← All notesMaths · Vectors⬇ Download PDF
A Level Maths 9709 · Paper 3 Pure Mathematics 3 · 3.7 · ~6 min read

Vectors

Vectors in two and three dimensions, magnitude and unit vectors, the scalar product and angles, the vector equation of a line, and deciding whether two lines are parallel, intersect or are skew, with the foot of a perpendicular and shortest distances.

Position & direction vectorsMagnitudeUnit vectorsScalar productAngle between vectorsVector equation of a lineParallel, intersecting, skewFoot of perpendicular
RSby Riyaz Sir
mentorriyaz.in
Paper tag = where it is examinedFORMULA LIST (MF19) given in the examM1A1B1 how marks are awardedhighlight = comes up again and againTest yourself at the end

In 30 seconds

  • AB = b − a (end minus start). Its length is ǀABǀ = √(x² + y² + z²).
  • Scalar product: a·b = a₁b₁ + a₂b₂ + a₃b₃ = ǀaǀǀbǀ cos θ. Zero means perpendicular.
  • A line is r = a + tb: a point on it plus any multiple of its direction.
  • Two lines in 3D are parallel (directions are multiples), intersect (a single solution for s and t), or are skew (neither).

By the end you can

  • Use vector notation in 2D and 3D: column vectors, i, j, k, position vectors and magnitudes.
  • Use the scalar product to find angles and test for perpendicularity.
  • Form the vector equation of a line through two points, or through a point with a given direction.
  • Decide whether two lines are parallel, intersect or are skew, and find the point of intersection.
  • Find the foot of the perpendicular from a point to a line and the shortest distance.

1Vectors and the scalar product

Basics

  • Notation: {(3;−2;2)} = 3i − 2j + 2k. A position vector OA (written a) goes from the origin to A.
  • Unit vector in the direction of a: a/ǀaǀ. Collinear points: AB is a multiple of BC.
a·b = a₁b₁ + a₂b₂ + a₃b₃ = ǀaǀ ǀbǀ cos θ FORMULA LIST (MF19)
a·b = 0 ⇔ perpendicular (non-zero vectors)
Worked example

Find the angle between a = i + 2j + 2k and b = 2i − j + 2k.

  1. a·b = 2 − 2 + 4 = 4; ǀaǀ = 3, ǀbǀ = 3 M1
  2. cos θ = 4/9 M1
  3. θ = 63.6° A1
  • Angle at a vertex: for angle ABC use BA and BC, both starting at B. Using AB and BC gives the supplementary angle.
Common mistakeVectors pointing the wrong way. For angle ABC, both vectors must point away from B.

2Lines

r = a + tb

intersectingskew: not parallel, never meetLines in 3D: parallel, intersecting, or skew
A line in 3D: a fixed point, and a direction you can travel along any distance.
  • Through A and B: r = a + t(b − a). Any point on the line has coordinates written in terms of t.
  • Different-looking equations can describe the same line: a different point or a multiple of the direction.
Worked example

Find a vector equation of the line through A(1, 2, 3) and B(4, 0, 5).

  1. AB = b − a = 3i − 2j + 2k B1
  2. r = i + 2j + 3k + t(3i − 2j + 2k) B1

Parallel, intersecting or skew

  • Step 1: are the directions multiples of each other? Yes → parallel (or the same line).
  • Step 2: set the x, y, z components equal. Solve two equations for s and t; check the third. Consistent → they intersect; inconsistent → skew.
Worked example

l₁: r = (1, 0, 2) + s(1, 1, −1) and l₂: r = (0, −3, 1) + t(1, 2, 0). Show they intersect, and find the point and the angle between them.

  1. 1 + s = t, s = −3 + 2t, 2 − s = 1 M1
  2. third equation: s = 1; then t = 2, check s = −3 + 4 = 1 ✓ A1
  3. point (2, 1, 1) A1
  4. directions: (1·1 + 1·2 + 0)/(√3 √5) = 3/√15 ⇒ θ = 39.2° M1 A1
Worked example

Show that l₁ above and l₃: r = (2, 3, −1) + t(0, 1, 1) are skew.

  1. directions (1, 1, −1) and (0, 1, 1) are not multiples, so not parallel B1
  2. 1 + s = 2 ⇒ s = 1; s = 3 + t ⇒ t = −2 M1
  3. z: 2 − 1 = 1 but −1 + (−2) = −3; 1 ≠ −3, so no intersection: skew A1
Common mistakeUsing the same parameter for both lines. Each line needs its own (s and t).

3Perpendicular distance

The foot of the perpendicular

direction bPN (foot)Ar = a + tbN = a + tb for some tPN · b = 0 (perpendicular)solve for t, then find NThe shortest distance from P to the line is PN, where PN is perpendicular to the line.
PN is the shortest distance from P to the line: it meets the line at right angles.
  • Method: write N = a + tb (a general point), form PN = n − p, then solve PN·b = 0 for t.
  • Shortest distance = ǀPNǀ. You can also find it using sin θ after finding the angle between AP and the line.
Worked example

Find the foot of the perpendicular from P(6, 0, 2) to the line r = (1, 2, 3) + t(3, −2, 2), and the perpendicular distance.

  1. N = (1 + 3t, 2 − 2t, 3 + 2t); PN = (3t − 5, 2 − 2t, 1 + 2t) M1
  2. PN·(3, −2, 2) = 9t − 15 − 4 + 4t + 2 + 4t = 17t − 17 = 0 ⇒ t = 1 M1 A1
  3. N = (4, 0, 5), PN = (−2, 0, 3), distance √13 ≈ 3.61 A1
Common mistakeDotting with the wrong vector. PN must be dotted with the direction b, not with the position vector a.
Linking question2D vectors are used for forces in Forces and equilibrium; complex numbers in Complex numbers behave like 2D vectors on an Argand diagram.

Test yourself — cover the answers first

Quick check

  1. Find ǀ2i − j + 2kǀ.
  2. a·b = 0 for non-zero a, b. What does this mean?
  3. Write the line through (1, 0, 4) parallel to 2i + k.
  4. Two lines are not parallel and do not meet. What are they?
  5. For angle ABC, which two vectors do you use?

Exam-style

  1. The points A(2, −1, 3), B(5, 1, 2), C(3, 4, k). Find k so that AB is perpendicular to BC.
  2. Find the unit vector in the direction of 4i − 4j + 2k.
  3. Show that the point (7, −2, 7) lies on r = (1, 2, 3) + t(3, −2, 2).
Show answers
  1. 3
  2. a and b are perpendicular
  3. r = i + 4k + t(2i + k)
  4. skew
  5. BA and BC
  6. AB = (3, 2, −1), BC = (−2, 3, k − 2); −6 + 6 − (k − 2) = 0 ⇒ k = 2
  7. ǀvǀ = 6, so (2/3)i − (2/3)j + (1/3)k
  8. t = 2 gives (7, −2, 7) in all three components
Practise this chapter
A Level practice papers
Next chapterDifferential equations →
Stuck on a question?

Send it to me on WhatsApp. I read every message.

Want regular help? Send this to your parent →
Share with a friendAsk Riyaz Sir

Independent notes written by Riyaz Sir. Not produced, endorsed or approved by Cambridge University Press & Assessment. Syllabus codes 9709 and 9702 refer to the Cambridge International AS & A Level syllabuses.