How can 2 lines are skew in $\mathbb{P}^{3}$

Question

So the two lines in $\mathbb{P}^{3}$ can be written as: $$d_1: \begin{cases} u_0x_0+u_1x_1+u_2x_2+u_3x_3=0 \\ u'_0x_0+u'_1x_1+u'_2x_2+u'_3x_3=0. \end{cases} $$ $$ d_2: \begin{cases} v_0x_0+v_1x_1+v_2x_2+v_3x_3=0 \\ v'_0x_0+v'_1x_1+v'_2x_2+v'_3x_3=0. \end{cases} $$ So I think these lines will be skew if the set of equations will have no solution. That means the matrix \begin{bmatrix} u_0 & u_1 & u_2 & u_3\\ u'_0 & u'_1 & u'_2 & u'_3\\ v_0 & v_1 & v_2 & v_3\\ v'_0 & v'_1 & v'_2 & v'_3\\ \end{bmatrix} has the smaller rank than the augmented matrix $$ \begin{bmatrix} u_0 & u_1 & u_2 & u_3 &\bigm| & 0 \\ u'_0 & u'_1 & u'_2 & u'_3 &\bigm| & 0 \\ v_0 & v_1 & v_2 & v_3 &\bigm| & 0 \\ v'_0 & v'_1 & v'_2 & v'_3 &\bigm| & 0 \\ \end{bmatrix} $$ Which I find is never the case since the last collumn is only zero? I'm so confused right now

Answer 1

The two lines are skew if the system of equations has no solutions in projective space. Recall that $(0:0:0:0)$ is not a point in projective space.

Passing to the underlying vector space; any two planes will intersect in the point $0$. So the lines are skew if the underlying planes intersect only at $0$.