# A nice question on basis.

Let $x=(x_1,x_2,x_3), y=(y_1,y_2,y_3)\in \mathbb{R}^3$ be linearly independent. Let $$\delta_1 =x_2y_3-y_2x_3$$ $$\delta_2 =x_1y_3-y_1x_3$$ $$\delta_3 =x_1y_2-y_1x_2$$ If V is the span of $x,y$ then,

• $V=\{(u,v,w):\delta_1u-\delta_2v+\delta_3w=0\}$
• $V=\{(u,v,w):-\delta_1u+\delta_2v+\delta_3w=0\}$
• $V=\{(u,v,w):\delta_1u+\delta_2v-\delta_3w=0\}$
• $V=\{(u,v,w):\delta_1u+\delta_2v+\delta_3w=0\}$

What I have understood so far that all deltas are actually co-factors of the determinant.

\begin{vmatrix} u & v & w \\ x_1 &x_2&x_3 \\ y_1&y_2&y_3 \\ \notag \end{vmatrix}

So, since $\{u,v,w\}$ are independent so, 1st option is correct.

Am I correct?

• You can simply write $(u,v,w)=c_1\cdot (x_1,x_2,x_3)+c_2\cdot (y_1,y_2,y_3)$ and then insert them inside the equations and then see which one is correct. – Levent Jul 5 '16 at 18:14

The first option is correct, since the expression $\delta_1u-\delta_2v+\delta_3w$ correctly expresses the determinant
$$\begin{vmatrix} u & v & w \\ x_1 &x_2&x_3 \\ y_1&y_2&y_3 \\ \notag \end{vmatrix}.$$
Since $x$ and $y$ are assumed to be linearly independent, a vector $(u,v,w)$ is in the plane spanned by $x$ and $y$ if and only if the above determinant is zero.