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Year 12 Specialist (Unit 3 & 4) Vectors in two and three dimensions

Intersection of lines and skew lines

20 practice questions 0 video lessons Theory + worked examples

Master the intersection of lines and skew lines for Year 12 Specialist Mathematics in Queensland (QCAA). Working from the vector equation of a line, you decide whether two lines in space intersect, run parallel, or are skew — the 3D case of lines that never meet yet never run parallel.

You will learn to equate the parametric forms, solve for the two parameters, and test the remaining component for consistency, then find the point of intersection and the angle between two lines — core skills for planes, spheres and vector geometry later in Unit 3.

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Theory

Intersection of lines and skew lines is where the vector equation of a line meets three dimensions in Year 12 Specialist Mathematics (QCAA, Queensland, Unit 3). Two lines in space either intersect at a point, run parallel, or are skew — never meeting yet never parallel. This page shows how to equate the parametric forms, solve for the two parameters, and test consistency to classify any pair.

A line in space is written in vector form \(\mathbf{r}=\mathbf{a}+\lambda\mathbf{d}\): a fixed point \(\mathbf{a}\) plus a scalar \(\lambda\) times a direction vector \(\mathbf{d}\). Each line uses its own parameter, so give the second line a different letter, \(\mu\).

Two lines intersect if there is one pair \((\lambda,\mu)\) making the two position vectors equal — a single common point. They are parallel when their direction vectors are scalar multiples of one another, and coincident (the same line) when they are parallel and share a point.

Two lines are skew when they are not parallel and do not intersect. Skew lines never lie in a single plane, so this can only happen in three dimensions — a genuinely 3D idea with no 2D analogue.

The deciding test is consistency. Equate the position vectors component by component, solve two of the equations for \(\lambda\) and \(\mu\), then substitute into the remaining component: if it holds the lines intersect, if it fails (and they are not parallel) the lines are skew.

Two intersecting lines in space Two straight lines drawn on isometric axes cross at one marked point, the point of intersection. x y z ℓ₁ ℓ₂ P
Intersecting lines: one pair \((\lambda,\mu)\) gives a single common point \(P\).
Two skew lines in space Two straight lines on isometric axes appear to cross, but the back line is broken where the front line passes over it, so the two share no common point. x y z ℓ₁ ℓ₂ no common point
Skew lines: not parallel, yet the back line passes under the front — no common point.

Each line carries its own parameter. For \(\ell_1\) and \(\ell_2\):

\[ \ell_1:\ \mathbf{r}=\mathbf{a}_1+\lambda\mathbf{d}_1\qquad \ell_2:\ \mathbf{r}=\mathbf{a}_2+\mu\mathbf{d}_2 \]
r1=a1+λd1

They meet where the position vectors are equal, giving one equation per component:

\[ \mathbf{a}_1+\lambda\mathbf{d}_1=\mathbf{a}_2+\mu\mathbf{d}_2 \]
a1+λd1=a2+μd2

The acute angle \(\theta\) between two intersecting lines comes from the directions:

\[ \cos\theta=\dfrac{|\mathbf{d}_1\cdot\mathbf{d}_2|}{|\mathbf{d}_1|\,|\mathbf{d}_2|} \]
cosθ=|d1d2||d1||d2|
Two solve, one to check. A pair of lines has only two unknowns \((\lambda,\mu)\), so two components fix them. The third (in 3D) is the consistency test: it holds for intersecting lines and fails for skew lines. Take the absolute value in the angle formula to keep \(\theta\) acute.

Classify a pair of lines

  1. Compare directions. If \(\mathbf{d}_2\) is a scalar multiple of \(\mathbf{d}_1\) the lines are parallel; if not, they are either intersecting or skew.
  2. If parallel, test a point. Substitute one line's base point into the other: on it means coincident (same line), off it means parallel and distinct.
  3. If not parallel, equate and solve. Set \(\mathbf{a}_1+\lambda\mathbf{d}_1=\mathbf{a}_2+\mu\mathbf{d}_2\) and solve two component equations for \(\lambda\) and \(\mu\).
  4. Check the last component. Substitute \(\lambda,\mu\) into the remaining equation: consistent gives an intersection (find the point), inconsistent gives skew.
Example 1 — Intersection in 2D
Find where \(\mathbf{r}_1=\begin{pmatrix}2\\1\end{pmatrix}+\lambda\begin{pmatrix}1\\2\end{pmatrix}\) meets \(\mathbf{r}_2=\begin{pmatrix}6\\3\end{pmatrix}+\mu\begin{pmatrix}-1\\1\end{pmatrix}\).
Solution

Equate the two position vectors component by component:

\(x:\ 2+\lambda\)\(=\)\(6-\mu\)
\(y:\ 1+2\lambda\)\(=\)\(3+\mu\)

Rearrange each into a neat pair, then add to eliminate \(\mu\):

\(\lambda+\mu\)\(=\)\(4\)
\(2\lambda-\mu\)\(=\)\(2\)
\(\text{add}:\ 3\lambda\)\(=\)\(6\)
\(\lambda\)\(=\)\(2\)
\(\mu\)\(=\)\(4-2=2\)

Substitute \(\lambda=2\) into \(\mathbf{r}_1\) for the point:

\((x,y)\)\(=\)\((2+2,\ 1+2(2))\)
\(=\)\((4,5)\)

The lines meet at \((4,5)\).

Example 2 — Intersection in 3D (check the third component)
Show that \(\ell_1:\ \mathbf{r}=\mathbf{i}+\mathbf{j}+\lambda(2\mathbf{i}+\mathbf{j}+\mathbf{k})\) and \(\ell_2:\ \mathbf{r}=2\mathbf{i}+\mathbf{j}-\mathbf{k}+\mu(\mathbf{i}+\mathbf{j}+2\mathbf{k})\) intersect, and find the point.
Solution

Equate \(\mathbf{r}_1=\mathbf{r}_2\) and write one equation per component:

\(x:\ 1+2\lambda\)\(=\)\(2+\mu\)
\(y:\ 1+\lambda\)\(=\)\(1+\mu\)
\(z:\ \lambda\)\(=\)\(-1+2\mu\)

Solve the \(x\)- and \(y\)-equations for \(\lambda\) and \(\mu\):

\(\text{from }y:\ \lambda\)\(=\)\(\mu\)
\(\text{in }x:\ 1+2\lambda\)\(=\)\(2+\lambda\)
\(\lambda\)\(=\)\(1,\quad \mu=1\)

Check the \(z\)-equation with these values:

\(z:\ \lambda\)\(=\)\(-1+2\mu\)
\(1\)\(=\)\(-1+2(1)\)
\(1\)\(=\)\(1\ \checkmark\)

The check holds, so substitute \(\lambda=1\) into \(\ell_1\):

\((x,y,z)\)\(=\)\((1+2,\ 1+1,\ 0+1)\)
\(=\)\((3,2,1)\)

The \(z\)-components agree, so the lines intersect at \((3,2,1)\).

Example 3 — Showing lines are skew
Show that \(\ell_1:\ \mathbf{r}=\mathbf{i}+\mathbf{k}+\lambda(\mathbf{i}+\mathbf{j})\) and \(\ell_2:\ \mathbf{r}=2\mathbf{j}+\mu(\mathbf{i}+\mathbf{k})\) are skew.
Solution

First test the directions for parallelism:

\(\mathbf{d}_1\)\(=\)\((1,1,0)\)
\(\mathbf{d}_2\)\(=\)\((1,0,1)\)
\(\mathbf{d}_2\)\(\neq\)\(k\,\mathbf{d}_1\ \Rightarrow\ \text{not parallel}\)

Equate \(\mathbf{r}_1=\mathbf{r}_2\) component by component:

\(x:\ 1+\lambda\)\(=\)\(\mu\)
\(y:\ \lambda\)\(=\)\(2\)
\(z:\ 1\)\(=\)\(\mu\)

Solve the \(y\)- and \(z\)-equations, then test \(x\):

\(\lambda\)\(=\)\(2,\quad \mu=1\)
\(x:\ 1+\lambda\)\(=\)\(\mu\)
\(1+2\)\(=\)\(1\)
\(3\)\(\neq\)\(1\ \times\)

Not parallel and the system is inconsistent, so the lines are skew.

Example 4 — Angle between intersecting lines
The lines \(\ell_1:\ \mathbf{r}=\lambda(\mathbf{i}+\mathbf{j})\) and \(\ell_2:\ \mathbf{r}=\mathbf{j}-\mathbf{k}+\mu(\mathbf{i}+\mathbf{k})\) intersect. Find the acute angle between them.
Solution

Read off the direction vectors and use \(\cos\theta=\dfrac{|\mathbf{d}_1\cdot\mathbf{d}_2|}{|\mathbf{d}_1||\mathbf{d}_2|}\):

\(\mathbf{d}_1\cdot\mathbf{d}_2\)\(=\)\((1)(1)+(1)(0)+(0)(1)\)
\(=\)\(1\)
\(|\mathbf{d}_1|\)\(=\)\(\sqrt{1+1+0}=\sqrt{2}\)
\(|\mathbf{d}_2|\)\(=\)\(\sqrt{1+0+1}=\sqrt{2}\)

Divide and take the inverse cosine:

\(\cos\theta\)\(=\)\(\dfrac{1}{\sqrt{2}\times\sqrt{2}}=\dfrac{1}{2}\)
\(\theta\)\(=\)\(\cos^{-1}\!\left(\dfrac{1}{2}\right)\)
\(=\)\(60^\circ\)

The acute angle between the lines is \(60^\circ\).

Common pitfalls

Using one parameter for both lines. Each line needs its own scalar. Writing \(\lambda\) in both equations forces a false relationship — always give the second line \(\mu\).
Forgetting the third-component check. Two components can always be solved for \(\lambda,\mu\); it is the leftover \(z\)-equation that decides intersect versus skew. Skipping it makes skew lines look as if they meet.
Calling non-parallel non-meeting lines "parallel". In 3D, lines that never meet are usually skew, not parallel. Only scalar-multiple directions are parallel.
Dropping the absolute value in the angle formula. Without \(|\mathbf{d}_1\cdot\mathbf{d}_2|\) a negative dot product gives an obtuse angle; the acute angle needs the absolute value.

Frequently asked questions

What are skew lines?

Skew lines are two straight lines in three dimensions that are not parallel and do not intersect. Because they never meet and never run parallel, they cannot lie in the same plane, so skew is only possible in 3D.

How do you find where two lines intersect?

Set the two position vectors equal, giving one equation per component. Solve two of them for \(\lambda\) and \(\mu\), then substitute into the remaining component: if it is consistent the lines meet, and you get the point by putting \(\lambda\) back into the first line.

How do you tell if two 3D lines are skew or intersecting?

Check they are not parallel, then equate components and solve two equations for \(\lambda,\mu\). Substitute into the third equation: consistent means they intersect, inconsistent means they are skew.

What is the difference between parallel and skew lines?

Parallel lines have direction vectors that are scalar multiples of each other; skew lines have directions that are not multiples. Both never meet, but only parallel lines keep a constant distance and lie in a common plane.

How do you find the angle between two intersecting lines?

Use the direction vectors: \(\cos\theta=\dfrac{|\mathbf{d}_1\cdot\mathbf{d}_2|}{|\mathbf{d}_1||\mathbf{d}_2|}\). The absolute value keeps the answer as the acute angle.

When are two lines the same line (coincident)?

When their directions are scalar multiples (parallel) and a point of one line also lies on the other. Then every point is shared, so they are one and the same line.