Further Maths/Lessons/Vectors/Intersections & Perpendiculars
Core Pure · Vectors · Year 1

Intersections & Perpendiculars

Do two lines meet, and where — and what is the shortest possible distance from a point to a line that misses it entirely?

15 min read 3 worked examples

Testing whether two lines intersect

Method 3.1 — do two 3D lines intersect?

Set the two line equations equal, componentwise, giving three equations in the two parameters $s,t$. Solve two of the equations simultaneously, then check the solution in the third. If it's consistent, the lines meet at that point; if not, the lines are skew (they miss each other without being parallel).

Two straight lines in a 3D coordinate box that do not touch and are not parallel, drawn to show that they pass over and under each other without meeting, labelled as skew lines
Fig. 1 — skew lines: two lines in 3D that are not parallel yet never meet, because they lie in different planes.
Example 1 · Testing two lines for intersection
Determine whether the lines $\mathbf{r}_1=(1,2,0)+s(1,-1,2)$ and $\mathbf{r}_2=(0,3,1)+t(2,1,-1)$ intersect.

Equate components: $x$: $1+s=2t$. $y$: $2-s=3+t$. $z$: $2s=1-t$.

From the $y$-equation: $s=-1-t$. Substitute into the $x$-equation: $1+(-1-t)=2t \Rightarrow -t=2t \Rightarrow t=0$, so $s=-1$.

Check in the $z$-equation: LHS $=2(-1)=-2$; RHS $=1-0=1$. These don't match, so the lines are skew — they do not intersect.

Perpendicular distance from a point to a line

Method 3.2 — foot of perpendicular from a point to a line

To find the shortest distance from point $P$ to line $\mathbf{r}=\mathbf{a}+t\mathbf{d}$: let $F=\mathbf{a}+t\mathbf{d}$ be a general point on the line. The vector $\vec{PF}$ is perpendicular to the line exactly when $\vec{PF}\cdot\mathbf{d}=0$. Solve this equation for $t$ to find $F$, the foot of the perpendicular, then the distance is $|\vec{PF}|$.

A straight line in 3D with a point P not on the line, and a dashed perpendicular segment drawn from P down to a labelled point F on the line, showing the shortest distance from the point to the line
Fig. 2 — the foot of the perpendicular F is the unique point on the line closest to P, where PF meets the line at a right angle.
Example 2 · Shortest distance from a point to a line
Find the shortest distance from $P(1,0,2)$ to the line $\mathbf{r}=(2,1,0)+t(1,1,-1)$.

General point on line: $F=(2+t,1+t,-t)$. Then $\vec{PF}=F-P=(1+t,1+t,-t-2)$.

$\vec{PF}\cdot(1,1,-1)=0$: $(1+t)+(1+t)-(-t-2)=0 \Rightarrow 1+t+1+t+t+2=0 \Rightarrow 3t+4=0 \Rightarrow t=-\tfrac43$.

$F=(2-\tfrac43,1-\tfrac43,\tfrac43)=(\tfrac23,-\tfrac13,\tfrac43)$. $\vec{PF}=(\tfrac23-1,-\tfrac13,\tfrac43-2)=(-\tfrac13,-\tfrac13,-\tfrac23)$.

$$|\vec{PF}|=\sqrt{\tfrac19+\tfrac19+\tfrac49}=\sqrt{\tfrac69}=\sqrt{\tfrac23}=\frac{\sqrt6}{3}$$
Example 3 · Finding the point of intersection
Show that the lines $\mathbf{r}_1=(0,0,0)+s(1,1,1)$ and $\mathbf{r}_2=(4,0,0)+t(-1,1,1)$ intersect, and find the point of intersection.

Equate components: $x$: $s=4-t$. $y$: $s=t$. $z$: $s=t$.

From $y$: $s=t$. Substituting into the $x$-equation: $t=4-t \Rightarrow 2t=4 \Rightarrow t=2$, so $s=2$.

Check in the $z$-equation: LHS $=s=2$; RHS $=t=2$. These match, confirming consistency across all three equations, so the lines intersect.

Substituting $s=2$ into $\mathbf{r}_1$: the point of intersection is $(2,2,2)$.

Common mistake

Using the same parameter letter for both lines is a common error — always use a different letter ($s$ and $t$) for each line, since they are independent variables that generally take different values at the point of intersection.

Exam-style questions

Question 1 7 marks

Show that the lines $\mathbf{r}_1=(1,0,2)+s(1,1,-1)$ and $\mathbf{r}_2=(3,2,0)+t(-1,1,1)$ intersect, and find the coordinates of the point of intersection.

$x$: $1+s=3-t \Rightarrow s+t=2$. $y$: $s=2+t \Rightarrow s-t=2$. (M1, A1)

Adding the two equations: $2s=4 \Rightarrow s=2$, so $t=0$. (M1, A1)

Check $z$: LHS $=2-s=2-2=0$; RHS $=0+t=0$. These match, confirming the lines intersect. (M1)

Substituting $s=2$ into $\mathbf{r}_1$: point of intersection is $(1+2,0+2,2-2)=(3,2,0)$. (A1)

Question 2 6 marks

Find the shortest distance from the point $Q(0,0,0)$ to the line $\mathbf{r}=(1,2,1)+t(1,-1,1)$.

General point: $F=(1+t,2-t,1+t)$. $\vec{QF}=F-Q=(1+t,2-t,1+t)$. (M1)

$\vec{QF}\cdot(1,-1,1)=0$: $(1+t)-(2-t)+(1+t)=0 \Rightarrow 3t=0 \Rightarrow t=0$ (M1, A1)

$F=(1,2,1)$, so $\vec{QF}=(1,2,1)$, and $|\vec{QF}|=\sqrt{1+4+1}=\sqrt6$ (M1, A1, A1)

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