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 \[\overrightarrow{PQ} = x \overrightarrow{PR} + y \overrightarrow{PS} . \]
\[10 \hat{i} - 12 \hat{j} - 4 \hat{k} = x \left( - 6 \hat{i} + 10 \hat{j} - 6 \hat{k} \right) + y \left( - 4 \hat{i} + 2 \hat{j} + 10 \hat{k} \right) . \]
\[ \Rightarrow 10 \hat{i} - 12 \hat{j} - 4 \hat{k} = \hat{i} \left( - 6x - 4y \right) + \hat{j} \left( 10x + 2y \right) + \hat{k} \left( - 6x + 10y \right) . \]
\[\Rightarrow - 6x - 4y = 10, 10x + 2y = - 12\] 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.