Find the point where the two lines intersect.

Question

I have two lines that intersect each other at a specific point. The equation of these lines is :

$$g_1: x = b_1 + sr_1= \begin{bmatrix}1\\6\\1\end{bmatrix} + s\begin{bmatrix}2\\0\\1\end{bmatrix}, s \in \mathbb{R}$$

$$g_2: x = b_2 + tr_1= \begin{bmatrix}6\\8\\9\end{bmatrix} + t\begin{bmatrix}9\\6\\9\end{bmatrix}, t \in \mathbb{R}$$

To solve for $(s,t)^T$, the point where $g_1$ and $g_2$ intersect, I made $g_1=g_2$, which is:

$$\begin{bmatrix}1\\6\\1\end{bmatrix} + s\begin{bmatrix}2\\0\\1\end{bmatrix}=\begin{bmatrix}6\\8\\9\end{bmatrix} + t\begin{bmatrix}9\\6\\9\end{bmatrix}$$ from which I got three equations:

$$ 1. 9t-2s = -5 $$

$$ 2. 6+0s=8+6t $$

$$ 3. 9t-s=-8 $$

From the second equation I got that $t = \frac{-1}{3}$ which I then substituted back into equation $1$ to get that $s=1$. However when I put $t$ into the 3rd equation, I get $s=-5$. I did this several times and got the same answer, which doesn't make sense unless I either don't understand the concept or my calculations are wrong.

Answer 1

The lines do not intersect. If they did, they would be coplanar, in which case $$\begin{vmatrix}1&6&1&1\\2&0&1&0\\6&8&9&1\\9&6&9&0\end{vmatrix}$$ would vanish, but its value is actually $48$. The rows of the above matrix are the homogeneous coordinates of the points and direction vectors that define the two lines.

Answer 2

These lines do not intersect. As you pointed out, there is no solution for the system of equations. Your method is correct.

Answer 3

A way to check whether or not a solution exists is to consider:

• $v_1=r_1=(2,0,1)$ parallel to $g_1$
• $v_2=r_2=(9,6,9)$ parallel to $g_2$
• $v_3=b_2-b_1=(5,2,8)$ from $g_1$ to $g_2$

and evaluate

$$\begin{vmatrix}2&0&1\\9&6&9\\5&2&8\end{vmatrix} \neq 0$$

which should be equal to zero if the two lines were intesecting.