Question

Show that the four points with position vectors $4\hat{i}+8\hat{j}+12\hat{k},2\hat{i}+4\hat{j}+6\hat{k},3\hat{i}+5\hat{j}+4\hat{k}$ and $5\hat{i}+8\hat{j}+5\hat{k}$ are coplanar.

Answer 1

Given $A(4,8,12),B(2,4,6),C(3,5,4),D(5,8,5)$

If $A,B,C,D$ are coplanar

$\overrightarrow{AB},\overrightarrow{AC},\overrightarrow{AD}$ lie in the same plane

$\overrightarrow{AB}=\overrightarrow{OB}-\overrightarrow{OA}=(2-4)\hat{i}+(4-8)\hat{j}+(6-12)\hat{k}$

$=-2\hat{i}-4\hat{j}-6\hat{k}$

$\overrightarrow{AC}=\overrightarrow{OC}-\overrightarrow{OA}=(3-4)\hat{i}+(5-8)\hat{j}+(4-12)\hat{k}$

$=-\hat{i}-3\hat{j}-8\hat{k}$

$\overrightarrow{AD}=\overrightarrow{OD}-\overrightarrow{OA}=(5-4)\hat{i}+(8-8)\hat{j}+(5-12)\hat{k}$

$=\hat{i}-7\hat{k}$

If $\overrightarrow{AB},\overrightarrow{AC},\overrightarrow{AD}$ are coplanar

$\left[\overrightarrow{AB}\overrightarrow{AC}\overrightarrow{AD}\right]=0$

$\left|\begin{array}{ccc}-2& -4& -6\\ -1& -3& -8\\ 1& 0& -7\end{array}\right|=-2\left(21\right)+4\left(15\right)-6\left(3\right)=-42+60-12=0$

So they are coplanar