Editing Plücker coordinates (section)

=== Quadratic relation ===
The image of {{math|α}} is not the complete set of points in {{tmath|\mathbb P^5}}; the Plücker coordinates of a line {{mvar|L}} satisfy the quadratic Plücker relation

: <math>
\begin{align}
0 & = p_{01}p^{01}+p_{02}p^{02}+p_{03}p^{03} \\
& = p_{01}p_{23}+p_{02}p_{31}+p_{03}p_{12}.
\end{align}
</math>

For proof, write this homogeneous polynomial as determinants and use [[Laplace expansion]] (in reverse).

: <math>
\begin{align}
0 & = \begin{vmatrix}x_0&y_0\\x_1&y_1\end{vmatrix}\begin{vmatrix}x_2&y_2\\x_3&y_3\end{vmatrix}+
\begin{vmatrix}x_0&y_0\\x_2&y_2\end{vmatrix}\begin{vmatrix}x_3&y_3\\x_1&y_1\end{vmatrix}+
\begin{vmatrix}x_0&y_0\\x_3&y_3\end{vmatrix}\begin{vmatrix}x_1&y_1\\x_2&y_2\end{vmatrix} \\[5pt]
& = (x_0 y_1-y_0 x_1)\begin{vmatrix}x_2&y_2\\x_3&y_3\end{vmatrix}-
(x_0 y_2-y_0 x_2)\begin{vmatrix}x_1&y_1\\x_3&y_3\end{vmatrix}+
(x_0 y_3-y_0 x_3)\begin{vmatrix}x_1&y_1\\x_2&y_2\end{vmatrix} \\[5pt]
& = x_0 \left(y_1\begin{vmatrix}x_2&y_2\\x_3&y_3\end{vmatrix}-
y_2\begin{vmatrix}x_1&y_1\\x_3&y_3\end{vmatrix}+
y_3\begin{vmatrix}x_1&y_1\\x_2&y_2\end{vmatrix}\right)
-y_0 \left(x_1\begin{vmatrix}x_2&y_2\\x_3&y_3\end{vmatrix}-
x_2\begin{vmatrix}x_1&y_1\\x_3&y_3\end{vmatrix}+
x_3\begin{vmatrix}x_1&y_1\\x_2&y_2\end{vmatrix}\right) \\[5pt]
& = x_0 \begin{vmatrix}x_1&y_1&y_1\\x_2&y_2&y_2\\x_3&y_3&y_3\end{vmatrix}
-y_0 \begin{vmatrix}x_1&x_1&y_1\\x_2&x_2&y_2\\x_3&x_3&y_3\end{vmatrix}
\end{align}
</math>

Since both 3×3 determinants have duplicate columns, the right hand side is identically zero.

Another proof may be done like this:
Since vector

: <math> d = \left( p_{01}, p_{02}, p_{03} \right) </math>

is perpendicular to vector

: <math> m = \left( p_{23}, p_{31}, p_{12} \right) </math>

(see above), the scalar product of {{mvar|d}} and {{mvar|m}} must be zero. q.e.d.