Mathematics 9709 · Pure Mathematics 3

Intersection of two lines

Two lines in 3D do one of three things: they cross, run parallel, or are skew (never meet, but not parallel either: this is new in 3D).
Working out which is just algebra. Set the two lines equal, match the rows, and solve two of the three equations.
The third equation is the check. It tells you whether they cross or are skew.

Set $\mathbf{a}+\lambda\mathbf{d_1}=\mathbf{b}+\mu\mathbf{d_2}$ , solve two rows for $\lambda,\mu$ , then test the third: it works → they cross; it fails → they miss (parallel if the directions match, otherwise skew).

Three outcomes, checked in order: are the directions a multiple of each other? → parallel. Otherwise solve two rows and ask does the third agree? → yes, they cross; no, skew. “Skew” means not parallel and never meeting, the case that only shows up in 3D.

Set equal, match components, then CHECK

Give each line its own letter ( $\lambda$ for one, $\mu$ for the other). Sharing one letter is the classic mistake.
Set them equal and you get three equations in two unknowns:

Check parallel first: if the directions are scalar multiples, the lines are parallel (and either identical or never meet) — no need to solve anything.
Not parallel? Solve two of the component equations for $\lambda$ and $\mu$ .
Substitute into the third. Fits → they intersect (put $\lambda$ back for the point); contradiction → they are skew.

At a crossing P, both lines hit the same point: a + λd₁ = b + μd₂.

Why the third equation matters

Two unknowns $(\lambda,\mu)$ , three equations. You can almost always solve any two of them. But that only finds the $\lambda,\mu$ where two coordinates agree. A real crossing needs all three coordinates to agree at once. So the third equation is the test: if the values that fix $x$ and $y$ also fix $z$ , the lines really meet. If not, they pass at different heights and are skew.

Worked examples

Worked example

Real question: lines $\mathbf{r}=(1,2,1)+\lambda(a,2,-1)$ and $\mathbf{r}=(2,1,-1)+\mu(2,-1,1)$ intersect: find $a$ and the meeting point

9709/31 O/N 2020 Q11(a). A twist: you are told they meet and asked for the unknown $a$ . So solve the two rows without $a$ first, then the third row gives you $a$ .

The $a$ sits only in the $x$ -row, so start with $y$ and $z$ : $y:\ 2+2\lambda=1-\mu$ , $z:\ 1-\lambda=-1+\mu$ .
Solve those two: $\lambda=-3,\ \mu=5$ .
Now the $x$ -row gives $a$ : $1+a\lambda=2+2\mu\Rightarrow1-3a=12\Rightarrow a=-\tfrac{11}{3}$ .
Put $\mu=5$ into line 2 for the point: $(2,1,-1)+5(2,-1,1)=(12,-4,4)$ .

Answer

a=-\tfrac{11}{3},\quad \text{meet at }(12,\,-4,\,4)

Worked example

Intersect $\mathbf{r}=(1,-2,-4)+\lambda(-1,-1,1)$ and $\mathbf{r}=(2,-2,1)+\mu(1,2,-7)$

Directions $(-1,-1,1)$ and $(1,2,-7)$ aren't multiples, so not parallel — worth checking for a crossing.
Match two components. From $y$ : $-2-\lambda=-2+2\mu\Rightarrow-\lambda=2\mu$ ; from $x$ : $1-\lambda=2+\mu$ . Solving: $\mu=1,\ \lambda=-2$ .
Check in $z$ : line 1 gives $-4+\lambda=-6$ , line 2 gives $1-7\mu=-6$ . They agree, so the lines meet.
Put $\lambda=-2$ back into line 1 for the point.

Answer

\text{meet at }(3,\,0,\,-6)

Worked example

Real question: show the lines are skew, directions $(-1,1,2)$ and $(5,-3,2)$

9709/33 O/N 2023 Q11(c): line $l=(1,-2,-3)+\lambda(-1,1,2)$ and line $m=(-2,2,-1)+\mu(5,-3,2)$ .

Not parallel: $(-1,1,2)$ and $(5,-3,2)$ are not multiples of each other.
Match $x$ and $y$ and solve: $\lambda=\tfrac{11}{2},\ \mu=-\tfrac12$ .
Check $z$ : line $l$ gives $-3+2\lambda=8$ , line $m$ gives $-1+2\mu=-2$ . They disagree, so no crossing.

Answer

\text{not parallel and never meet}\ \Rightarrow\ \text{skew}

Where the marks go

Common mistake

Always put

\lambda,\mu

back into the third equation. If it fails, the lines are skew, not crossing. Two matching equations on their own never prove a crossing in 3D. (9709/33 Oct/Nov 2023 Q11)

Common mistake

Check parallel first (one direction a multiple of the other). Two different parallel lines never meet, but don't call them skew. Skew means not parallel and not meeting.

Common mistake

Give each line its own letter. Reusing one letter forces the lines to move together, which breaks the method.

Common mistake

When an unknown constant hides in one row (find

a

when you are told the lines meet), solve the two rows without it first to find

\lambda,\mu

, then put those into the last row to get the constant. Don't try to juggle all three unknowns at once. (9709/31 Oct/Nov 2020 Q11)

Now you try

Line $l$ is $\mathbf{r}=(1,-2,-3)+s(-1,1,2)$ and line $m$ is $\mathbf{r}=(-2,2,-1)+u(5,-3,2)$ . Are they parallel, do they cross, or are they skew? (9709/33 Oct/Nov 2023 Q11)

Your turn— tap to reveal the worked answer (9709/33 Oct/Nov 2023 Q11(c))

Directions $(-1,1,2)$ and $(5,-3,2)$ aren't multiples, so not parallel. From $x$ and $y$ : $s=\tfrac{11}{2},\ u=-\tfrac12$ .

Now check $z$ : line $l$ gives $-3+2s=8$ , line $m$ gives $-1+2u=-2$ . They disagree, so no crossing:

Answer

\text{the lines are skew}

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Intersection of two lines

Two lines in 3D do one of three things: they cross, run parallel, or are skew (never meet, but not parallel either: this is new in 3D).
Working out which is just algebra. Set the two lines equal, match the rows, and solve two of the three equations.
The third equation is the check. It tells you whether they cross or are skew.

Set equal, match components, then CHECK

Give each line its own letter ( $\lambda$ for one, $\mu$ for the other). Sharing one letter is the classic mistake.
Set them equal and you get three equations in two unknowns:

Check parallel first: if the directions are scalar multiples, the lines are parallel (and either identical or never meet) — no need to solve anything.
Not parallel? Solve two of the component equations for $\lambda$ and $\mu$ .
Substitute into the third. Fits → they intersect (put $\lambda$ back for the point); contradiction → they are skew.

At a crossing P, both lines hit the same point: a + λd₁ = b + μd₂.

Why the third equation matters

Worked examples

Worked example

Real question: lines $\mathbf{r}=(1,2,1)+\lambda(a,2,-1)$ and $\mathbf{r}=(2,1,-1)+\mu(2,-1,1)$ intersect: find $a$ and the meeting point

9709/31 O/N 2020 Q11(a). A twist: you are told they meet and asked for the unknown $a$ . So solve the two rows without $a$ first, then the third row gives you $a$ .

The $a$ sits only in the $x$ -row, so start with $y$ and $z$ : $y:\ 2+2\lambda=1-\mu$ , $z:\ 1-\lambda=-1+\mu$ .
Solve those two: $\lambda=-3,\ \mu=5$ .
Now the $x$ -row gives $a$ : $1+a\lambda=2+2\mu\Rightarrow1-3a=12\Rightarrow a=-\tfrac{11}{3}$ .
Put $\mu=5$ into line 2 for the point: $(2,1,-1)+5(2,-1,1)=(12,-4,4)$ .

Answer

a=-\tfrac{11}{3},\quad \text{meet at }(12,\,-4,\,4)

Worked example

Intersect $\mathbf{r}=(1,-2,-4)+\lambda(-1,-1,1)$ and $\mathbf{r}=(2,-2,1)+\mu(1,2,-7)$

Directions $(-1,-1,1)$ and $(1,2,-7)$ aren't multiples, so not parallel — worth checking for a crossing.
Match two components. From $y$ : $-2-\lambda=-2+2\mu\Rightarrow-\lambda=2\mu$ ; from $x$ : $1-\lambda=2+\mu$ . Solving: $\mu=1,\ \lambda=-2$ .
Check in $z$ : line 1 gives $-4+\lambda=-6$ , line 2 gives $1-7\mu=-6$ . They agree, so the lines meet.
Put $\lambda=-2$ back into line 1 for the point.

Answer

\text{meet at }(3,\,0,\,-6)

Worked example

Real question: show the lines are skew, directions $(-1,1,2)$ and $(5,-3,2)$

9709/33 O/N 2023 Q11(c): line $l=(1,-2,-3)+\lambda(-1,1,2)$ and line $m=(-2,2,-1)+\mu(5,-3,2)$ .

Not parallel: $(-1,1,2)$ and $(5,-3,2)$ are not multiples of each other.
Match $x$ and $y$ and solve: $\lambda=\tfrac{11}{2},\ \mu=-\tfrac12$ .
Check $z$ : line $l$ gives $-3+2\lambda=8$ , line $m$ gives $-1+2\mu=-2$ . They disagree, so no crossing.

Answer

\text{not parallel and never meet}\ \Rightarrow\ \text{skew}

Where the marks go

Common mistake

Always put

\lambda,\mu

back into the third equation. If it fails, the lines are skew, not crossing. Two matching equations on their own never prove a crossing in 3D. (9709/33 Oct/Nov 2023 Q11)

Common mistake

Check parallel first (one direction a multiple of the other). Two different parallel lines never meet, but don't call them skew. Skew means not parallel and not meeting.

Common mistake

Give each line its own letter. Reusing one letter forces the lines to move together, which breaks the method.

Common mistake

When an unknown constant hides in one row (find

a

when you are told the lines meet), solve the two rows without it first to find

\lambda,\mu

, then put those into the last row to get the constant. Don't try to juggle all three unknowns at once. (9709/31 Oct/Nov 2020 Q11)

Now you try

Line $l$ is $\mathbf{r}=(1,-2,-3)+s(-1,1,2)$ and line $m$ is $\mathbf{r}=(-2,2,-1)+u(5,-3,2)$ . Are they parallel, do they cross, or are they skew? (9709/33 Oct/Nov 2023 Q11)

Your turn— tap to reveal the worked answer (9709/33 Oct/Nov 2023 Q11(c))

Directions $(-1,1,2)$ and $(5,-3,2)$ aren't multiples, so not parallel. From $x$ and $y$ : $s=\tfrac{11}{2},\ u=-\tfrac12$ .

Now check $z$ : line $l$ gives $-3+2s=8$ , line $m$ gives $-1+2u=-2$ . They disagree, so no crossing:

Answer

\text{the lines are skew}