360\mathbf{j}-180\mathbf{k}|\\ =& {\displaystyle \frac{1}{2}}\sqrt{(-360{)}^2+(-360{)}^2+(-180{)}^2}\\ =& 270\end{array}$

3. This is out of syllabus after 2022.

   Sketch a graph according to the question.

   

   The required volume

   $\begin{array}{cl}=& {\displaystyle \frac{1}{6}}|(\overrightarrow{AB}\times \overrightarrow{AC})\cdot \overrightarrow{AD}|\\ =& {\displaystyle \frac{1}{6}}|(-360\mathbf{i}-360\mathbf{j}-180\mathbf{j})\cdot (\overrightarrow{OD}-\overrightarrow{OA})|\\ =& {\displaystyle \frac{1}{6}}|(-360\mathbf{i}-360\mathbf{j}-180\mathbf{j})\cdot (18\mathbf{i}+12\mathbf{j}+14\mathbf{k}+18\mathbf{i}-14\mathbf{j}-10\mathbf{k})|\\ =& {\displaystyle \frac{1}{6}}|(-360\mathbf{i}-360\mathbf{j}-180\mathbf{j})\cdot (36\mathbf{i}-2\mathbf{j}+4\mathbf{k})|\\ =& {\displaystyle \frac{1}{6}}|(-360)(36)+(-360)(-2)+(-180)(4)|\\ =& {\displaystyle \frac{1}{6}}|-12960|\\ =& 2160\end{array}$

4. Let $d$ be the shortest distance from $D$ to $\mathrm{\Pi }$.

   $\begin{array}{rcl}{\displaystyle \frac{1}{3}}\times \text{area of }\mathrm{\Delta }ABC\times d& =& \text{volume of the tetrahedron }ABCD\\ {\displaystyle \frac{1}{3}}\times 270\times d& =& 2160\\ d& =& 24\end{array}$

   Therefore, the shortest distance from $D$ to $\mathrm{\Pi }$ is $24$.

Consider $\overrightarrow{DE}$, we have

$\begin{array}{cl}& \overrightarrow{DE}\\ =& \left({\displaystyle \frac{\overrightarrow{AD}\cdot (\overrightarrow{AB}\times \overrightarrow{AC})}{|\overrightarrow{AB}\times \overrightarrow{AC}|}}\right)\times {\displaystyle \frac{\overrightarrow{AB}\times \overrightarrow{AC}}{|\overrightarrow{AB}\times \overrightarrow{AC}|}}\\ =& d\times {\displaystyle \frac{-360\mathbf{i}-360\mathbf{j}-180\mathbf{k}}{\sqrt{(-360{)}^2+(-360{)}^2+(-180{)}^2}}}\\ =& 24\times {\displaystyle \frac{-360\mathbf{i}-360\mathbf{j}-180\mathbf{k}}{540}}\\ =& -16\mathbf{i}-16\mathbf{j}-8\mathbf{k}\end{array}$

Hence, we have

$\begin{array}{rcl}\overrightarrow{EA}& =& \overrightarrow{DA}-\overrightarrow{DE}\\ \overrightarrow{EA}& =& -36\mathbf{i}+2\mathbf{j}-4\mathbf{k}+16\mathbf{i}+16\mathbf{j}+8\mathbf{k}\\ \overrightarrow{EA}& =& -20\mathbf{i}+18\mathbf{j}+4\mathbf{k}\end{array}$

Also, we have

$\begin{array}{rcl}\overrightarrow{EB}& =& \overrightarrow{EA}+\overrightarrow{AB}\\ \overrightarrow{EB}& =& -20\mathbf{i}+18\mathbf{j}+4\mathbf{k}+30\mathbf{i}-24\mathbf{j}-12\mathbf{k}\\ \overrightarrow{EB}& =& 10\mathbf{i}-6\mathbf{j}-8\mathbf{k}\end{array}$

Consider $\mathrm{\Delta }ABE$, by the section formula, we have

$\begin{array}{rcl}\overrightarrow{EM}& =& {\displaystyle \frac{1}{2}}(\overrightarrow{EA}+\overrightarrow{EB})\\ \overrightarrow{EM}& =& {\displaystyle \frac{1}{2}}(-20\mathbf{i}+18\mathbf{j}+4\mathbf{k}+10\mathbf{i}-6\mathbf{j}-8\mathbf{k})\\ \overrightarrow{EM}& =& {\displaystyle \frac{1}{2}}(-10\mathbf{i}+12\mathbf{j}-4\mathbf{k})\\ \overrightarrow{EM}& =& -5\mathbf{i}+6\mathbf{j}-2\mathbf{k}\end{array}$

Hence, we have

$\begin{array}{cl}& \overrightarrow{EM}\cdot \overrightarrow{AB}\\ =& (-5\mathbf{i}+6\mathbf{j}-2\mathbf{k})\cdot (30\mathbf{i}-24\mathbf{j}-12\mathbf{k})\\ =& (-5)(30)+(6)(-24)+(-2)(-12)\\ =& -270\\ \ne & 0\end{array}$

Therefore, $EM$ is not perpendicular to $AB$.

Hence, $EM$ is not the perpendicular bisector of $AB$.

Hence, $E$ is not the circumcentre of $\mathrm{\Delta }ABC$.