3.7 Vectors

Cambridge International AS & A Level Mathematics 9709 (2026-2027) · Paper 3: Pure Mathematics 3 · official syllabus

VECTORS LAB

Learn it by playing it

Four short missions on magnitude, unit vectors, the dot product and intersecting lines in 3D. In each one you read the diagram, predict the rule, and slide your answer onto the exact value - when a mission glows green, read why it worked, then take the next one.

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Mission 1 of 4

Size the vector

A vector's magnitude is its length, and in 3D Pythagoras still does the work: square the three components, add, square root. The white arrow is v = (2, -1, 2), drawn from the origin; the teal ruler is the length you set. Slide the length until the ruler matches the arrow.

Before you slide: the magnitude of (a, b, c) is...

Your length 1.5 against the true 3.0 - square all three components before adding.

Study notes

Played the lab above? These notes are the reference version of what you just felt - use them to check exact formulas and the exam wording.

Magnitude and distance in 3D 3.7

  • The magnitude of (a, b, c) is sqrt(a^2 + b^2 + c^2) - Pythagoras with a third square, and negative components square away.
  • Worked example: |(2, -1, 2)| = sqrt(4 + 1 + 4) = 3 exactly.
  • The distance between points A and B is the magnitude of the displacement vector AB = b - a, found by subtracting position vectors component by component.
  • AB = b - a, never a - b: the displacement points from A to B, and swapping the order reverses the direction.

Unit vectors and direction 3.7

  • A unit vector keeps the direction and forces the length to 1: divide every component of v by |v|.
  • Worked example: (6, -2, 3) has magnitude sqrt(36 + 4 + 9) = 7, so its unit vector is (6/7, -2/7, 3/7), with first component 6/7 = 0.8571 (rounded to 4 d.p.).
  • Dividing by a component instead of the magnitude changes the direction - the scaling must be the same for all three components.
  • Check a unit-vector answer by squaring and adding the components: the total must come back to 1.

The dot product and the angle between vectors 3.7

  • a . b = a1 b1 + a2 b2 + a3 b3 = |a| |b| cos theta, so cos theta = (a . b)/(|a| |b|) - the angle comes from pairing components, multiplying and adding.
  • Worked example: (1, 2, 2) . (2, -1, 2) = 2 - 2 + 4 = 4 with |a| = |b| = 3, so cos theta = 4/9 and theta = 63.61 degrees (2 d.p.).
  • Watch the signs inside the dot product: a negative component subtracts its term, and one missed minus changes the angle.
  • Perpendicular vectors have a . b = 0 exactly - the standard proof of a right angle, and the test for it.

The vector equation of a line 3.7

  • A line through point A with direction d is r = a + t d: the position vector of any point on it, with the parameter t measuring how far along the direction you travel.
  • The direction vector is any multiple of the line's direction - scaling d by a non-zero constant describes the same line with a rescaled parameter.
  • To build a line through two points A and B, use direction d = b - a and either point as the base.
  • A point P lies on the line only if one single value of t satisfies all three coordinates of p = a + t d.

Intersecting and skew lines 3.7

  • To find where two lines meet, equate the coordinates with separate parameters s and t, solve any two coordinate equations for s and t, then check the third equation - it must hold with the same values.
  • Worked example: (1, -3, 0) + s(1, 2, 1) and (4, 0, 0) + t(-1, 1, 2) give s = 2t and -3 + 2s = t, so t = 1 and s = 2; the third coordinate confirms 1 + 2 = 4 - 1 = 3, and the lines meet at (3, 1, 2), at distance sqrt(14) = 3.7417 (4 d.p.) from the origin.
  • If the third equation fails, the lines do not meet - in 3D that makes them skew, not parallel: skew lines are neither parallel nor intersecting.
  • Parallel lines have direction vectors that are multiples of each other; check that first, before spending effort on the intersection algebra.

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