The Intersection of Two Distinct Lines in the Real Projective Plane

The Intersection of Two Distinct Lines in the Real Projective Plane

On The Existence of a Unique Line Through Two Distinct Points in the Real Projective Plane page we saw that if $[p_1, p_2, p_3]$ and $[q_1, q_2, q_3]$ are two distinct points in $\mathbb{P}^2(\mathbb{R})$ then there exists a unique line containing both of these points.

On The Equation of the Line Passing Through Two Distinct Points in the Real Projective Plane page we obtained the following formula for the line $<a_1, a_2, a_3>$ containing the points $[p_1, p_2, p_3]$ and $[q_1, q_2, q_3]$:

(1)
\begin{align} \quad <a_1, a_2, a_3> = \left < \begin{vmatrix} p_2 & p_3\\ q_2 & q_3 \end{vmatrix}, \begin{vmatrix} p_3 & p_1\\ q_3 & q_1 \end{vmatrix}, \begin{vmatrix} p_1 & p_2\\ q_1 & q_2 \end{vmatrix} \right > \end{align}

Recall from The Existence of a Unique Point Through Two Distinct Lines in the Real Projective Plane page that if $<a_1, a_2, a_3>$ and $<b_1, b_2, b_3>$ are two distinct lines in $\mathbb{P}^2(\mathbb{R})$ then there exists a unique point of intersection for these two lines. This the dual to the theorem regarding the existence of a unique line passing through two distinct points.

As a consequence of the duality, we see that if $<a_1, a_2, a_3>$ and $<b_1, b_2, b_3>$ are two distinct lines, then the coordinates for the point of intersection $[p_1, p_2, p_3]$ are given by:

(2)
\begin{align} \quad [p_1,p_2, p_3] = \left [ \begin{vmatrix} a_2 & a_3\\ b_2 & b_3 \end{vmatrix}, \begin{vmatrix} a_3 & a_1\\ b_3 & b_1 \end{vmatrix}, \begin{vmatrix} a_1 & a_2\\ b_1 & b_2 \end{vmatrix} \right ] \end{align}

Let's look at an example.

Example 1

Consider the projective plane $\mathbb{P}^2(\mathbb{R})$. Find the point of intersection of the lines $<1, 1, 4>$ and $<2, 3, 1>$.

Applying the formula above yields:

(3)
\begin{align} \quad [p_1,p_2, p_3] &= \left [ \begin{vmatrix} 1 & 4\\ 3 & 1 \end{vmatrix}, \begin{vmatrix} 4 & 1\\ 1 & 2 \end{vmatrix}, \begin{vmatrix} 1 & 1\\ 2 & 3 \end{vmatrix} \right ] \\ \quad [p_1, p_2, p_3] &= [1 - 12, 8 - 1, 3 - 2] \\ \quad [p_1, p_2, p_3] &= [-11, 7, 1] \end{align}

Similarly, we can verify that $[-11, 7, 1]$ lies on both $<1, 1, 4>$ and $<2, 3, 1>$ by verifyin that $[-11, 7, 1] \cdot <1, 1, 4> = 0$ and $[-11, 7, 1] \cdot <2, 3, 1> = 0$. We have that:

(4)
\begin{align} \quad [-11, 7, 1] \cdot <1, 1, 4> = (-11)(1) + (7)(1) + (1)(4) = 0 \end{align}

And also:

(5)
\begin{align} \quad [-11, 7, 1] \cdot <2, 3, 1> = (-11)(2) + (7)(3) + (1)(1) = 0 \end{align}
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