Points in a plane
Task number: 4535
Decide whether the real vectors \((1, 0, 2)^\mathsf T , (2, 2, 1)^\mathsf T , (2, 1, 3)^\mathsf T , (3, 3, 2)^\mathsf T\) lie in the same plane.
Solution
We denote \(\boldsymbol u= (1, 0, 2)^\mathsf T\) and check that the vectors given by the differences of the other three vectors \(\boldsymbol v_1= (2, 2, 1)^\mathsf T,\boldsymbol v_2= (2, 1, 3)^\mathsf T\) and \(\boldsymbol v_3= (3, 3, 2)^\mathsf T\) with the vector \(\boldsymbol u\) generate a subspace of dimension 2, i.e. a plane.
\( (\boldsymbol v_1-\boldsymbol u, \boldsymbol v_2-\boldsymbol u, \boldsymbol v_3-\boldsymbol u)= \begin{pmatrix} 2-1 & 2-1 & 3-1 \\ 2-0 & 1-0 & 3-0 \\ 1-2 & 3-2 & 2-2 \end{pmatrix} = \begin{pmatrix} 1 & 1 & 2 \\ 2 & 1 & 3 \\ -1 & 1 & 0 \end{pmatrix} \sim\sim \begin{pmatrix} 1 & -1 & 0 \\ 0 & 2 & 2 \\ 0 & 0 & 0 \end{pmatrix} \)The resulting matrix has rank 2 which corresponds to the dimension of the corresponding affine space.
Answer
The four vectors lie in the same plane.
