Question

Find x such that the four points A(4,1,2),B(5, x,6),C(5,1,-1) and D(7,4,0)are coplanar.

Answer 1

Position vector of $\overline{OA}=4\hat{i}+\hat{j}+2\hat{k}$

Position vector of $\overline{OB}=5\hat{i}+x\hat{j}+6\hat{k}$

Position vector of $\overline{OC}=5\hat{i}+\hat{j}-\hat{k}$

Position vector of $\overline{OD}=7\hat{i}+4\hat{j}+0\hat{k}$

$\overline{AB}=\overline{OB}-\overline{0A}=5\hat{i}+x\hat{j}+6\hat{k}-4\hat{i}-\hat{j}-2\hat{k}=\hat{i}+(x-1)\hat{j}+4\hat{k}\overline{AD}=\overline{OC}-\overline{0A}=5\hat{i}+\hat{j}-\hat{k}-4\hat{i}-\hat{j}-2\hat{k}=3\hat{i}+3\hat{j}-2\hat{k}$

$\overline{AD}=\overline{OD}-\overline{OA}=7\hat{i}+4\hat{j}+0\hat{k}-4\hat{i}-\hat{j}-2\hat{k}=3\hat{i}+3\hat{j}-2\hat{k}$

The above three vectors are coplanar

$\Rightarrow A\overline{B}.(A\overline{C}\times A\overline{D})=0\Rightarrow \left|\begin{array}{ccc}1& x-1& 4\\ 1& 0& -3\\ 3& 3& -2\end{array}\right|=0\Rightarrow 1(10+9)-(x-1)(-2+9)+4(3-0)=0\Rightarrow 9-7(x-1)+12=0\Rightarrow -7(x-1)=-21\Rightarrow x-1=3\therefore x=4$