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Cambridge IGCSE0580

Vector geometry

Mathematics 0580 Chapter Notes

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Vector geometry
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Vector geometry notes

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In this chapter

  1. 1Vectors as directed line segments
  2. 2Position vectors
  3. 3Sums and differences: the route method
  4. 4Parallel vectors and collinear points
  5. 5Ratio and similarity

1. Vectors as directed line segments

A vector has size and direction. It is drawn as a directed line segment: an arrow from one point to another. overrightarrowAB goes from A to B.

overrightarrowBA has the same length but the opposite direction. A vector parallel to a is a multiple of a, and a negative multiple points the other way.

overrightarrowBA = -overrightarrowAB

Examiner insight

“Thus, for instance, some candidates assumed that the vector along each side of the hexagon was y because the sides are of equal length, without taking into account the change of direction of each side.” “A few candidates did not appreciate the reverse direction of the vector to AB and gave the answer q.”

Worked example 13 marks

ABCD is a parallelogram. overrightarrowAB = m and overrightarrowAD = n. Write down, in terms of m and n,(i) overrightarrowDC,(ii) overrightarrowCB,(iii) overrightarrowAC.

  1. 1

    DC is parallel to AB, same length, same direction: overrightarrowDC = m [1]

  2. 2

    overrightarrowBC = n, so overrightarrowCB is the opposite direction: overrightarrowCB = -n [1]

  3. 3

    overrightarrowAC = overrightarrowAB + overrightarrowBC = m + n [1]

2. Position vectors

A position vector is the vector from the origin O to a point. The position vector of P is overrightarrowOP.

A vector such as overrightarrowAM or overrightarrowPQ is not a position vector, because it does not start at O. A route for a position vector always starts at O.

position vector of P = overrightarrowOP

Examiner insight

“Part (b) was often better answered than part (a), but some candidates did not know the meaning of position vector and it was very regularly left blank.”

Worked example 13 marks

O is the origin, overrightarrowOR = r and overrightarrowOS = s. P is a point on RS and RP : PS = 2 : 5. Find the position vector of P. Give your answer, in terms of r and s, in its simplest form.

  1. 1

    overrightarrowRS = -r + s = s - r [1]

  2. 2

    overrightarrowRP = 2/7(s - r), since RP is 2 parts of 7 [1]

  3. 3

    overrightarrowOP = r + 2/7(s -r) = 5/7r + 2/7s [1]

3. Sums and differences: the route method

To find a vector, write a route along known vectors from its start to its end. Add a vector when you travel along its arrow. Subtract it when you travel against it.

Bracket every vector you subtract, then collect like terms for the final answer in its simplest form.

overrightarrowAB = overrightarrowAO + overrightarrowOB = b - a

Examiner insight

“Giving a correct route is a mark accessible to most candidates and so this should be encouraged as a first step.” “Many candidates did not use brackets and some of those who used brackets only applied the minus sign to the first term in the brackets instead of to both terms.”

Worked example 13 marks

OABC is a parallelogram. overrightarrowOA = a and overrightarrowOC = c. P is a point on AC and AP : PC = 1 : 4. Find, in terms of a and c, in its simplest form,(i) overrightarrowAP,(ii) overrightarrowOP.

  1. 1

    overrightarrowAC = overrightarrowAO + overrightarrowOC = -a + c [1]

  2. 2

    overrightarrowAP = 1/5overrightarrowAC = 1/5(c -a) [1]

  3. 3

    overrightarrowOP = overrightarrowOA + overrightarrowAP = a + 1/5(c -a) = 1/5(4a +c) [1]

4. Parallel vectors and collinear points

Two vectors are parallel when one is a multiple of the other. To show three points lie on a straight line, find two vectors between them and show they share a common factor.

Parallel is not enough. The vectors must also share a common point. Then the points are collinear.

overrightarrowOQ = k overrightarrowOP ⇒ O, P, Q collinear

Examiner insight

“The statement that was often omitted was that the three points are collinear.” “A significant number thought that the two vectors were perpendicular or that the three points formed a triangle.”

Worked example 12 marks

OABC is a parallelogram. overrightarrowOA = a and overrightarrowOC = c. P is a point on AC and AP : PC = 1 : 4, so overrightarrowOP = 1/5(4a + c). Q is a point on AB and overrightarrowAQ = 1/4c. Show that O, P and Q lie in a straight line.

  1. 1

    overrightarrowOQ = overrightarrowOA + overrightarrowAQ = a + 1/4c = 1/4(4a + c). Both overrightarrowOP and overrightarrowOQ have the common factor (4a +c) [1]

  2. 2

    overrightarrowOQ is a multiple of overrightarrowOP, so they are parallel. They share the point O, so O, P and Q are collinear [1]

5. Ratio and similarity

A point that divides a line in a ratio takes a fraction of that line. Add the ratio parts to get the denominator.

If AP : PB = 4 : 1, then overrightarrowAP = 4/5overrightarrowAB and overrightarrowPB = 1/5overrightarrowAB. Write each answer in its simplest form.

In similar triangles, matching sides are in the same ratio. Use that ratio in the same way to take a fraction of a line.

AP : PB = a : b ⇒ overrightarrowAP = a/(a+b)overrightarrowAB

Examiner insight

“Those who were most successful clearly identified a route from O to S.”

Worked example 14 marks

ABC is a triangle. P lies on AB and AP : PB = 4 : 1. Q lies on BC and overrightarrowBQ = 2/5overrightarrowBC. overrightarrowAB = m and overrightarrowAC = 5n. In each part, give your answer in terms of m and n in its simplest form.(i) Find overrightarrowBC.(ii) Find overrightarrowPQ.

  1. 1

    overrightarrowBC = overrightarrowBA + overrightarrowAC = -m + 5n = 5n - m [1]

  2. 2

    overrightarrowPB = 1/5m and overrightarrowBQ = 2/5(5n -m) [1]

  3. 3

    overrightarrowPQ = overrightarrowPB + overrightarrowBQ = 1/5m + 2n - 2/5m [1]

  4. 4

    = 2n - 1/5m [1]

Recap

  • A vector has size and direction. overrightarrowBA = -overrightarrowAB.
  • Position vector of P = overrightarrowOP, from the origin.
  • Find a vector by writing a route. Add along an arrow, subtract against it.
  • Bracket every subtracted vector and collect like terms.
  • AP : PB = a : b means overrightarrowAP = a/(a+b)overrightarrowAB.
  • Collinear: one vector is a multiple of the other and they share a common point.

End-of-chapter exercise

Test yourself on the whole chapter. Work through these before moving on.

  1. Do not use a calculator. overrightarrowPQ = a. Write down overrightarrowQP. R is a point such that overrightarrowPR is in the same direction as overrightarrowPQ and twice as long. Write down overrightarrowPR.2 marks
  2. overrightarrowAB = p and overrightarrowCD = -3p. Describe fully how overrightarrowCD is related to overrightarrowAB.2 marks
  3. Do not use a calculator. O is the origin. overrightarrowOA = a and overrightarrowOB = b. Find overrightarrowAB in terms of a and b.2 marks
  4. Do not use a calculator. Point A has position vector 3a + b and point B has position vector a - 2b. Find overrightarrowAB in its simplest form.3 marks
  5. O is the origin. overrightarrowOA = a and overrightarrowOB = b. M is the midpoint of AB. Find the position vector of M.3 marks
  6. Do not use a calculator. The points P and Q have position vectors p and q. R lies on PQ and PR : RQ = 3 : 1. Find the position vector of R in its simplest form.3 marks
  7. In triangle OAB, overrightarrowOA = a and overrightarrowOB = b. P lies on OA and OP : PA = 2 : 1. Q is the midpoint of OB. Find overrightarrowPQ.3 marks
  8. Do not use a calculator. overrightarrowAB = 6a - 4b and overrightarrowCD = 9a - 6b. Show that AB and CD are parallel and find the ratio AB : CD.3 marks
  9. Do not use a calculator. On a chart with origin O, a harbour A has position vector 4a and a lighthouse B has position vector 6b. A buoy M lies on the straight line AB with AM : MB = 3 : 1. The line OB is extended to a rock C so that OB : BC = 2 : 1. A marker N is the midpoint of OA. Show that N, M and C lie on a straight line, and find the ratio NM : MC.6 marks
  10. A roof frame OABC is a trapezium with vertices in that order. O is the origin, overrightarrowOA = 4a, overrightarrowOC = 3c and overrightarrowCB = 2a. The diagonals OB and AC cross at X. The sides OC and AB are extended to meet at G. M is the midpoint of OA. Use similar triangles to find the position vectors of X and G in their simplest form. Then show that G, X and M lie on a straight line and find the ratio GX : XM.7 marks

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