High School Maths · Year 2 · Unit 12 · Lesson 2 of 5

Vector Geometry: Prove Lines Are Parallel Without Measuring

About 15 minutes · position vectors · finding a vector by a route (AB = b − a) · midpoints and points that divide a line in a ratio · showing that lines are parallel · showing that three points are collinear · vectors in parallelograms and regular hexagons · no calculator · exit quiz

Hook

Kai and Lena start at the same point O. Kai walks along the vector a and then along b. Lena walks along b and then along a.

Do they end up in the same place?

Yes. Their two routes are the sides of a parallelogram, so both end at the far corner: a + b = b + a. Following vectors round a shape like this is how you can prove facts about lines without measuring anything.

Learn

Master the essentials

From A to B: back along a, then along b. So AB is b − a.

Position vector

The position vector of a point A is OA: the vector from the origin O to A.

A(3, −1) has position vector 3−1

Routes

To find a vector, follow a route of vectors you know. Going backwards along a vector changes its sign.

AB = AO + OB = −a + b

End minus start

If A and B have position vectors a and b, then AB = b − a.

PQ = q − p

Midpoint

Go to A, then half of the way along AB.

OM = 12 (a + b)

Ratio on a line

If AP : PB = m : n, then AP is mm + n of AB.

AP : PB = 2 : 5 gives AP = 27 AB

Parallel

Two vectors are parallel when one is a number times the other. Take out a common factor to see it.

10a − 15b = 5(2a − 3b)

Collinear

Three points lie on one straight line when two of the vectors between them are parallel AND share a point.

AC = 3AB, and A is on both

Shapes

Opposite sides of a parallelogram are equal vectors. In a regular hexagon, each side matches a vector from the centre.

In parallelogram OABC, CB = OA

Worked examples

See it done

1. In triangle OAB, OA = a and OB = b. M is the midpoint of AB. Find AB and OM in terms of a and b.

  1. AB = AO + OB = −a + b = b − a.
  2. M is halfway, so AM = 12 (b − a).
  3. OM = OA + AM = a + 12 b − 12 a.
  4. Collect: OM = 12 a + 12 b.

AB = b − a and OM = 12 a + 12 b

2. OA = a and OB = b. Point C has OC = 3b − 2a. Show that A, B and C lie on a straight line.

  1. AB = b − a.
  2. BC = OC − OB = 3b − 2a − b = 2b − 2a.
  3. Take out the factor 2: BC = 2(b − a) = 2AB.
  4. So AB and BC are parallel, and B is on both lines.
  5. They are parts of one straight line: A, B and C are collinear ✓.

BC = 2AB and B is shared, so A, B and C are collinear

3. OA = a and OB = b. P is the point on AB with AP : PB = 3 : 1. Find OP in terms of a and b.

  1. 3 : 1 splits AB into 3 + 1 = 4 equal parts. AP is 3 of them.
  2. AP = 34 AB = 34 (b − a).
  3. OP = a + 34 b − 34 a.
  4. Collect: OP = 14 a + 34 b.

OP = 14 a + 34 b

Common mistake: Writing AB = a − b. From A to B you go back along a, then along b: AB = −a + b = b − a. Think "end minus start". And parallel is not enough for collinear: the two vectors must also share a point.

PractiseNo calculator

10 questions with instant feedback

Choose the correct answer. Sketch the shape, mark the vectors you know, and find a route from the start of the vector to its end. Questions 1–3 are warm-ups, 4–7 are at exam standard and 8–10 are challenge questions. After you answer, the feedback shows the full solution.

Score: 0 / 10Answered: 0 / 10

Warm-up (The basic skill: one or two steps.)

1.In triangle OAB, OA = a and OB = b. Find BA in terms of a and b.

Correct.

Solution

  1. Route from B to A: B to O, then O to A.
  2. BO = −b and OA = a.
  3. BA = −b + a = a − b.

Not quite. This is the vector from A to B. From B to A is the reverse.

Solution

  1. Route from B to A: B to O, then O to A.
  2. BO = −b and OA = a.
  3. BA = −b + a = a − b.

Answer: A. a − b.

Not quite. This adds. From B you go back along b, so the b part is −b.

Solution

  1. Route from B to A: B to O, then O to A.
  2. BO = −b and OA = a.
  3. BA = −b + a = a − b.

Answer: A. a − b.

Not quite. This goes back along both vectors. From B to A: back along b, then forward along a.

Solution

  1. Route from B to A: B to O, then O to A.
  2. BO = −b and OA = a.
  3. BA = −b + a = a − b.

Answer: A. a − b.

2.OABC is a parallelogram with OA = a and OC = c. Find the diagonal AC in terms of a and c.

Correct.

Solution

  1. Route from A to C: A to O, then O to C.
  2. AO = −a and OC = c.
  3. AC = c − a.

Not quite. This is OB, the other diagonal.

Solution

  1. Route from A to C: A to O, then O to C.
  2. AO = −a and OC = c.
  3. AC = c − a.

Answer: A. c − a.

Not quite. This is CA, from C to A.

Solution

  1. Route from A to C: A to O, then O to C.
  2. AO = −a and OC = c.
  3. AC = c − a.

Answer: A. c − a.

Not quite. This is AB, a side. AB = OC = c.

Solution

  1. Route from A to C: A to O, then O to C.
  2. AO = −a and OC = c.
  3. AC = c − a.

Answer: A. c − a.

3.In triangle OAB, OA = a and OB = b. X is the midpoint of OB. Find AX in terms of a and b.

Not quite. This is XA, from X to A.

Solution

  1. Route from A to X: A to O, then O to X.
  2. AO = −a, and OX = 12 b because X is halfway along OB.
  3. AX = 12 b − a.

Answer: D. 12 b − a.

Not quite. This adds a. From A you go back along a, so the a part is −a.

Solution

  1. Route from A to X: A to O, then O to X.
  2. AO = −a, and OX = 12 b because X is halfway along OB.
  3. AX = 12 b − a.

Answer: D. 12 b − a.

Not quite. This goes all the way to B. X is only halfway along OB.

Solution

  1. Route from A to X: A to O, then O to X.
  2. AO = −a, and OX = 12 b because X is halfway along OB.
  3. AX = 12 b − a.

Answer: D. 12 b − a.

Correct.

Solution

  1. Route from A to X: A to O, then O to X.
  2. AO = −a, and OX = 12 b because X is halfway along OB.
  3. AX = 12 b − a.

Exam standard (Several steps, like real exam questions.)

4.In triangle OAB, OA = a and OB = b. P is the point on AB with AP : PB = 2 : 3. Find OP in terms of a and b.

Not quite. The fractions are the wrong way round. P is nearer to A, so there is more a.

Solution

  1. 2 : 3 splits AB into 5 parts, so AP is 2/5 of AB.
  2. AP = 25 (b − a).
  3. OP = a + 25 b − 25 a = 35 a + 25 b.

Answer: C. 35 a + 25 b.

Not quite. This uses 2/3. The ratio 2 : 3 splits AB into 2 + 3 = 5 parts, so AP = 2/5 of AB.

Solution

  1. 2 : 3 splits AB into 5 parts, so AP is 2/5 of AB.
  2. AP = 25 (b − a).
  3. OP = a + 25 b − 25 a = 35 a + 25 b.

Answer: C. 35 a + 25 b.

Correct.

Solution

  1. 2 : 3 splits AB into 5 parts, so AP is 2/5 of AB.
  2. AP = 25 (b − a).
  3. OP = a + 25 b − 25 a = 35 a + 25 b.

Not quite. This uses AB = a − b. From A to B is b − a.

Solution

  1. 2 : 3 splits AB into 5 parts, so AP is 2/5 of AB.
  2. AP = 25 (b − a).
  3. OP = a + 25 b − 25 a = 35 a + 25 b.

Answer: C. 35 a + 25 b.

5.The table shows four vectors. Which one is parallel to 3a − 2b?

VectorIn terms of a and b
p6a − 4b
q6a − 2b
r3a + 2b
s2a − 3b

Correct.

Solution

  1. A parallel vector is a number times 3a − 2b.
  2. 6a − 4b = 2(3a − 2b).
  3. No other vector in the table is a multiple of 3a − 2b.

Not quite. Only the a part is doubled. Both parts must be multiplied by the same number.

Solution

  1. A parallel vector is a number times 3a − 2b.
  2. 6a − 4b = 2(3a − 2b).
  3. No other vector in the table is a multiple of 3a − 2b.

Answer: A. p = 6a − 4b.

Not quite. The b part has the wrong sign. For a multiple, both signs change or neither does.

Solution

  1. A parallel vector is a number times 3a − 2b.
  2. 6a − 4b = 2(3a − 2b).
  3. No other vector in the table is a multiple of 3a − 2b.

Answer: A. p = 6a − 4b.

Not quite. The numbers are swapped. 2a − 3b is not a number times 3a − 2b.

Solution

  1. A parallel vector is a number times 3a − 2b.
  2. 6a − 4b = 2(3a − 2b).
  3. No other vector in the table is a multiple of 3a − 2b.

Answer: A. p = 6a − 4b.

6.OA = a, OB = b and OC = 4b − 3a. Which working shows that A, B and C are collinear?

Not quite. Both vectors are wrong. AB = b − a, and AC = OC − OA.

Solution

  1. AB = b − a.
  2. AC = OC − OA = 4b − 3a − a = 4b − 4a.
  3. = 4(b − a) = 4AB, so the lines are parallel.
  4. A is on both, so A, B and C are collinear.

Answer: D. AB = b − a and AC = 4b − 4a = 4AB. So AC is parallel to AB, and A is on both: collinear.

Not quite. The vectors are right, but AB and AC share the point A, so they are the same line.

Solution

  1. AB = b − a.
  2. AC = OC − OA = 4b − 3a − a = 4b − 4a.
  3. = 4(b − a) = 4AB, so the lines are parallel.
  4. A is on both, so A, B and C are collinear.

Answer: D. AB = b − a and AC = 4b − 4a = 4AB. So AC is parallel to AB, and A is on both: collinear.

Not quite. C would be on OB only if OC were a multiple of b alone.

Solution

  1. AB = b − a.
  2. AC = OC − OA = 4b − 3a − a = 4b − 4a.
  3. = 4(b − a) = 4AB, so the lines are parallel.
  4. A is on both, so A, B and C are collinear.

Answer: D. AB = b − a and AC = 4b − 4a = 4AB. So AC is parallel to AB, and A is on both: collinear.

Correct.

Solution

  1. AB = b − a.
  2. AC = OC − OA = 4b − 3a − a = 4b − 4a.
  3. = 4(b − a) = 4AB, so the lines are parallel.
  4. A is on both, so A, B and C are collinear.

7.ABCDEF is a regular hexagon with centre O. OA = a and OB = b. Find AC in terms of a and b.

Not quite. This is AB (or OC). C is two steps round from A.

Solution

  1. OABC is a parallelogram (a rhombus), so OC = AB = b − a.
  2. AC = OC − OA = b − a − a.
  3. AC = b − 2a.

Answer: D. b − 2a.

Not quite. This is CA, from C to A.

Solution

  1. OABC is a parallelogram (a rhombus), so OC = AB = b − a.
  2. AC = OC − OA = b − a − a.
  3. AC = b − 2a.

Answer: D. b − 2a.

Not quite. This uses OC = a + b. In a regular hexagon, OC = b − a.

Solution

  1. OABC is a parallelogram (a rhombus), so OC = AB = b − a.
  2. AC = OC − OA = b − a − a.
  3. AC = b − 2a.

Answer: D. b − 2a.

Correct.

Solution

  1. OABC is a parallelogram (a rhombus), so OC = AB = b − a.
  2. AC = OC − OA = b − a − a.
  3. AC = b − 2a.

Challenge (Harder problems that mix ideas or need reasoning.)

8.OP = a + 2b and OQ = 3a + kb. O, P and Q lie on a straight line. Find k.

Not quite. This adds 3 to the 2. The whole vector must be multiplied by 3.

Solution

  1. O is on both lines, so OQ must be a multiple of OP.
  2. The a parts: 3 ÷ 1 = 3, so OQ = 3OP.
  3. The b parts: k = 3 × 2 = 6.

Answer: B. 6.

Correct.

Solution

  1. O is on both lines, so OQ must be a multiple of OP.
  2. The a parts: 3 ÷ 1 = 3, so OQ = 3OP.
  3. The b parts: k = 3 × 2 = 6.

Not quite. This divides 2 by 3. Multiply: 2 × 3.

Solution

  1. O is on both lines, so OQ must be a multiple of OP.
  2. The a parts: 3 ÷ 1 = 3, so OQ = 3OP.
  3. The b parts: k = 3 × 2 = 6.

Answer: B. 6.

Not quite. This makes the differences equal: 3 − 1 = 2, so 2 + 2. Collinear vectors are multiples, not equal steps.

Solution

  1. O is on both lines, so OQ must be a multiple of OP.
  2. The a parts: 3 ÷ 1 = 3, so OQ = 3OP.
  3. The b parts: k = 3 × 2 = 6.

Answer: B. 6.

9.P lies on the line AB. OA = a, OB = b and OP = 27 a + 57 b. Find the ratio AP : PB.

Not quite. This reads the a part as AP. The more b there is, the nearer P is to B.

Solution

  1. OP = a + AP, so AP = 57 b − 57 a = 57 (b − a).
  2. So AP = 57 AB, which leaves PB = 2/7 of AB.
  3. AP : PB = 5 : 2.

Answer: C. 5 : 2.

Not quite. This compares AP with the whole of AB, not with PB.

Solution

  1. OP = a + AP, so AP = 57 b − 57 a = 57 (b − a).
  2. So AP = 57 AB, which leaves PB = 2/7 of AB.
  3. AP : PB = 5 : 2.

Answer: C. 5 : 2.

Correct.

Solution

  1. OP = a + AP, so AP = 57 b − 57 a = 57 (b − a).
  2. So AP = 57 AB, which leaves PB = 2/7 of AB.
  3. AP : PB = 5 : 2.

Not quite. This compares PB with the whole of AB.

Solution

  1. OP = a + AP, so AP = 57 b − 57 a = 57 (b − a).
  2. So AP = 57 AB, which leaves PB = 2/7 of AB.
  3. AP : PB = 5 : 2.

Answer: C. 5 : 2.

10.In triangle OAB, OA = a and OB = b. P is on OA with OP : PA = 1 : 2, and Q is on OB with OQ : QB = 1 : 2. Which statement is true?

Not quite. OP : PA = 1 : 2 means OP is 1/3 of OA, not 1/2.

Solution

  1. OP = 13 a and OQ = 13 b.
  2. PQ = OQ − OP = 13 b − 13 a = 13 (b − a).
  3. b − a = AB, so PQ = 13 AB: parallel, and one third as long.

Answer: B. PQ = 13 AB, so PQ is parallel to AB and one third as long.

Correct.

Solution

  1. OP = 13 a and OQ = 13 b.
  2. PQ = OQ − OP = 13 b − 13 a = 13 (b − a).
  3. b − a = AB, so PQ = 13 AB: parallel, and one third as long.

Not quite. This uses PA, which is 2/3 of OA. P is 1/3 of the way from O.

Solution

  1. OP = 13 a and OQ = 13 b.
  2. PQ = OQ − OP = 13 b − 13 a = 13 (b − a).
  3. b − a = AB, so PQ = 13 AB: parallel, and one third as long.

Answer: B. PQ = 13 AB, so PQ is parallel to AB and one third as long.

Not quite. This works out OP − OQ. Use end minus start: OQ − OP.

Solution

  1. OP = 13 a and OQ = 13 b.
  2. PQ = OQ − OP = 13 b − 13 a = 13 (b − a).
  3. b − a = AB, so PQ = 13 AB: parallel, and one third as long.

Answer: B. PQ = 13 AB, so PQ is parallel to AB and one third as long.

Apply

Prove it with vectors

OABC is a parallelogram with OA = a and OC = c. M is the midpoint of AB. N is the point on the diagonal OB with ON : NB = 2 : 1.

1. Find OB and CM in terms of a and c.

2. Find CN in terms of a and c.

3. Show that C, N and M lie on a straight line, and find the ratio CN : NM.

  1. OB = OA + AB = a + c (AB = OC = c).
  2. CM = CO + OA + AM = −c + a + 12 c = a − 12 c.
  1. ON = 23 (a + c).
  2. CN = CO + ON = −c + 23 a + 23 c = 23 a − 13 c.
  1. 23 CM = 23 (a − 12 c) = 23 a − 13 c = CN.
  2. So CN is parallel to CM, and C is on both: C, N and M are collinear ✓.
  3. CN is 23 of CM, so NM is 13 of CM: CN : NM = 2 : 1.

Next: the Lesson 2 exit quiz. Five questions, recorded on your dashboard. Aim for 4 out of 5.

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