Core Pure · Vectors · Year 1

Points of Intersection

Do two lines meet, or pass each other by in different planes entirely? Setting parametric equations equal — and always checking the leftover equation — gives the answer.

17 min read 4 worked examples

Intersection of two lines

Method 37.1 — do two 3D lines intersect?

Set the two line equations equal, componentwise, giving three equations in two parameters $\lambda,\mu$. 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 $l_1$ and $l_2$, drawn as if intersecting at a point, but really passing over and under one another in different planes.

Two straight lines in a 3D coordinate box, drawn so that neither their directions nor their positions ever bring them together, illustrating skew lines that plainly do not meet

Fig. 2 — skew lines $m_1$ and $m_2$: their projections don't even cross here, making it clear the lines never meet, wherever they're viewed from.

Example 1 · Testing two lines for intersection

Determine whether the lines $\mathbf{r}_1=(1,2,0)+\lambda(1,-1,2)$ and $\mathbf{r}_2=(0,3,1)+\mu(2,1,-1)$ intersect.

Equate components: $x$: $1+\lambda=2\mu$. $y$: $2-\lambda=3+\mu$. $z$: $2\lambda=1-\mu$.

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

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.

Example 2 · Finding the point of intersection

Show that the lines $\mathbf{r}_1=(0,0,0)+\lambda(1,1,1)$ and $\mathbf{r}_2=(4,0,0)+\mu(-1,1,1)$ intersect, and find the point of intersection.

Equate components: $x$: $\lambda=4-\mu$. $y$: $\lambda=\mu$. $z$: $\lambda=\mu$.

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

Check in the $z$-equation: LHS $=\lambda=2$; RHS $=\mu=2$. These match, confirming the lines intersect.

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

Intersection of a line and a plane

Method 37.2 — line meeting a plane

Substitute the parametric coordinates of the line ($x=x_0+\lambda l$, etc.) into the Cartesian equation of the plane, and solve for $\lambda$. Substituting this value of $\lambda$ back into the line's equation gives the point of intersection. If the resulting equation in $\lambda$ has no solution, the line is parallel to the plane and doesn't meet it (or, if the equation is true for all $\lambda$, the line lies entirely within the plane).

Example 3 · Finding where a line meets a plane

Find the point where the line $\mathbf{r}=(1,2,-1)+\lambda(2,-1,3)$ meets the plane $x+2y-z=9$.

Substitute $x=1+2\lambda$, $y=2-\lambda$, $z=-1+3\lambda$ into the plane equation:

$(1+2\lambda)+2(2-\lambda)-(-1+3\lambda)=9$

$1+2\lambda+4-2\lambda+1-3\lambda=9 \Rightarrow 6-3\lambda=9 \Rightarrow -3\lambda=3 \Rightarrow \lambda=-1$.

Substituting back: $x=1-2=-1$, $y=2+1=3$, $z=-1-3=-4$.

The line meets the plane at $(-1,3,-4)$.

Example 4 · A line parallel to a plane (no intersection)

Show that the line $\mathbf{r}=(2,1,0)+\lambda(1,1,1)$ does not meet the plane $x+y-2z=1$.

Substitute: $(2+\lambda)+(1+\lambda)-2(\lambda)=1$

$2+\lambda+1+\lambda-2\lambda=1 \Rightarrow 3+0\lambda=1 \Rightarrow 3=1$, which is never true.

Since this equation has no solution for $\lambda$, the line never meets the plane — it is parallel to it (and, checking the direction vector $(1,1,1)$ against the normal $(1,1,-2)$: their dot product is $1+1-2=0$, confirming the line's direction is indeed perpendicular to the normal, i.e. parallel to the plane).

Common mistake

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

Exam-style questions

Question 1 6 marks

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

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

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

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

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

Question 2 6 marks

Find the point where the line $\mathbf{r}=(0,-1,2)+\lambda(1,2,-1)$ meets the plane $2x-y+z=7$.

Substitute: $2(\lambda)-(-1+2\lambda)+(2-\lambda)=7$ (M1, A1)

$2\lambda+1-2\lambda+2-\lambda=7 \Rightarrow 3-\lambda=7 \Rightarrow \lambda=-4$ (M1, A1)

$x=-4$, $y=-1-8=-9$, $z=2+4=6$; point is $(-4,-9,6)$ (M1, A1)

Question 3 6 marks

Show that the lines $\mathbf{r}_1=(1,0,0)+\lambda(0,1,0)$ and $\mathbf{r}_2=(0,0,1)+\mu(1,0,0)$ are skew.

$x$: $1=\mu$. $y$: $\lambda=0$. $z$: $0=1$. (M1, A1)

The $x$- and $y$-equations give $\mu=1$, $\lambda=0$ directly, but the $z$-equation reads $0=1$, which is never true regardless of $\lambda$ or $\mu$. (M1, A1)

The direction vectors $(0,1,0)$ and $(1,0,0)$ are not parallel (not scalar multiples of one another), so the lines are not parallel; combined with the inconsistent $z$-equation, the lines never meet. (A1)

Since the lines are neither parallel nor intersecting, they are skew. (A1)

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Angles Between Lines & Planes

Finding Perpendiculars