0580

Vectors

Vectors and Transformations

Position vectors

  • A position vector gives the location of a point relative to the origin, so the position vector of A is OA
  • These papers use the phrase "position vector" constantly, and finding one is the standard question
  • The vector between two points is found by going backwards to the origin and forwards again: AB = OB − OA
  • Reading that as "end minus start" is the quickest way to remember it
Three points A(−4, 2), B(2, 4) and C(3, −4) plotted on axes and joined by arrows labelled AB, AC and CB, each vector found by taking the end point's coordinates and subtracting the start point's
Source: Vector diagrams by Save My Exams
A triangle OAB with the position vector of A written as OA = a and the position vector of B as OB = b, and the vector AB worked out by travelling back to O and out again: AB = −OA + OB = b − a
Source: Position & displacement vectors by Save My Exams

Finding a path

  • Any vector across a diagram can be built by travelling along vectors you already know, in any route that works
  • Travelling with an arrow adds that vector; travelling against it subtracts
  • Different routes give different-looking expressions that simplify to the same thing, and mark schemes accept any correct path
A grid of identical parallelograms with vertices labelled A to T, where AB is the vector a and AF is the vector b: the route from F to B is traced in red as going back along b and then along a, giving FB = −b + a
Source: Finding vector paths by Save My Exams
  • Collect like terms at the end, because "give your answer in its simplest form" appears on almost every one of these questions
Worked example

A point dividing a line in a ratio

In triangle OAB, OA = a and OB = b. The point P lies on AB so that AP : PB = 2 : 3. Find OP in terms of a and b, in its simplest form.

Triangle OAB with O at the bottom left. An arrow from O to A is labelled a and an arrow from O to B is labelled b. The point P is marked on the line AB, with two tick marks on the part from A to P and three tick marks on the part from P to B. The figure is marked NOT TO SCALE.

Solution:

  • Travel from O to P by going along OA and then part of AB
  • First find the whole of AB: AB = OB − OA = b − a
  • The ratio 2 : 3 splits AB into 5 equal parts, and P is 2 of them along from A
  • So AP = ⅖(b − a)
  • Now OP = OA + AP = a + ⅖(b − a)
  • Expand: OP = a + ⅖b − ⅖a
  • Collect the a terms: OP = ⅗a + ⅖b
  • So OP = ⅗a + ⅖b
  • Check: at the far end, O to B would give a + 1(b − a) = b, as it should
Worked example

Column vectors and a magnitude

a = (3, −2) and b = (−1, 5), written as column vectors. Find 2a + 3b as a single column vector, and find the magnitude of a − b.

Solution:

  • Deal with the scalars first: 2a = (6, −4) and 3b = (−3, 15)
  • Add the tops and the bottoms separately: 6 + (−3) = 3 and −4 + 15 = 11
  • So 2a + 3b = (3, 11)
  • For the magnitude, first find a − b: 3 − (−1) = 4 and −2 − 5 = −7
  • So a − b = (4, −7)
  • The magnitude is √(4² + (−7)²) = √(16 + 49) = √65
  • So |a − b| = √65, which is 8.06 to 3 significant figures