Line Intersection

Topics: Geometry - Line - 3D Space

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Let $l_1$ and $l_2$ be two lines in ${\mathbb{R}}^3$.

We say $l_1$ and $l_2$ intersect on a point if they are different lines if there exists a point $R\in {\mathbb{R}}^3$ such that:

$$l_1\cap l_2=\{R\}$$

Procedure §

Given two lines in its general vectorial representation:

$$l_1=\{(P_1,P_2,P_3)+t(u_1,u_2,u_3)\mid t\in \mathbb{R}\}$$

$$l_2=\{(Q_1,Q_2,Q_3)+s(v_1,v_2,v_3)\}$$

We can find their intersection by making them equal. This is the equation from where we start:

$$(P_1,P_2,P_3)+t(u_1,u_2,u_3)=(Q_1,Q_2,Q_3)+s(v_1,v_2,v_3)$$

With that, we get the system with two variables $t$ and $s$:

$$\{\begin{array}{l}{P}_1+t{u}_1=Q_1+s{v}_1\\ P_2+t{u}_2=Q_2+s{v}_2\\ P_3+t{u}_3=Q_3+s{v}_3\end{array}⟹\{\begin{array}{l}t{u}_1-s{v}_1=-P_1+Q_1\\ t{u}_2-s{v}_2=-P_2+Q_2\\ t{u}_3-s{v}_3=-P_3+Q_3\end{array}$$

Solving this system gives us the intersection of the two lines. We can get three cases:

1. There is one and only one solution $⟹$ the lines intersect (case 3)
2. There are no solutions $⟹$ the lines never intersect
   • The lines could be parallel or not (case 2 or case 4)
3. There are infinite solutions $⟹$ the lines are actually the same line (i.e. they intersect everywhere; case 1)

Example 1

Let:

$$l_1:\frac{x-3}{2}=\frac{y-4}{5}=\frac{z-7}{4}$$

$$l_2:\frac{x-4}{8}=\frac{y+9}{20}=\frac{z+11}{16}$$

With these two lines, we get the system:

$$\{\begin{array}{l}2t-8s=1\\ 5t-20s=-13\\ 4t-16s=-18\end{array}$$

To solve this system, we can remove one of the equations, and then see if the determinant of the corresponding matrix is $0$.

For instance, if we remove the second equation we get:

$$\{\begin{array}{l}2t-8s=1\\ 4t-16s=-18\end{array}$$

$$⟹\left|\begin{array}{cc}2& -8\\ 4& -16\end{array}\right|=-32+32=0$$

Since the determinant is $0$, then we can’t work with these two equations. We shall then choose another combination of two equations and try and see if the corresponding determinant is non-zero.

However, for these two lines, we get that all three possible determinants are $0$. In that case, then it follows that the two lines don’t intersect (i)

Example 2

Let:

$$l_1:\frac{x-3}{2}=\frac{y-4}{5}=\frac{z-7}{4}$$

$$l_2:\frac{x-4}{8}=\frac{y+9}{22}=\frac{z+11}{16}$$

Or, in their vectorial representation:

$$l_1=\{(3,4,7)+t(2,5,4)\mid t\in \mathbb{R}\}$$

$$l_2=\{(4,-9,-11)+s(8,22,16)\mid s\in \mathbb{R}\}$$

With these two lines, we get the system:

$$\{\begin{array}{l}2t-8s=1\\ 5t-22s=-13\\ 4t-16s=-18\end{array}$$

To solve this system, we can remove one of the equations, and then see if the determinant of the corresponding matrix is $0$.

For instance, if we remove the first equation we get:

$$\{\begin{array}{l}5t-22s=-13\\ 4t-16s=-18\end{array}$$

$$⟹\left|\begin{array}{cc}5& -22\\ 4& -16\end{array}\right|=-80+88=8\ne 0$$

Since the determinant is non-zero, we can work with this specific two-equation system.

Upon solving it with whichever method we find most convenient, we can find the solutions to these two equations are:

• $t=-\frac{47}{2}$
• $s=-\frac{19}{4}$

When plugging these values into the equation from where we started:

$$(P_1,P_2,P_3)+t(u_1,u_2,u_3)=(Q_1,Q_2,Q_3)+s(v_1,v_2,v_3)$$

…we should get the intersection point.

However, notice these two solutions won’t solve the other equation, the one we removed earlier. As such, the system has no solution, and thus, the lines never intersect.