F6 - Intersections in space | AQA A-Level Further Mathematics

Lines that cross on a drawing may miss each other in space. You will find line-line and line-plane intersections by solving for parameters, then use direction and consistency checks to distinguish a single intersection, parallel objects, coincident lines and skew lines.

Solving line-line and line-plane intersections

If two lines cross when viewed from above, must they meet in space? Only if their heights agree at the same position too. An intersection is a point satisfying every coordinate equation of both objects.

You already know that $\mathbf{r}=\mathbf{a}+s\mathbf{u}$ , with $\mathbf{u}\ne\mathbf{0}$ , describes a line. Each line needs its own parameter: the same meeting point can occur at different parameter values on the two lines.

For lines $\mathbf{r}=\mathbf{a}+s\mathbf{u}$ and $\mathbf{r}=\mathbf{b}+t\mathbf{v}$ , equating position vectors gives three scalar equations in $s$ and $t$ . Solve a useful pair, then check the third. A candidate from only two coordinates is not yet an intersection.

Worked example 1: intersecting two lines

Question. Find the intersection of

L:\ \mathbf{r}=(1,0,2)+s(1,2,-1),\qquad M:\ \mathbf{r}=(2,3,2)+t(0,1,1)

Plan. Equate the first two coordinates to find both parameters, then verify the third coordinate.

\begin{aligned} 1+s&=2 &&\text{equate first coordinates}\\ s&=1\\ 2s&=3+t &&\text{equate second coordinates}\\ 2&=3+t,\quad t=-1\\ 2-s&=2+t &&\text{check third coordinates}\\ 1&=1 \end{aligned}

Answer. The lines intersect at $(1,0,2)+(1)(1,2,-1)=(2,2,1)$ .

Check. Substituting $t=-1$ into the other line gives $(2,3,2)-(0,1,1)=(2,2,1)$ , the same complete point.

Keep both direction vectors fixed and move the second line's starting point by $(1,2,-1)$ , one direction step along $L$ . The intersection should move by the same step.

Worked example 2: translating the crossing line

Question. Find the intersection of

L:\ \mathbf{r}=(1,0,2)+s(1,2,-1),\qquad N:\ \mathbf{r}=(3,5,1)+t(0,1,1)

Plan. Repeat the coordinate comparison and check the unused coordinate before stating the point.

\begin{aligned} 1+s&=3,\quad s=2 &&\text{first coordinates}\\ 2s&=5+t,\quad t=-1 &&\text{second coordinates}\\ 2-s&=1+t &&\text{third-coordinate check}\\ 0&=0 \end{aligned}

Answer. The intersection is $(3,4,0)$ .

Check. $(3,4,0)-(2,2,1)=(1,2,-1)$ , the translation used. Direct substitution in $N$ also gives $(3,5,1)-(0,1,1)=(3,4,0)$ .

A line-plane intersection uses the same membership idea. A point on the line already has its coordinates linked by one parameter; substitute all three expressions into the plane equation. For a plane $\mathbf{n}\cdot\mathbf{r}=d$ , with $\mathbf{n}\ne\mathbf{0}$ , this gives

\mathbf{n}\cdot(\mathbf{a}+s\mathbf{u})=d \quad\Longrightarrow\quad s(\mathbf{n}\cdot\mathbf{u})=d-\mathbf{n}\cdot\mathbf{a}

When $\mathbf{n}\cdot\mathbf{u}\ne0$ , there is exactly one parameter value and therefore exactly one intersection. When this coefficient is zero, do not divide; the next part handles the possibilities.

Worked example 3: meeting a plane

Question. Find where $L:\ \mathbf{r}=(1,0,2)+s(1,2,-1)$ meets the plane $x+y+z=7$ .

Plan. Substitute the line coordinates into the plane, solve for $s$ , then recover the point.

\begin{aligned} (1+s)+2s+(2-s)&=7 &&\text{apply the plane condition}\\ 3+2s&=7 &&\text{collect like terms}\\ s&=2 &&\text{solve}\\ \mathbf{r}&=(1,0,2)+2(1,2,-1)=(3,4,0) &&\text{substitute in the line} \end{aligned}

Answer. $(3,4,0)$ .

Check. In the plane, $3+4+0=7$ . On the line, the displacement $(3,4,0)-(1,0,2)=(2,4,-2)$ is twice the direction vector.

Change only the plane's constant. Its normal stays $(1,1,1)$ , so it remains parallel to the previous plane.

Worked example 4: a parallel plane at a new position

Question. Find where the same line $L$ meets $x+y+z=3$ .

Plan. Use the new plane condition with the unchanged parametric coordinates.

\begin{aligned} (1+s)+2s+(2-s)&=3 &&\text{substitute into the new plane}\\ 3+2s&=3,\quad s=0 &&\text{solve}\\ \mathbf{r}&=(1,0,2) &&\text{use }s=0 \end{aligned}

Answer. $(1,0,2)$ .

Check. The line's given starting point satisfies $1+0+2=3$ . Also $\mathbf{n}\cdot\mathbf{u}=(1,1,1)\cdot(1,2,-1)=2\ne0$ , confirming that the line crosses the plane at just one point.

The line-line method has one extra safeguard: a third coordinate equation remains after solving a pair.

Watch out

Wrong: “The first two coordinate equations give $s$ and $t$ , so the lines meet.” Those values must satisfy the third equation too. Also, using one shared parameter for two independently parametrised lines can miss a real intersection.

Find a new line-plane intersection, giving coordinates rather than just a parameter.

Parallel, coincident and skew lines

Two lines in three dimensions can fail to meet even when their directions are not parallel. This is the possibility a flat drawing can hide.

Relationships between lines

Intersecting lines have exactly one common point. Distinct parallel lines have parallel direction vectors and no common points. Coincident lines are the same line, with infinitely many common points. Skew lines are nonparallel lines that do not intersect; they do not lie in one plane.

All line direction vectors must be nonzero. Compare them first using scalar multiples, including negative multiples. If they are parallel, test whether a starting point of one line belongs to the other: membership means coincidence; failure means distinct parallel lines. If the directions are nonparallel, solve for an intersection and check all coordinates: a consistent solution means one intersection; inconsistency means skew lines.

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F6 - Intersections in space

Solving line-line and line-plane intersections

Parallel, coincident and skew lines

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