Question

Show that the four points having position vectors $6\hat{i}-7\hat{j},16\hat{i}-19\hat{j}-4\hat{k},3\hat{j}-6\hat{k},2\hat{i}-5\hat{j}+10\hat{k}$ are coplanar.

Answer 1

Let the given four points be $P,Q,R$ and $S$ respectively. Three points are coplanar if the vectors $\overrightarrow{PQ},\overrightarrow{PR}$ and $\overrightarrow{PS}$ are coplanar. These vectors are coplanar iff one of them can be expressed as a linear combination of the other two. So, let
$$\begin{gathered} \overrightarrow{PQ}=x\overrightarrow{PR}+y\overrightarrow{PS}. \\ 10\hat{i}-12\hat{j}-4\hat{k}=x\left(-6\hat{i}+10\hat{j}-6\hat{k}\right)+y\left(-4\hat{i}+2\hat{j}+10\hat{k}\right). \\ \Rightarrow 10\hat{i}-12\hat{j}-4\hat{k}=\hat{i}\left(-6x-4y\right)+\hat{j}\left(10x+2y\right)+\hat{k}\left(-6x+10y\right). \end{gathered}$$
$\Rightarrow -6x-4y=10,10x+2y=-12$ and $-6x+10y=-4.$ [ Equating coefficients of $\hat{i},\hat{j},\hat{k}$ on both sides]
Solving the first of these three equations, we get $x=-1$ and $y=-1$. These values also satisfy the third equation.
Hence, the given four points are coplanar.