3.15 Coincident, parallel, intersecting and skew lines

Two straight lines in a plane either meet or are parallel. In three dimensions there is a fourth possibility: two lines can point in different directions and still never meet, like a road on a bridge and a railway passing underneath it at an angle. Deciding which of the four cases applies, and finding where lines meet or how close they come, is a basic task in navigation, engineering design and computer graphics.

The four possible relationships between two lines

Every line has a direction vector, and any two lines can be compared using their direction vectors and the points they pass through. Two questions settle the relationship completely: are the directions the same, and do the lines share a point?

Coincident lines often look different on paper, because the same line can be written with any of its points as the starting point and any non-zero multiple of its direction vector. Skew lines can only exist in three dimensions. Two lines that are parallel or intersecting always lie in a common plane (they are coplanar), while two skew lines never do.

The table summarises how each case is recognised from the vector equations $\boldsymbol{r} = \boldsymbol{a}_1 + \lambda\boldsymbol{b}_1$ and $\boldsymbol{r} = \boldsymbol{a}_2 + \mu\boldsymbol{b}_2$.

Lines in two dimensions

In a plane, two lines with non-parallel direction vectors always meet. So in two dimensions only three cases occur: coincident, parallel or intersecting. To find the intersection of $\boldsymbol{r} = \begin{pmatrix} 1 \\ 2 \end{pmatrix} + \lambda\begin{pmatrix} 3 \\ 1 \end{pmatrix}$ and $\boldsymbol{r} = \begin{pmatrix} 0 \\ 5 \end{pmatrix} + \mu\begin{pmatrix} 1 \\ -1 \end{pmatrix}$, equate the two components:

Substituting the first equation into the second gives $2 + \lambda = 4 - 3\lambda$, so $\lambda = \frac{1}{2}$ and $\mu = \frac{5}{2}$. The lines meet at $\left(\frac{5}{2}, \frac{5}{2}\right)$. With two unknowns and two equations there is nothing left over to check, which is exactly why skew lines cannot occur in a plane.

Parallel and coincident lines

Two lines are parallel (or coincident) exactly when one direction vector is a scalar multiple of the other. The multiple can be negative or a fraction: $\begin{pmatrix} -4 \\ 2 \\ -6 \end{pmatrix} = -2\begin{pmatrix} 2 \\ -1 \\ 3 \end{pmatrix}$, so these two vectors give parallel lines even though they point in opposite senses.

To separate parallel from coincident, test whether one point of the first line lies on the second. If it does, the lines share a point and the same direction, so they are the same line. If it does not, they are distinct parallel lines and never meet.

Intersecting and skew lines

When the direction vectors are not parallel, the lines either intersect at one point or are skew. A point of intersection must lie on both lines, so there must be a value of $\lambda$ and a value of $\mu$ that make the two position vectors equal:

Equating components

Writing this vector equation one component at a time gives three equations in only two unknowns, $\lambda$ and $\mu$. The method is:

1. Use a different parameter for each line (for example $\lambda$ and $\mu$).

2. Equate the $x$-, $y$- and $z$-components to get three equations.

3. Solve two of the equations simultaneously for $\lambda$ and $\mu$.

4. Substitute both values into the third equation. If it is satisfied, the lines intersect; substitute $\lambda$ into $L_1$ (or $\mu$ into $L_2$) to find the point. If it is not satisfied, the lines do not meet, and since they are not parallel they are skew.

Three equations in two unknowns usually have no solution, which is why skew lines are the typical case in three dimensions. Two lines drawn at random in space are almost never in a common plane.

The decision process

The checks combine into a short decision process. Comparing direction vectors comes first, because it decides which of the two follow-up questions to ask.

The check in the third equation is the step that proves the lines meet. Without it, the two values found only make two of the three components agree, and the "point of intersection" could be two different points that happen to share an $x$- and a $y$-coordinate.

Lines given in Cartesian form

When a line is given in Cartesian form, convert it to vector form before equating components. The line $\frac{x - x_0}{l} = \frac{y - y_0}{m} = \frac{z - z_0}{n}$ passes through $(x_0, y_0, z_0)$ with direction $\begin{pmatrix} l \\ m \\ n \end{pmatrix}$. Setting each fraction equal to a parameter gives the parametric equations directly. For example,

because $y - 5$ is the same as $\frac{y - 5}{1}$. The signs need care: $z + 2 = z - (-2)$, so the $z$-coordinate of the point is $-2$.

Finding a parameter so that lines intersect

Some questions give one line with an unknown constant and ask for the value that makes the lines meet. The method is the same: solve the two equations that do not contain the unknown, then use the third equation to find it.

Paths that cross and objects that collide

In kinematics, $\boldsymbol{r} = \boldsymbol{a} + t\boldsymbol{v}$ gives the position of a moving object at time $t$ (see 3.14). Two different questions can be asked about two such objects. Whether their paths cross is a question about the lines, so it uses a separate parameter for each line. Whether the objects collide asks whether they are at the same point at the same time, so it uses the same $t$ in both equations. Paths can cross without a collision when the objects reach the crossing point at different times.

Shortest distance from a point to a line

The distance from a point to a line means the shortest distance. Of all the segments joining a point $A$ to points of a line $L$, the shortest one meets $L$ at a right angle. Its end on the line is the foot of the perpendicular, usually labelled $F$.

The segment $AF$ is shortest because, for any other point $P$ on the line, triangle $AFP$ has a right angle at $F$, so $AP$ is its hypotenuse and is longer than $AF$.

Finding the foot of the perpendicular

The method uses the scalar product to express "perpendicular" as an equation (3.13):

1. Write a general point $P$ on the line in terms of the parameter: $\overrightarrow{OP} = \boldsymbol{a} + \lambda\boldsymbol{b}$.

2. Find $\overrightarrow{AP}$ in terms of $\lambda$.

3. Set $\overrightarrow{AP} \cdot \boldsymbol{b} = 0$ and solve for $\lambda$. This value gives the foot $F$

4. The shortest distance is $|\overrightarrow{AF}|$.

The equation in step 3 is always linear in $\lambda$, so it has exactly one solution: there is only one foot of the perpendicular.

A formula using the vector product

Once the vector product is known (3.16), there is a quicker way to find the distance alone. If $P_0$ is any point on the line and $\theta$ is the angle between $\overrightarrow{P_0A}$ and $\boldsymbol{b}$, then the distance is $d = |\overrightarrow{P_0A}|\sin\theta$. Since $|\overrightarrow{P_0A} \times \boldsymbol{b}| = |\overrightarrow{P_0A}||\boldsymbol{b}|\sin\theta$,

In the example, $P_0 = (2, -1, 0)$, $\overrightarrow{P_0A} = \begin{pmatrix} 3 \\ 5 \\ -1 \end{pmatrix}$ and $\overrightarrow{P_0A} \times \boldsymbol{b} = \begin{pmatrix} 11 \\ -7 \\ -2 \end{pmatrix}$, so $d = \frac{\sqrt{174}}{\sqrt{6}} = \sqrt{29}$, as before. This formula does not give the foot $F$, so use the scalar product method whenever the point itself is asked for.

Distance between parallel lines

Parallel lines stay the same distance apart everywhere. So the distance between them is the distance from any point on one line to the other line, found with the foot of the perpendicular.

Shortest distance between skew lines

Two skew lines come closest at one pair of points, $P$ on $L_1$ and $Q$ on $L_2$. The segment $PQ$ is perpendicular to both lines; it is called the common perpendicular. If $PQ$ were not perpendicular to $L_1$, sliding $P$ a little along $L_1$ would shorten it, and the same is true for $L_2$.

Method 1: general points and two perpendicular conditions

This method finds the points $P$ and $Q$ as well as the distance.

1. Write general points $P$ on $L_1$ (parameter $\lambda$) and $Q$ on $L_2$ (parameter $\mu$).

2. Find $\overrightarrow{PQ}$ in terms of $\lambda$ and $\mu$.

3. Set $\overrightarrow{PQ} \cdot \boldsymbol{b}_1 = 0$ and $\overrightarrow{PQ} \cdot \boldsymbol{b}_2 = 0$

4. Substitute to find $P$, $Q$ and $|\overrightarrow{PQ}|$.

Method 2: projecting onto the common normal

A vector perpendicular to both lines is $\boldsymbol{n} = \boldsymbol{b}_1 \times \boldsymbol{b}_2$ (3.16). The segment $PQ$ is parallel to $\boldsymbol{n}$. For any point $A_1$ on $L_1$ and any point $A_2$ on $L_2$, the length of $PQ$ is the component of $\overrightarrow{A_1A_2}$ in the direction of $\boldsymbol{n}$:

This works because moving $A_1$ along $L_1$ or $A_2$ along $L_2$ only adds multiples of $\boldsymbol{b}_1$ or $\boldsymbol{b}_2$, and both are perpendicular to $\boldsymbol{n}$, so the dot product does not change. If $d = 0$, the lines are coplanar, so they intersect (or are parallel, in which case $\boldsymbol{n} = \boldsymbol{0}$ and the formula cannot be used).

A modelling application

Engineers planning underground services check the clearance between straight pipes, cables and tunnels, which are usually skew. A calculator handles the arithmetic once the method is set up.

Exam-style questions on lines in three dimensions

The questions combine every skill on lines: classifying, intersecting, angles (3.14) and distances. They often move from a "show that" part to a part that uses its result.

The command term matters in the last part of questions like this. "Find the shortest distance" accepts either method; "find the points on each line that are closest" needs Method 1, because Method 2 gives only the length.

Self-check

• To do:

  I can test whether two direction vectors are parallel and decide between coincident and parallel lines by testing a point.

• To do:

  I can find the point where two lines intersect, solving two equations and checking the third.

• To do:

  I can prove that two lines are skew, stating both conditions.

• To do:

  I can convert a line from Cartesian form to vector form before comparing it with another line.

• To do:

  I can find an unknown constant that makes two lines intersect.

• To do:

  I can tell the difference between paths crossing and objects colliding.

• To do:

  I can find the foot of the perpendicular from a point to a line and the shortest distance to the line.

• To do:

  I can find the shortest distance between parallel lines and between skew lines.

Now test whether the point $(5, 0, 5)$ on $L_2$ lies on $L_1$. Each component gives a value of $\lambda$:

All three components give the same value, $\lambda = 2$, so $(5, 0, 5)$ lies on $L_1$. The lines have parallel directions and a common point, so they are coincident.

(b) $\begin{pmatrix} 4 \\ -2 \\ 6 \end{pmatrix} = 2\begin{pmatrix} 2 \\ -1 \\ 3 \end{pmatrix}$, so $L_1$ and $L_3$ are parallel or coincident. Test the point $(4, 0, 1)$ on $L_3$:

The values disagree, so no single value of $\lambda$ gives $(4, 0, 1)$. The point is not on $L_1$, so $L_1$ and $L_3$ are parallel and distinct.

Equate components:

From the $x$-equation, $\lambda = 2\mu$. Substitute into the $y$-equation: $4\mu = 1 + 3\mu$, so $\mu = 1$ and $\lambda = 2$.

Check in the $z$-equation: left side $2 - 2 = 0$, right side $-2 + 2(1) = 0$. Both sides are equal, so the lines intersect.

Substitute $\lambda = 2$ into $L_1$:

As a check, $\mu = 1$ in $L_2$ gives $(1 + 2, 1 + 3, -2 + 2) = (3, 4, 0)$ as well. The point of intersection is $(3, 4, 0)$.

No common point: equate components.

Substitute $\mu = 1 + \lambda$ from the $x$-equation into the $y$-equation: $2\lambda = 1 - (1 + \lambda) = -\lambda$, so $3\lambda = 0$, giving $\lambda = 0$ and $\mu = 1$.

Check in the $z$-equation: left side $2 - 0 = 2$, right side $1 + 2(1) = 3$. Since $2 \neq 3$, there are no values of $\lambda$ and $\mu$ satisfying all three equations, so the lines do not intersect.

The lines are neither parallel nor intersecting, so they are skew.

The $x$- and $y$-equations do not involve $k$, so solve them first. Subtracting the $x$-equation from the $y$-equation:

Then $\mu = 2\lambda - 1 = -5$.

The lines intersect, so the $z$-equation must also hold: $2 - (-2) = k + (-5)$, which gives $k = 9$.

The point is $\lambda = -2$ on $L_1$: $(1 - 2, \ 0 - 4, \ 2 + 2) = (-1, -4, 4)$. Check with $\mu = -5$ on $L_4$: $(4 - 5, \ 1 - 5, \ 9 - 5) = (-1, -4, 4)$.

The $z$-equation gives $0.4s = 2$, so $s = 5$. Then the $y$-equation gives $10 - 2t = 2$, so $t = 4$.

Check in the $x$-equation: left side $-20 + 32 = 12$, right side $-3 + 15 = 12$. The equation holds, so the paths cross. The crossing point is $\boldsymbol{r}_A$ at $t = 4$: $(12, 2, 9)$.

(b) Aircraft $A$ is at the crossing point when $t = 4$ (12:04) and aircraft $B$ when $t = 5$ (12:05). They are never at the same place at the same time, so they do not collide.

(c) At $t = 4$, $\boldsymbol{r}_B = \begin{pmatrix} -3 + 12 \\ -18 + 16 \\ 7 + 1.6 \end{pmatrix} = \begin{pmatrix} 9 \\ -2 \\ 8.6 \end{pmatrix}$. The distance between the aircraft is

For the foot of the perpendicular, $\overrightarrow{AP}$ is perpendicular to the direction vector:

So $\lambda = 1$, and the foot is $F(3, 0, 2)$.

(b) With $\lambda = 1$, $\overrightarrow{AF} = \begin{pmatrix} -2 \\ -4 \\ 3 \end{pmatrix}$. As a check, $\begin{pmatrix} -2 \\ -4 \\ 3 \end{pmatrix} \cdot \begin{pmatrix} 1 \\ 1 \\ 2 \end{pmatrix} = -2 - 4 + 6 = 0$. The shortest distance is

Set $\overrightarrow{AP} \cdot \begin{pmatrix} 2 \\ -1 \\ 3 \end{pmatrix} = 0$:

The foot is $F(3, 1, 2)$ and $\overrightarrow{AF} = \begin{pmatrix} -1 \\ 1 \\ 1 \end{pmatrix}$. The distance between the lines is

Perpendicular to $L_1$:

Perpendicular to $L_3$:

So $6\lambda + 3\mu = 2$ and $3\lambda + 6\mu = 4$. Doubling the second equation and subtracting the first gives $9\mu = 6$, so $\mu = \frac{2}{3}$, and then $6\lambda = 2 - 2 = 0$, so $\lambda = 0$.

Substitute: $P = (1, 0, 2)$ and $Q = \left(\frac{2}{3}, \frac{1}{3}, \frac{7}{3}\right)$, so $\overrightarrow{PQ} = \begin{pmatrix} -\frac{1}{3} \\ \frac{1}{3} \\ \frac{1}{3} \end{pmatrix}$.

Check with Method 2: $\boldsymbol{b}_1 \times \boldsymbol{b}_2 = \begin{pmatrix} 3 \\ -3 \\ -3 \end{pmatrix}$, and $\boldsymbol{a}_2 - \boldsymbol{a}_1 = \begin{pmatrix} -1 \\ 1 \\ -1 \end{pmatrix}$, so

Both methods agree. Notice also that $\overrightarrow{PQ}$ is a multiple of $\boldsymbol{b}_1 \times \boldsymbol{b}_2$, as it must be.

The vector between the two given points is $\begin{pmatrix} 3 \\ -1 \\ -2 \end{pmatrix}$, and

So the shortest distance is

Since $2.52 > 2$, the plan meets the rule. In this model the pipe and duct are lines with no thickness; a real check would also subtract their radii from the distance between the centre lines.

The first gives $\mu = 2\lambda$. Substituting into the second: $-2 + \lambda = -7 + 6\lambda$, so $\lambda = 1$ and $\mu = 2$. Check the third: $3 - 1 = 2$ and $-2 + 4 = 2$, which agree. The point is $\lambda = 1$ on $L_1$: $(3, -1, 2)$. Writing "the third equation is satisfied, so the lines intersect" earns the reasoning mark.

(b) Use the direction vectors, not the position vectors:

With a GDC, this gives $\theta = 70.9°$.

(c) Write $L_3$ in vector form. With $z = t$, $x = t - 1$ and $y = 1$, so $L_3: \boldsymbol{r} = \begin{pmatrix} -1 \\ 1 \\ 0 \end{pmatrix} + t\begin{pmatrix} 1 \\ 0 \\ 1 \end{pmatrix}$. The directions $\begin{pmatrix} 2 \\ 1 \\ -1 \end{pmatrix}$ and $\begin{pmatrix} 1 \\ 0 \\ 1 \end{pmatrix}$ are not parallel, so the lines are not parallel. Equate components:

The second gives $\lambda = 3$, then the third gives $t = 0$. Check the first: $1 + 6 = 7$ but $-1 + 0 = -1$. The equations are inconsistent, so the lines do not meet, and since they are not parallel, they are skew.

(d) $\boldsymbol{n} = \begin{pmatrix} 2 \\ 1 \\ -1 \end{pmatrix} \times \begin{pmatrix} 1 \\ 0 \\ 1 \end{pmatrix} = \begin{pmatrix} 1 \times 1 - (-1) \times 0 \\ (-1) \times 1 - 2 \times 1 \\ 2 \times 0 - 1 \times 1 \end{pmatrix} = \begin{pmatrix} 1 \\ -3 \\ -1 \end{pmatrix}$. With $\boldsymbol{a}_3 - \boldsymbol{a}_1 = \begin{pmatrix} -2 \\ 3 \\ -3 \end{pmatrix}$:

$(11, 7, 3)$ A1

Note: Award M0 for using different parameters for the two drones, unless it is then shown that both parameters are equal.

Show that $L_1$ and $L_2$ intersect, and find the coordinates of their point of intersection.

[5]5 marks

1. (a)

   Find a vector equation of $L_1$.

   [2]2 marks

2. (b)

   The line $L_2$ has vector equation

   $$\mathbf{r} = \begin{pmatrix} 0 \\ 1 \\ 5 \end{pmatrix} + \mu \begin{pmatrix} 3 \\ 1 \\ -2 \end{pmatrix}, \quad \mu \in \mathbb{R}.$$

   Show that $L_1$ and $L_2$ are skew.

   [5]5 marks