MathematicsExam code: 0580

Vectors

Vectors and Transformations

PractiseFlashcards

Parallel vectors

  • Two vectors are parallel exactly when one is a scalar multiple of the other, so b = ka for some number k
  • To prove it for two expressions, factorise both and show the same bracket falls out of each
    • 6a + 4b factorises to 2(3a + 2b), and 9a + 6b factorises to 3(3a + 2b)
    • So 9a + 6b = 1.5(6a + 4b), which makes them parallel
  • A negative k means parallel but pointing the opposite way
  • Say explicitly that one is a multiple of the other, since that statement is what earns the mark

Collinear points

  • Three points are collinear when they all lie on one straight line
  • Show it by finding two vectors joining the points, showing they are parallel, and noting that they share a common point
  • The shared point is essential: parallel alone only means the lines have the same direction, not that they are the same line
Three points A, B and C shown twice: in the first they lie on one straight line, so AB is parallel to AC and to BC and the points are collinear; in the second they form a triangle, so none of those vectors is parallel and the points are not collinear
Source: Vector problem-solving by Save My Exams
  • The syllabus names showing vectors parallel, showing three points collinear, and ratio and similarity problems as the things to be able to do
A not-to-scale figure with O, A, B and C joined by several lines, where OA is the vector 2a and OC is the vector c, the kind of diagram on which a path such as AB or the ratio in which two lines cut each other has to be worked out
Source: Vector problem-solving by Save My Exams

Worked example

Showing two vectors are parallel

In a diagram, AB = 2a + 3b and CD = 4a + 6b. Show that AB and CD are parallel.

Solution:

  • Two vectors are parallel exactly when one is a scalar multiple of the other
  • Look for a common factor in CD
  • 4a + 6b = 2(2a + 3b)
  • The bracket is AB, so CD = 2 × AB
  • CD is a scalar multiple of AB, therefore AB and CD are parallel
  • Saying only that "the letters are the same" is not a proof: the statement that must appear is that one vector is a multiple of the other
  • If the two also shared a point, that same working would prove the points collinear: parallel plus a common point is what collinear needs

Worked example

Where two lines meet, from a parallel side

O is the origin, OA = a and OC = c. B is the point with CB = ⅔a, so CB is parallel to OA. The lines OC and AB are extended to meet at D. Find the position vector of D, and the vector BD, in terms of a and c.

Triangle OAD with O at the bottom left, A at the bottom right and D at the top. C lies on OD and B lies on AD, and the line CB is drawn across the triangle, parallel to OA. An arrow along OA is labelled a and an arrow along OC is labelled c. No other lengths or vectors are marked. The figure is marked NOT TO SCALE.

Solution:

  • CB is parallel to OA, so triangles DCB and DOA are similar: they share the angle at D, and the parallel sides make their other angles equal
  • The scale factor from triangle DOA to triangle DCB is CB ÷ OA = ⅔, so DC is ⅔ of DO
  • That leaves OC as ⅓ of OD, so OD = 3 × OC
  • The position vector of D is 3c
  • Reach B from O through C: OB = OC + CB = c + ⅔a
  • BD = OD − OB = 3c − (c + ⅔a)
  • So BD = 2c − ⅔a
  • Check: DB is ⅔ of DA, so BD should be ⅔ of AD; AD = 3c − a, and ⅔(3c − a) = 2c − ⅔a, which matches

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