Question

Show that the four points whose position vectors are $6\hat{i}-7\hat{j},16\hat{i}-29\hat{j}-4\hat{k},3\hat{j}-6\hat{k}$ and $2\hat{i}+5\hat{j}+10\hat{k}$ are coplanar.

Answer 1

Let the four points be $A,B,C,D$

then,

$\overrightarrow{OA}=6\hat{i}-7\hat{j}+0\hat{k}$

$\overrightarrow{OB}=16\hat{i}-29\hat{j}-4\hat{k}$

$\overrightarrow{OC}=0\hat{i}+3\hat{j}-6\hat{k}$

$\overrightarrow{OD}=2\hat{i}+5\hat{j}+10\hat{k}$

$\overrightarrow{AB}=\overrightarrow{OA}-\overrightarrow{OA}=10\hat{i}-22\hat{j}-4\hat{k}$

$\overrightarrow{AC}=\overrightarrow{OC}-\overrightarrow{OA}=-6\hat{i}+10\hat{j}-6\hat{k}$

$\overrightarrow{AD}=-4\hat{i}+12\hat{j}+10\hat{k}$

The given points are coplanar , If

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

Now,

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

$\begin{array}{c}=\left|\begin{array}{ccc}10& -22& -4\\ -6& 10& 6\\ -4& 12& 10\end{array}\right|\\ \\ =10\left(100+72\right)+22\left(-60-24\right)-4\left(-72+40\right)\\ \\ =1720+22\times -84+4\times 32\\ \\ =1848-1848\\ \\ =0\end{array}$

Hence,

Given four points are coplanar.