The intersection between straight lines

Consider two straight lines gg and hh in a three dimensional space. Assume that gg passes through point AA and has direction vector v⃗\vec v, and hh passes through point BB and has direction vector w⃗\vec w. The two lines can have the following position relative to each other:

The lines intersect at one point SS

Two straight lines intersect if they have exactly one point SS in common. This point is called the point of intersection.

Often we want to find the coordinates of SS. To do so, note that for SS to be the intersection point it has to be on gg as well as on hh. For no other point in the 3d-space is this true. Using vector language, this condition translates to the following two conditions for point SS:

AS→∥v⃗ and BS→∥w⃗\boxed{\overrightarrow{AS} \parallel \vec v \text{ and } \overrightarrow{BS} \parallel \vec w}

These two conditions hide a system of three equations from which the coordinates for SS can be found (see example below). Note that the two conditions above can also be rephrased as

there is a scalar c with S⃗=A⃗+c⋅v⃗ and a scalar d with S⃗=B⃗+d⋅w⃗\boxed{\text{there is a scalar $c$ with } \vec S=\vec A+c\cdot \vec v \text{ and a scalar $d$ with } \vec S=\vec B+d\cdot \vec w}
Example 1

The straight line gg passes through point A(1∣0∣0)A(1 \vert 0 \vert 0) and has direction v⃗=(2−11)\vec v=\left(\begin{array}{r} 2\\ -1\\ 1 \end{array}\right), and the straight line hh passes through point B(13∣−5∣−5)B(13 \vert -5 \vert -5) and has direction w⃗=(−314)\vec w =\left(\begin{array}{r} -3 \\ 1\\ 4 \end{array}\right). They intersect at point S(x∣y∣z)S(x\vert y\vert z). Let us find its coordinates.

Solution

Let us denote the coordinates of the unknown intersection point SS by S(x∣y∣z)S(x\vert y\vert z). For it to be on gg and on hh, there must exist two scalars cc and dd such that

S⃗=A⃗+c⋅v⃗\vec S = \vec A+ c\cdot \vec{v}S⃗=B⃗+d⋅w⃗\vec S = \vec B+ d\cdot \vec{w}

Expressed with components, we get

(xyz)=(100)+c⋅(2−11)=(1+2c−cc)\left(\begin{array}{r} x\\ y\\ z \end{array}\right) = \left(\begin{array}{r} 1\\ 0\\ 0 \end{array}\right) +c\cdot\left(\begin{array}{r} 2\\ -1\\ 1 \end{array}\right) = \left(\begin{array}{c} 1+2c\\ -c\\ c \end{array}\right)(xyz)=(13−5−5)+d⋅(−314)=(13−3d−5+d−5+4d)\left(\begin{array}{r} x\\ y\\ z \end{array}\right) = \left(\begin{array}{r} 13\\ -5\\ -5 \end{array}\right) +d\cdot\left(\begin{array}{r} -3\\ 1\\ 4 \end{array}\right) = \left(\begin{array}{c} 13-3d\\ -5+d\\ -5+4d \end{array}\right)

and thus we get the system of three equations

1+2c=13−3d−c=−5+dc=−5+4d\begin{array}{rll} 1+2c&=& 13-3d\\ -c&=&-5+d\\ c &=& -5+4d \end{array}

Using two equations, we get c=3c=3 and d=2d=2. Using c=3c=3, we get S(7∣−3∣3)S(7\vert -3 \vert 3), using d=2d=2 we also get S(7∣−3∣3)S(7\vert -3 \vert 3).

Warning

Always calculate SS twice, once using the value for cc and once using the value for dd. If there is an intersection point you will get twice the same coordinates for SS. However, if the lines do not intersect, we could still make the same calculations above, but in this case the coordinates calculated using cc will differ from the coordinates calculated using dd. This fact will be relevant for detecting skew lines (see below).

The lines are parallel

Two straight lines are parallel, if they have collinear direction vectors.

From the figure it follows that gg and hh are parallel if

v⃗∥w⃗\boxed{\vec v\parallel \vec w}

And they form the same line if, and only if

AB→∥v⃗ or AB→∥w⃗\boxed{\overrightarrow{AB} \parallel \vec v \text{ or } \overrightarrow{AB} \parallel \vec w}

The lines are skew

Two straight lines are said to be skew, if they neither intersect nor are they parallel. Skew lines are not possible in a 2d-space, because they have to be on two different, parallel planes:

To find out if the two lines are skew or do intersect:

  1. verify that they are not parallel (which is easy, and quickly done)
  2. if not parallel, assume the lines intersect and calculate the point of intersection twice, once with cc and once with dd. If we get two different points SS, we know that there is no intersection point, and the lines have to be skew.
Example 2

The straight line gg passes through point A(1∣0∣0)A(1 \vert 0 \vert 0) and has direction v⃗=(2−11)\vec v=\left(\begin{array}{r} 2\\ -1\\ 1 \end{array}\right), and the straight line hh passes through point B(3∣−5∣−5)B(3 \vert -5 \vert -5) and has direction w⃗=(−314)\vec w =\left(\begin{array}{r} -3 \\ 1\\ 4 \end{array}\right). Are they skew?

Solution

They are not parallel and not identical, because v⃗∦w⃗\vec v \not \parallel \vec w, so they might be skew. So assume there is an intersection point S(x∣y∣z)S(x\vert y\vert z), and let us find its coordinates: Thus, find values cc and dd with

(xyz)=(100)+c⋅(2−11)=(1+2c−cc)\left(\begin{array}{r} x\\ y\\ z \end{array}\right) = \left(\begin{array}{r} 1\\ 0\\ 0 \end{array}\right) +c\cdot\left(\begin{array}{r} 2\\ -1\\ 1 \end{array}\right) = \left(\begin{array}{c} 1+2c\\ -c\\ c \end{array}\right)(xyz)=(3−5−5)+d⋅(−314)=(3−3d−5+d−5+4d)\left(\begin{array}{r} x\\ y\\ z \end{array}\right) = \left(\begin{array}{r} 3\\ -5\\ -5 \end{array}\right) +d\cdot\left(\begin{array}{r} -3\\ 1\\ 4 \end{array}\right) = \left(\begin{array}{c} 3-3d\\ -5+d\\ -5+4d \end{array}\right)

We get the system of three equations

1+2c=3−3d−c=−5+dc=−5+4d\begin{array}{rll} 1+2c&=& 3-3d\\ -c&=&-5+d\\ c &=& -5+4d \end{array}

Using the last two equations, we get c=3c=3 and d=2d=2. For c=3c=3 we get S(7∣−3∣3)S(7\vert -3 \vert 3), for d=2d=2 we get S(−3∣−3∣3)S(-3\vert -3 \vert 3). We therefore get different points, which also means that the system of equations has no solution. The conclusion is that there is no intersection point, and the lines must be skew!

Exercise 1
Q1

The straight line gg passes through AA and has direction v⃗\vec{v}, the straight line hh passes though the point BB and has direction w⃗\vec{w}. Determine the relative position of gg and hh (parallel, identical, skew, intersecting). If they intersect, determine the coordinates of the intersection point SS.

  1. A(−2∣1∣3)A(-2\vert 1 \vert 3), B(1∣6∣2)B(1\vert 6\vert 2), v⃗=(−0.6−10.2)\vec{v}=\left(\begin{array}{r} -0.6\\ -1\\ 0.2 \end{array}\right), w⃗=(35−1)\vec{w}=\left(\begin{array}{r} 3\\ 5\\ -1 \end{array}\right)

  2. A(3∣2∣−1)A(3\vert 2\vert -1), B(2∣6∣1)B(2\vert 6\vert 1), v⃗=(0.80.2−1)\vec{v}=\left(\begin{array}{r} 0.8\\ 0.2\\ -1 \end{array}\right), w⃗=(−4−15)\vec{w}=\left(\begin{array}{r} -4\\ -1\\ 5 \end{array}\right)

  3. A(2∣−1∣1)A(2\vert -1\vert 1), B(−3∣5∣4)B(-3\vert 5\vert 4), v⃗=(−121)\vec{v}=\left(\begin{array}{r} -1\\ 2\\ 1 \end{array}\right), w⃗=(3−21)\vec{w}=\left(\begin{array}{r} 3\\ -2\\ 1 \end{array}\right)

  4. A(0∣0∣1)A(0\vert 0\vert 1), B(−2∣−2∣5)B(-2\vert -2\vert 5), v⃗=(323)\vec{v}=\left(\begin{array}{r} 3\\ 2\\ 3 \end{array}\right), w⃗=(431)\vec{w}=\left(\begin{array}{r} 4\\ 3\\ 1 \end{array}\right)

Q2

Consider the triangle ABCABC with the vertices A(0∣0∣0)A(0\vert 0\vert 0), B(4∣0∣0)B(4\vert 0\vert 0), and C(2∣3∣9)C(2\vert 3\vert 9). Show that all three medians ("Seitenhalbierende") intersect at the same point SS. Determine the coordinates of this point.

This, by the way, is a general fact about triangles.

Q3

Assume that the xyxy-plane represents the ground, and the zz-axis is the height. Aircraft 1 starts at airport A(0km∣0km∣0km)A(0km \vert 0km \vert 0km), moving along the direction (−5km2.5km0.1km)\left(\begin{array}{r} -5km\\ 2.5km\\ 0.1km \end{array}\right). Aircraft 2 starts at airport B(200km∣300km∣3km)B(200km \vert 300km \vert 3km), moving along the direction (−1.2km−0.2km0.01km)\left(\begin{array}{r} -1.2km\\ -0.2km\\ 0.01km \end{array}\right). Is it theoretically possible for the two planes to collide? If so, and given that they take off at the same time, at what speed would the planes have to travel for a collision to happen after exactly two hours?

Solution
A1
  1. The lines are parallel (overlapping).
  2. The lines are parallel (not overlapping).
  3. The lines are skew.
  4. The lines intersect, S(6∣4∣7)S(6\vert 4\vert 7).
A2

The intersection point is S(2∣1∣3)S(2\vert 1 \vert 3).

A3

The collision point is at S(−400km∣200km∣8km)S(-400km\vert 200km\vert 8km). For a collision to happen, aircraft 1 needs to travel at a speed of 223.6km/h223.6km/h, aircraft 2 at a speed of 304.1km/h304.1km/h.