How to yield the system of equation of this plane defined by two 4-dimensional vectors?

Question

I'm stuck at a problem : Find the system of equations whose solution set is the plane : I'm not sure how to fetch the system of equations and I guess c1 and c2 are free variables but I don't know how to use them. ( Sorry for posting pic I didn't know how to write those matrices. )

Answer 1

A plane $\pi$ in four dimensions is completely determined by a point $Q$ (through which it passes) and a pair $\{u,v\}$ of linearly independent vectors (to which it is orthogonal). $\pi$ is then formed by the points $P$ that satisfy the system of equations: \begin{equation} (P-Q)\cdot u=(P-Q)\cdot v=0. \end{equation} You may expand these equations in components, with $P=(x,y,z,w)$, and then you'll find a system of two equations over the variables $\{x,y,z,w\}$.

To be able to write the equations you should find the data $\{Q,u,v\}$. The easiest point $Q$ in $\pi$ to calculate is the one for $c_1=c_2=0$. Finding vectors orthogonal to $\pi$ is the same as finding the ones orthogonal to the vectors $\{w_1,w_2\}$ that multiply $c_1$ and $c_2$ in the definition of $\pi$.

A straightforward method for doing this last part is: take $u$ to have its third and fourth components to be zero, and then you can easily determine that the first two must be equal and of opposite sign. Then take $v$ to have the first two components to be equal and of the same sign (to make sure that it is linearly independent of $u$) and the third and fourth components will follow. The modulus of both vectors is, of course, arbitrary.

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Therefore, the resulting final system of equations is: \begin{equation} x-y=x+y-6z+2w+10=0 \end{equation}