Salam Academy
Sign inCreate account
Contents · 0% complete
  1. Learning objectives
  2. Core idea
  3. Key formulae
  4. Worked examples
  5. Explore the graph
  6. Common mistakes
  7. Exam tip
  8. Guided practice
  9. Practice questions
  10. Challenge
  11. Summary
  12. Quick quiz
  13. Same topic, different board
  14. Past paper questions
Learn9FM0Option · Further Pure Mathematics 1VectorsShortest distances and triple products
Vectors · Subtopic 2 of 2

Shortest Distances and the Triple Product

The shortest distance is always along a perpendicular. From point PP to plane r⋅n=d\mathbf r\cdot\mathbf n=d: ∣p⋅n−d∣∣n∣\frac{|\mathbf p\cdot\mathbf n-d|}{|\mathbf n|}. Between skew lines: project the join of two points onto n=d1×d2\mathbf n=\mathbf d_1\times\mathbf d_2. The triple product a⋅(b×c)\mathbf a\cdot(\mathbf b\times\mathbf c) is the volume of the parallelepiped.

Cambridge 9231 Paper 1Edexcel YFM01 WFM03Edexcel 9FM0 Paper 1Edexcel 9FM0 Paper 2Edexcel 9FM0 OptionAQA 7367 Paper 2≈ 45 minRevision 1 · updated 2026-10-27
Practise this topicTimed testAsk Salam Tutor
Section 1 of 14

Learning objectives

  • Find the distance from a point to a plane and to a line.
  • Find the shortest distance between skew lines.
  • Use the scalar triple product for volumes and coplanarity.
Pause and think

From a point above a floor, what's the shortest way down to it?

Section 2 of 14

Core idea

Start hereHow close do two flight paths come?

Two aircraft fly straight lines at different heights and directions; their paths never meet. Air traffic control needs the closest approach. The answer lies along the unique line perpendicular to both paths, found with a cross product.

The big idea

Shortest distances lie along perpendiculars. Point–plane: ∣p⋅n−d∣∣n∣\frac{|\mathbf p\cdot\mathbf n-d|}{|\mathbf n|}. Skew lines: ∣(b−a)⋅(d1×d2)∣∣d1×d2∣\frac{|(\mathbf b-\mathbf a)\cdot(\mathbf d_1\times\mathbf d_2)|}{|\mathbf d_1\times\mathbf d_2|}. Volume: ∣a⋅(b×c)∣|\mathbf a\cdot(\mathbf b\times\mathbf c)|.

1. Point to plane

Projecting onto the unit normal
  1. n^=n∣n∣\hat{\mathbf n}=\frac{\mathbf n}{|\mathbf n|}
    Unit normal.
  2. distance=∣(p−a)⋅n^∣\text{distance}=|(\mathbf p-\mathbf a)\cdot\hat{\mathbf n}|
    Component of AP→\overrightarrow{AP} along the normal (AA any point in the plane).
  3. =∣p⋅n−a⋅n∣∣n∣=∣p⋅n−d∣∣n∣=\frac{|\mathbf p\cdot\mathbf n-\mathbf a\cdot\mathbf n|}{|\mathbf n|}=\frac{|\mathbf p\cdot\mathbf n-d|}{|\mathbf n|}
    a⋅n=d\mathbf a\cdot\mathbf n=d.
Worked example(1,1,1)(1,1,1) to x+2y+2z=14x+2y+2z=14

Distance.

Goal
Number.
Strategy
Formula.
  1. ∣1+2+2−14∣=9|1+2+2-14|=9
    Substitute the point.
  2. ∣n∣=1+4+4=3|\mathbf n|=\sqrt{1+4+4}=3
    Normal length.
Answer33.
✓ CheckFoot of perpendicular (1,1,1)+1(1,2,2)=(2,3,3)(1,1,1)+1(1,2,2)=(2,3,3): 2+6+6=142+6+6=14 ✓, length 33 ✓.

2. Skew lines

The common perpendicular
  1. n=d1×d2\mathbf n=\mathbf d_1\times\mathbf d_2
    Perpendicular to both lines.
  2. distance=∣(b−a)⋅n∣∣n∣\text{distance}=\frac{|(\mathbf b-\mathbf a)\cdot\mathbf n|}{|\mathbf n|}
    Project any join AB→\overrightarrow{AB} onto n\mathbf n.
If this is 00 the lines intersect.
Worked exampleTwo skew lines

r=s(1,0,0)\mathbf r=s(1,0,0) and r=(0,1,2)+t(0,1,1)\mathbf r=(0,1,2)+t(0,1,1).

Goal
Shortest distance.
Strategy
Common perpendicular.
  1. n=(1,0,0)×(0,1,1)=(0,−1,1)\mathbf n=(1,0,0)\times(0,1,1)=(0,-1,1)
    Cross product.
  2. (0,1,2)⋅(0,−1,1)=1(0,1,2)\cdot(0,-1,1)=1
    Join from (0,0,0)(0,0,0) to (0,1,2)(0,1,2).
  3. 12\frac1{\sqrt2}
    Divide by ∣n∣|\mathbf n|.
Answer12≈0.707\frac1{\sqrt2}\approx0.707.
✓ CheckNearest points (0,0,0)(0,0,0) and (0,−12,12)(0,-\frac12,\frac12) (at s=0s=0, t=−32t=-\frac32): distance 12\frac1{\sqrt2} ✓.

3. The scalar triple product

Volume of a parallelepiped
  1. base area=∣b×c∣\text{base area}=|\mathbf b\times\mathbf c|
    Parallelogram on b\mathbf b, c\mathbf c.
  2. height=∣a⋅(b×c)∣∣b×c∣\text{height}=\frac{|\mathbf a\cdot(\mathbf b\times\mathbf c)|}{|\mathbf b\times\mathbf c|}
    Component of a\mathbf a along the base's normal.
  3. V=∣a⋅(b×c)∣V=|\mathbf a\cdot(\mathbf b\times\mathbf c)|
    Base × height.
Tetrahedron: 16\frac16 of this. Zero: the vectors are coplanar.
Worked exampleCoplanar vectors

Find kk so (1,2,k)(1,2,k), (0,1,1)(0,1,1), (1,0,1)(1,0,1) are coplanar.

Goal
kk.
Strategy
Triple product (determinant) =0=0.
  1. 1(1−0)−2(0−1)+k(0−1)1(1-0)-2(0-1)+k(0-1)
    Expand along the first row.
  2. =3−k=0=3-k=0
    Coplanar.
Answerk=3k=3.
✓ Check(1,2,3)=(1,0,1)+2(0,1,1)(1,2,3)=(1,0,1)+2(0,1,1): a combination of the other two ✓.
Common misconceptionDistance between two given points The distance between the lines' base points is not the shortest distance, unless the join happens to be perpendicular to both.

Summary

  • Project onto a normal.
  • Skew lines: d1×d2\mathbf d_1\times\mathbf d_2.
  • Triple product: volume; tetrahedron ÷6.
Key words
Skew lines
Non-parallel lines that never meet.
Scalar triple product
a⋅(b×c)\mathbf a\cdot(\mathbf b\times\mathbf c).
HistoryLagrange. Joseph-Louis Lagrange used what we now call the triple product in 1773 to compute volumes of tetrahedra, long before vector notation existed.
Fun factSkew lines everywhere. A road on a bridge and the railway beneath it are skew: they never meet but aren't parallel. Engineers need exactly their shortest distance, the clearance.
Section 3 of 14

Key formulae

Point to plane∣n1x0+n2y0+n3z0−d∣n12+n22+n32\frac{|n_1x_0+n_2y_0+n_3z_0-d|}{\sqrt{n_1^2+n_2^2+n_3^2}}
Skew lines∣(b−a)⋅(d1×d2)∣∣d1×d2∣\frac{|(\mathbf b-\mathbf a)\cdot(\mathbf d_1\times\mathbf d_2)|}{|\mathbf d_1\times\mathbf d_2|}
Triple productVparallelepiped=∣a⋅(b×c)∣,Vtetrahedron=16∣a⋅(b×c)∣V_{\text{parallelepiped}}=|\mathbf a\cdot(\mathbf b\times\mathbf c)|,\qquad V_{\text{tetrahedron}}=\tfrac16|\mathbf a\cdot(\mathbf b\times\mathbf c)|
TrickTriple product as a determinant. a⋅(b×c)\mathbf a\cdot(\mathbf b\times\mathbf c) is the 3×33\times3 determinant with rows a\mathbf a, b\mathbf b, c\mathbf c. Zero means the three vectors are coplanar.
Section 4 of 14

Worked examples

Example 1 · Point to plane0 / 2 steps
Find the distance from (1,1,1)(1,1,1) to x+2y+2z=14x+2y+2z=14.
    Example 2 · Skew lines0 / 3 steps
    Find the shortest distance between r=s(1,0,0)\mathbf r=s(1,0,0) and r=(0,1,2)+t(0,1,1)\mathbf r=(0,1,2)+t(0,1,1).
      Section 5 of 14

      Explore the graph

      Distance from a point to a line (2D view)
      Loading the interactive graph…
      Try this
      1. The second point is the foot of the perpendicular from (p,q)(p,q).
      2. Move the line: the shortest segment always meets it at right angles.

      The shortest distance runs along the perpendicular.

      In the real worldRobot collision checks. Robot arms are modelled as line segments; controllers compute shortest distances between them thousands of times a second to avoid collisions.
      Section 6 of 14

      Common mistakes

      IncorrectForgetting the modulus
      CorrectDistances are positive
      The dot product can be negative depending on direction.
      IncorrectNot dividing by ∣n∣|\mathbf n|
      CorrectDivide by ∣n∣|\mathbf n|
      Unless n\mathbf n is a unit vector, the dot product is scaled.
      IncorrectTetrahedron =∣a⋅(b×c)∣=|\mathbf a\cdot(\mathbf b\times\mathbf c)|
      Correct16\frac16 of it
      A tetrahedron is one-sixth of the parallelepiped.

      Two skew lines: we'll never meet

      Cross product: but I can tell you exactly how close you got.

      Section 7 of 14

      Exam tip

      Exam tipName the perpendicular. Say "the common perpendicular has direction d1×d2\mathbf d_1\times\mathbf d_2" before computing: it earns the method mark even if arithmetic slips.
      Section 8 of 14

      Guided practice

      SA-FP-VD-G012 marksL1 FoundationNon-calculatorPoint distances
      Find the distance from the origin to the plane 2x+y+2z=92x+y+2z=9.

      Sign in or create a free account to check answers, get hints and track your progress.

      Salam Academy Original Practice · Proposed mark scheme
      Section 9 of 14

      Practice questions

      SA-FP-VD-0012 marksL2 StandardNon-calculatorPoint distances
      Find the distance from (1,1,1)(1,1,1) to the plane x+2y+2z=14x+2y+2z=14.

      Sign in or create a free account to check answers, get hints and track your progress.

      Salam Academy Original Practice · Proposed mark scheme
      SA-FP-VD-0022 marksL2 StandardNon-calculatorTriple product
      Find the volume of the parallelepiped with edges (100)\begin{pmatrix}1\\0\\0\end{pmatrix}, (020)\begin{pmatrix}0\\2\\0\end{pmatrix}, (113)\begin{pmatrix}1\\1\\3\end{pmatrix}.

      Sign in or create a free account to check answers, get hints and track your progress.

      Salam Academy Original Practice · Proposed mark scheme
      SA-FP-VD-0034 marksL3 ExamCalculatorSkew lines
      Find the shortest distance between r=s(100)\mathbf r=s\begin{pmatrix}1\\0\\0\end{pmatrix} and r=(012)+t(011)\mathbf r=\begin{pmatrix}0\\1\\2\end{pmatrix}+t\begin{pmatrix}0\\1\\1\end{pmatrix} (to 3 s.f.).

      Sign in or create a free account to check answers, get hints and track your progress.

      Salam Academy Original Practice · Proposed mark scheme
      SA-FP-VD-0042 marksL2 StandardNon-calculatorPoint distances
      Find the distance from (2,3,4)(2,3,4) to the xx-axis.

      Sign in or create a free account to check answers, get hints and track your progress.

      Salam Academy Original Practice · Proposed mark scheme

      More questions on every skill in this subtopic are in the question bank under Practise.

      Section 10 of 14

      Challenge

      SA-FP-VD-0053 marksL4 ChallengingNon-calculatorTriple product
      The vectors (12k)\begin{pmatrix}1\\2\\k\end{pmatrix}, (011)\begin{pmatrix}0\\1\\1\end{pmatrix}, (101)\begin{pmatrix}1\\0\\1\end{pmatrix} are coplanar. Find kk.

      Sign in or create a free account to check answers, get hints and track your progress.

      Salam Academy Original Practice · Proposed mark scheme
      Section 11 of 14

      Summary

      • Point–plane: ∣p⋅n−d∣∣n∣\frac{|\mathbf p\cdot\mathbf n-d|}{|\mathbf n|}.
      • Skew lines: project onto d1×d2\mathbf d_1\times\mathbf d_2.
      • Triple product: volume; zero means coplanar.
      Section 12 of 14

      Quick quiz

      Q1 · SA-FP-VD-Q011 markL1 FoundationNon-calculatorTriple product
      If a⋅(b×c)=0\mathbf a\cdot(\mathbf b\times\mathbf c)=0, the three vectors are
      Choose one option

      Sign in or create a free account to check answers, get hints and track your progress.

      Salam Academy Original Practice · Proposed mark scheme
      Section 13 of 14

      Same topic, different board

      Specification
      IAL Further Mathematics (YFM01) and A level Further Mathematics (9FM0)
      Where
      Core Pure · Vectors; FP1 Further vectors (IAL: WFM03)
      Expected
      Shortest distances; scalar triple product for volumes of parallelepipeds and tetrahedra (FP1/WFM03).
      Calculator
      Permitted; show enough working to earn the method marks.
      Typical style
      Edexcel often opens with a routine part and builds to a "show that" or "prove" part.
      Open in
      YFM01 · WFM039FM0 · Paper 19FM0 · Paper 29FM0 · Option
      Section 14 of 14

      Past paper questions

      No past paper questions are tagged to this subtopic yet. We add papers every week.

      Browse past papers

      Salam Academy original content · Content status: reviewed and published. Syllabus mapping: Pearson Edexcel 9FM0, First teaching 2017 (official specification).