To check if two or more vectors are Coplanar. I have written my own understand for this. Please correct me if I’m right or wrong.
To get two or more vectors to be coplanar , I have noticed that there is a kind of series which forms between the coordinates.
For example : (1,1,1) , (2,2,2) , (3,3,3) .
Similarly , when I made it a little difficult. I still got the same correct result.
If the Q is like : If vector $2i + 2j - 2k$ , $5i + yj + k $and $ -i + 2j + 2k $. Check if the vectors are co planar and then find the value of y?
Then , my approach can be a little too time consuming .
I would like to know if you have any different kind of approach to these kind of questions.
Maybe intuitively or using formula of vectors . Anything.
Thank you.
 A: Hint:
If three vectors are coplanar, cross product of any two is perpendicular to the third one.
A: A little nudge in the right direction; -What property does two or more vectors that lie in the same plane have in common?
A: Let four points are given. (When you are given three points, there always exist a plane such that the points are on the plane. I think 'three vectors are coplanar' means that origin and the given three points are on one plane.)
Subtract one vector to every given vectors (translation doesn't harm the coplanarity). Say this set of vectors are $ S = \{0, (a_1, a_2, a_3), (b_1, b_2, b_3), (c_1, c_2, c_3)\}$. Then, there exist a plane $\pi : \{\alpha x + \beta y + \gamma z = 0\}$ such that $S \subset \pi$. The equation of $\pi$ is homogeneous (which means the right hand side of its defining equation is 0) since $0 \in S$. This implies
$
\begin{pmatrix} a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3  \\ c_1 & c_2 & c_3 \end{pmatrix} \begin{pmatrix} \alpha \\ \beta \\ \gamma \end{pmatrix} = 0
$ has a nontrivial solution. This implies that $\begin{pmatrix} a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3  \\ c_1 & c_2 & c_3 \end{pmatrix}$ is singular. So what to check is the singularity of this matrix.
If you were given $m+1$ points for $m\ge3$, then the matrix which we want to check the singularity would be $m$ by $3$ matrix.
To check the singularity, finding determinant is one way, but it is computationally heavy and not extended to the situation of that five or more points were given. (you can find the determinants of $3$ by $3$ submatrices. It is somewhat similar to the situation of choosing all $4$-tuples of points and find the coplanarity) Another way to do this is finding the RREF (row reduced echelon form) of this matrix. Refer to any basic linear algebra textbook.
