GCSE Maths: Vector geometry and proof (Higher)
Part of the GCSE Maths loci, bearings and vectors chapter (Higher). Chapter overview · whole chapter on one page · Download worksheet (PDF) · Answers (PDF)
Vector geometry and proof
Vectors are a powerful way to prove things about shapes. Instead of numbers, the vectors are written with letters. The usual set-up is a triangle OAB with OA = a and OB = b. The vectors a and b are called position vectors: they give the positions of A and B relative to the origin O. Every other vector in the diagram is made from a and b.
The route method. To find a vector, travel from the start point to the end point along vectors you already know, and add them. Moving along an arrow gives that vector; moving against it gives the negative. For example, to find →AB go from A back to O (that is −a) and then from O to B (that is b): AB = −a + b = b − a.
Midpoints and ratios. If M is the midpoint of AB then AM = ½AB. If P is on AB with AP : PB = m : n, then AP is m/(m + n) of the whole AB, because AB is made of m + n equal parts and AP is m of them. To find OP, go from O to A, then m/(m + n) of the way along AB: OP = a + m/(m + n) (b − a).
Parallel vectors. Vectors are parallel if one is a scalar multiple of the other. If MN = ½AB then MN is parallel to AB and half as long. The multiple can be any number, including a negative (opposite direction) or a fraction.
Collinear points. Points are collinear if they lie on one straight line. To prove that P, Q and R are collinear, show that PQ and QR are parallel (one is a multiple of the other). They share the point Q, so they lie on the same line. You must say both things: they are parallel and they share a point.
How to write a proof. (1) State what you are given. (2) Write the vectors you need by the route method. (3) Simplify, collecting the a terms and the b terms. (4) Finish with a sentence that says what you have shown, such as “so MN is parallel to AB and half its length”.
Key ideas
- AB = −a + b = b − a (go back along a, then along b).
- Midpoint of AB: AM = ½AB, and OM = ½(a + b).
- AP : PB = m : n means AP = m/(m + n) of AB.
- Parallel means one vector is a scalar multiple of the other.
- Collinear: show PQ is a multiple of QR (parallel) and say they share the point Q.
- Finish every proof with a conclusion in words.
Key words. A vector has a size (its length) and a direction. A column vector writes it as two numbers: how far across (right is +) over how far up (up is +). A scalar is an ordinary number.
Method. Find a route from A to B using vectors you know. Moving against an arrow gives the negative of that vector.
- Choose a route. Go from A to B along paths whose vectors we know or can work out. Add the vectors on the way.→AB = →AO + →OB
- Work out AO. Going from A to O is against the arrow OA, so it is the negative of OA.→AO = −a
- Work out OB. Going from O to B is along the arrow OB, which is given as b.→OB = b
- Add the parts and simplify. Collect the a terms together and the b terms together.→AB = (−a) + (b)
= b − a
Check. Try a different route from A to B, or use numbers (say a = (2, 0) and b = (0, 3)): both routes give the same vector. ✓
Watch out. Going the opposite way along a vector reverses its sign: AO = −OA. Always say which way each arrow points.
Key words. A vector has a size (its length) and a direction. A column vector writes it as two numbers: how far across (right is +) over how far up (up is +). A scalar is an ordinary number.
Method. Find a route from O to M using vectors you know. Moving against an arrow gives the negative of that vector.
- Choose a route. Go from O to M along paths whose vectors we know or can work out. Add the vectors on the way.→OM = →OA + →AM
- Work out OA. OA is given as a.→OA = a
- Work out AM. M is the midpoint of AB, so AM is half of AB. First AB = b − a (route A, O, B).→AM = 12b − 12a
- Add the parts and simplify. Collect the a terms together and the b terms together.→OM = (a) + (12b − 12a)
= 12a + 12b
Check. Try a different route from O to M, or use numbers (say a = (2, 0) and b = (0, 3)): both routes give the same vector. ✓
Watch out. Going the opposite way along a vector reverses its sign: AO = −OA. Always say which way each arrow points.
Key words. A vector has a size (its length) and a direction. A column vector writes it as two numbers: how far across (right is +) over how far up (up is +). A scalar is an ordinary number.
Method. Find AP as a fraction of AB, then go from O to P by way of A.
- Find AB. Go from A to B via O: first back to O, then on to B.→AB = →AO + →OB = b − a
- Find AP as a fraction of AB. AP : PB = 1 : 2, so the whole line AB is 3 equal parts and AP is 1 of them.→AP = 13 →AB = 13(b − a) = 13b − 13a
- Go from O to P via A. →OP = →OA + →AP→OP = a + (13b − 13a) = 23a + 13b
Check. Put in numbers, say a = (6, 0) and b = (0, 6). Then A = (6, 0) and B = (0, 6). A + 13(B − A) = (4, 2), and the formula gives (4, 2). They match. ✓
Watch out. AP : PB = 3 : 1 means P is three quarters of the way along (3 out of 4 parts), not three times as far.
Key words. A vector has a size (its length) and a direction. A column vector writes it as two numbers: how far across (right is +) over how far up (up is +). A scalar is an ordinary number.
Method. Find a route from M to N using vectors you know. Moving against an arrow gives the negative of that vector.
- Choose a route. Go from M to N along paths whose vectors we know or can work out. Add the vectors on the way.→MN = →MO + →ON
- Work out MO. M is the midpoint of OA, so MO is the negative of half of OA: −½a.→MO = −12a
- Work out ON. N is the midpoint of OB, so ON is half of OB: ½b.→ON = 12b
- Add the parts and simplify. Collect the a terms together and the b terms together.→MN = (−12a) + (12b)
= 12b − 12a - Compare with AB. From the first example, AB = b − a. Look for a multiple.→MN = 12b − 12a = 12(b − a) = ½→AB
- Write the conclusion. MN is a scalar multiple of AB, so they are parallel. The multiple is ½.MN is parallel to AB and half as long.
Check. Try a different route from M to N, or use numbers (say a = (2, 0) and b = (0, 3)): both routes give the same vector. ✓
Watch out. Going the opposite way along a vector reverses its sign: AO = −OA. Always say which way each arrow points.
Key words. A vector has a size (its length) and a direction. A column vector writes it as two numbers: how far across (right is +) over how far up (up is +). A scalar is an ordinary number.
Method. Points are collinear if the vectors between them are parallel and share a point. Show that QR is a multiple of PQ.
- Write the vector PQ. A vector between two points is the end position vector minus the start position vector.→PQ = →OQ − →OP = (3a + 3b) − (a + 2b) = 2a + b
- Write the vector QR. Do the same for the second pair of points.→QR = →OR − →OQ = (7a + 5b) − (3a + 3b) = 4a + 2b
- Compare the two vectors. If one is a scalar multiple of the other, they are parallel.→QR = 2(2a + b) = 2 →PQ
- Write the conclusion. The vectors are parallel, and they share the point Q. Parallel vectors through a common point lie on the same straight line.P, Q and R are collinear (on one straight line).
Check. With a = (1, 0) and b = (0, 1): P = (1, 2), Q = (3, 3), R = (7, 5). The gradients PQ and QR are both 0.5. ✓
Watch out. Parallel is not enough: the two vectors must also share a point (here Q), otherwise they could be two separate parallel lines.
Key words. A vector has a size (its length) and a direction. A column vector writes it as two numbers: how far across (right is +) over how far up (up is +). A scalar is an ordinary number.
Method. Find the midpoint of OB and the midpoint of AC using position vectors. If they are the same point, the diagonals cross at their midpoints.
- Find OB. In a parallelogram, AB is equal to OC (opposite sides), so go from O to A, then along AB = c.→OB = →OA + →AB = a + c
- Find the midpoint M of OB. The midpoint is half way along OB.→OM = ½(a + c) = 12a + 12c
- Find AC. Go from A back to O, then to C.→AC = −a + c = c − a
- Find the midpoint N of AC. Go from O to A, then half of AC.→ON = →OA + ½→AC = a + ½(c − a) = ½a + ½c
- Compare. Both midpoints have the same position vector 12a + 12c.→OM = →ON, so M and N are the same point
- Write the conclusion. The midpoint of OB is also the midpoint of AC.The diagonals bisect each other.
Check. Test with numbers: a = (4, 0), c = (1, 3). B = (5, 3), midpoint of OB = (2.5, 1.5). A = (4, 0), C = (1, 3), midpoint of AC = (2.5, 1.5). ✓
Watch out. Finish with a sentence. A proof must say what has been shown.
Key words. A vector has a size (its length) and a direction. A column vector writes it as two numbers: how far across (right is +) over how far up (up is +). A scalar is an ordinary number.
Method. Write OX in two different ways (once along each line), then equate the coefficients of a and b to find the two unknown fractions.
- Write AN and BM. Use the route A → O → N for AN, and B → O → M for BM.→AN = −a + 13b = 13b − a
→BM = −b + 12a = 12a − b - Describe X using AN. X is on AN, so AX is some multiple λ of AN. Go from O to A and then along AN.→OX = →OA + λ→AN = a + λ(13b − a) = (1 − λ)a + 13λb
- Describe X using BM. X is also on BM, so BX is some multiple μ of BM. Go from O to B and then along BM.→OX = →OB + μ→BM = b + μ(12a − b) = 12μa + (1 − μ)b
- Compare the coefficients. a and b are not parallel, so the a parts must match and the b parts must match.1 − λ = 12μ (1)
13λ = 1 − μ (2) - Solve the simultaneous equations. Rearrange (1) to get λ = 1 − 12μ, then substitute into (2) and collect the μ terms.13(1 − 12μ) = 1 − μ
13 − 16μ = 1 − μ
μ − 16μ = 1 − 13
56μ = 23
μ = 45
λ = 1 − 12 × 45 = 35 - Find OX. Substitute λ = 35 back into the expression from AN.→OX = (1 − 35)a + 13 × 35b = 25a + 15b
Check. Use the BM route: 12 × 45 = 25 for a and 1 − 45 = 15 for b. This agrees with 25a + 15b. ✓
Watch out. Both routes must give the same vector. Equating coefficients only works because a and b are not parallel.
Watch out
Always say which way each arrow goes. AB and BA are opposite vectors: BA = −AB. When you add vectors, they must join nose to tail, so check the route is continuous. In a “show that” question you must simplify to a form that makes the result obvious and then write a conclusion.Exercise 22L Vector routes, midpoints and ratios
Non-calculatorKey words. A vector has a size (its length) and a direction. A column vector writes it as two numbers: how far across (right is +) over how far up (up is +). A scalar is an ordinary number.
Method. Find a route from A to B using vectors you know. Moving against an arrow gives the negative of that vector.
- Choose a route. Go from A to B along paths whose vectors we know or can work out. Add the vectors on the way.→AB = →AO + →OB
- Work out AO. Against the arrow OA gives −a.→AO = −a
- Work out OB. Along OB gives b.→OB = b
- Add the parts and simplify. Collect the a terms together and the b terms together.→AB = (−a) + (b)
= b − a
Check. Try a different route from A to B, or use numbers (say a = (2, 0) and b = (0, 3)): both routes give the same vector. ✓
Watch out. Going the opposite way along a vector reverses its sign: AO = −OA. Always say which way each arrow points.
Key words. A vector has a size (its length) and a direction. A column vector writes it as two numbers: how far across (right is +) over how far up (up is +). A scalar is an ordinary number.
Method. Find a route from B to A using vectors you know. Moving against an arrow gives the negative of that vector.
- Choose a route. Go from B to A along paths whose vectors we know or can work out. Add the vectors on the way.→BA = →BO + →OA
- Work out BO. Against the arrow OB gives −b.→BO = −b
- Work out OA. Along OA gives a.→OA = a
- Add the parts and simplify. Collect the a terms together and the b terms together.→BA = (−b) + (a)
= a − b
Check. Try a different route from B to A, or use numbers (say a = (2, 0) and b = (0, 3)): both routes give the same vector. ✓
Watch out. Going the opposite way along a vector reverses its sign: AO = −OA. Always say which way each arrow points.
Key words. A vector has a size (its length) and a direction. A column vector writes it as two numbers: how far across (right is +) over how far up (up is +). A scalar is an ordinary number.
Method. Find a route from O to M using vectors you know. Moving against an arrow gives the negative of that vector.
- Choose a route. Go from O to M along paths whose vectors we know or can work out. Add the vectors on the way.→OM = →OM
- Work out OM. M is the midpoint of OA, so OM is half of OA: ½a.→OM = 12a
- Add the parts and simplify. Collect the a terms together and the b terms together.→OM = (12a)
= 12a
Check. Try a different route from O to M, or use numbers (say a = (2, 0) and b = (0, 3)): both routes give the same vector. ✓
Watch out. Going the opposite way along a vector reverses its sign: AO = −OA. Always say which way each arrow points.
Key words. A vector has a size (its length) and a direction. A column vector writes it as two numbers: how far across (right is +) over how far up (up is +). A scalar is an ordinary number.
Method. Find a route from M to B using vectors you know. Moving against an arrow gives the negative of that vector.
- Choose a route. Go from M to B along paths whose vectors we know or can work out. Add the vectors on the way.→MB = →MO + →OB
- Work out MO. MO is the opposite of OM, which is −½a.→MO = −12a
- Work out OB. OB is b.→OB = b
- Add the parts and simplify. Collect the a terms together and the b terms together.→MB = (−12a) + (b)
= b − 12a
Check. Try a different route from M to B, or use numbers (say a = (2, 0) and b = (0, 3)): both routes give the same vector. ✓
Watch out. Going the opposite way along a vector reverses its sign: AO = −OA. Always say which way each arrow points.
Key words. A vector has a size (its length) and a direction. A column vector writes it as two numbers: how far across (right is +) over how far up (up is +). A scalar is an ordinary number.
Method. Find a route from M to A using vectors you know. Moving against an arrow gives the negative of that vector.
- Choose a route. Go from M to A along paths whose vectors we know or can work out. Add the vectors on the way.→MA = →MO + →OA
- Work out MO. Against OM: −½a.→MO = −12a
- Work out OA. OA is a.→OA = a
- Add the parts and simplify. Collect the a terms together and the b terms together.→MA = (−12a) + (a)
= 12a
Check. Try a different route from M to A, or use numbers (say a = (2, 0) and b = (0, 3)): both routes give the same vector. ✓
Watch out. Going the opposite way along a vector reverses its sign: AO = −OA. Always say which way each arrow points.
Key words. A vector has a size (its length) and a direction. A column vector writes it as two numbers: how far across (right is +) over how far up (up is +). A scalar is an ordinary number.
Method. M is the midpoint of AB, so AM = ½AB. First find AB.
- Find AB. Go from A back to O, then to B.→AB = −a + b = b − a
- Halve it. AM is half of AB.→AM = ½(b − a) = 12b − 12a
Check. MB = AM, so M is half way along AB. ✓
Watch out. The midpoint splits the vector into two equal halves.
Key words. A vector has a size (its length) and a direction. A column vector writes it as two numbers: how far across (right is +) over how far up (up is +). A scalar is an ordinary number.
Method. Find a route from O to M using vectors you know. Moving against an arrow gives the negative of that vector.
- Choose a route. Go from O to M along paths whose vectors we know or can work out. Add the vectors on the way.→OM = →OA + →AM
- Work out OA. OA is given: a.→OA = a
- Work out AM. AM is half of AB = b − a.→AM = 12b − 12a
- Add the parts and simplify. Collect the a terms together and the b terms together.→OM = (a) + (12b − 12a)
= 12a + 12b
Check. Try a different route from O to M, or use numbers (say a = (2, 0) and b = (0, 3)): both routes give the same vector. ✓
Watch out. Going the opposite way along a vector reverses its sign: AO = −OA. Always say which way each arrow points.
Key words. A vector has a size (its length) and a direction. A column vector writes it as two numbers: how far across (right is +) over how far up (up is +). A scalar is an ordinary number.
Method. M is the midpoint of AB, so MB = AM = ½AB.
- Find AB. Go from A to O to B.→AB = b − a
- Halve it. MB is the other half of AB, in the same direction as AB.→MB = ½(b − a) = 12b − 12a
Check. AM + MB = AB: (−½a + ½b) + (−½a + ½b) = b − a. ✓
Watch out. MB is in the same direction as AB (from A towards B).
Key words. A vector has a size (its length) and a direction. A column vector writes it as two numbers: how far across (right is +) over how far up (up is +). A scalar is an ordinary number.
Method. Find AP as a fraction of AB, then go from O to P by way of A.
- Find AB. Go from A to B via O: first back to O, then on to B.→AB = →AO + →OB = b − a
- Find AP as a fraction of AB. AP : PB = 2 : 1, so the whole line AB is 3 equal parts and AP is 2 of them.→AP = 23 →AB = 23(b − a) = 23b − 23a
- Go from O to P via A. →OP = →OA + →AP→OP = a + (23b − 23a) = 13a + 23b
Check. Put in numbers, say a = (6, 0) and b = (0, 6). Then A = (6, 0) and B = (0, 6). A + 23(B − A) = (2, 4), and the formula gives (2, 4). They match. ✓
Watch out. AP : PB = 3 : 1 means P is three quarters of the way along (3 out of 4 parts), not three times as far.
Key words. A vector has a size (its length) and a direction. A column vector writes it as two numbers: how far across (right is +) over how far up (up is +). A scalar is an ordinary number.
Method. Find AP as a fraction of AB, then go from O to P by way of A.
- Find AB. Go from A to B via O: first back to O, then on to B.→AB = →AO + →OB = b − a
- Find AP as a fraction of AB. AP : PB = 3 : 2, so the whole line AB is 5 equal parts and AP is 3 of them.→AP = 35 →AB = 35(b − a) = 35b − 35a
- Go from O to P via A. →OP = →OA + →AP→OP = a + (35b − 35a) = 25a + 35b
Check. Put in numbers, say a = (6, 0) and b = (0, 6). Then A = (6, 0) and B = (0, 6). A + 35(B − A) = (2.4, 3.6), and the formula gives (2.4, 3.6). They match. ✓
Watch out. AP : PB = 3 : 1 means P is three quarters of the way along (3 out of 4 parts), not three times as far.
Key words. A vector has a size (its length) and a direction. A column vector writes it as two numbers: how far across (right is +) over how far up (up is +). A scalar is an ordinary number.
Method. Find a route from O to B using vectors you know. Moving against an arrow gives the negative of that vector.
- Choose a route. Go from O to B along paths whose vectors we know or can work out. Add the vectors on the way.→OB = →OA + →AB
- Work out OA. OA is a.→OA = a
- Work out AB. AB is opposite OC in the parallelogram, so AB = c.→AB = c
- Add the parts and simplify. Collect the a terms together and the b terms together.→OB = (a) + (c)
= a + c
Check. Try a different route from O to B, or use numbers (say a = (2, 0) and b = (0, 3)): both routes give the same vector. ✓
Watch out. Going the opposite way along a vector reverses its sign: AO = −OA. Always say which way each arrow points.
Key words. A vector has a size (its length) and a direction. A column vector writes it as two numbers: how far across (right is +) over how far up (up is +). A scalar is an ordinary number.
Method. Find a route from A to C using vectors you know. Moving against an arrow gives the negative of that vector.
- Choose a route. Go from A to C along paths whose vectors we know or can work out. Add the vectors on the way.→AC = →AO + →OC
- Work out AO. Against OA: −a.→AO = −a
- Work out OC. OC is c.→OC = c
- Add the parts and simplify. Collect the a terms together and the b terms together.→AC = (−a) + (c)
= c − a
Check. Try a different route from A to C, or use numbers (say a = (2, 0) and b = (0, 3)): both routes give the same vector. ✓
Watch out. Going the opposite way along a vector reverses its sign: AO = −OA. Always say which way each arrow points.
Key words. A vector has a size (its length) and a direction. A column vector writes it as two numbers: how far across (right is +) over how far up (up is +). A scalar is an ordinary number.
Method. Find a route from C to A using vectors you know. Moving against an arrow gives the negative of that vector.
- Choose a route. Go from C to A along paths whose vectors we know or can work out. Add the vectors on the way.→CA = →CO + →OA
- Work out CO. Against OC: −c.→CO = −c
- Work out OA. OA is a.→OA = a
- Add the parts and simplify. Collect the a terms together and the b terms together.→CA = (−c) + (a)
= a − c
Check. Try a different route from C to A, or use numbers (say a = (2, 0) and b = (0, 3)): both routes give the same vector. ✓
Watch out. Going the opposite way along a vector reverses its sign: AO = −OA. Always say which way each arrow points.
Key words. A vector has a size (its length) and a direction. A column vector writes it as two numbers: how far across (right is +) over how far up (up is +). A scalar is an ordinary number.
Method. Find a route from A to B using vectors you know. Moving against an arrow gives the negative of that vector.
- Choose a route. Go from A to B along paths whose vectors we know or can work out. Add the vectors on the way.→AB = →AO + →OC + →CB
- Work out AO. Against OA: −a.→AO = −a
- Work out OC. OC is c.→OC = c
- Work out CB. CB = 2a (given).→CB = 2a
- Add the parts and simplify. Collect the a terms together and the b terms together.→AB = (−a) + (c) + (2a)
= a + c
Check. Try a different route from A to B, or use numbers (say a = (2, 0) and b = (0, 3)): both routes give the same vector. ✓
Watch out. Going the opposite way along a vector reverses its sign: AO = −OA. Always say which way each arrow points.
Key words. A vector has a size (its length) and a direction. A column vector writes it as two numbers: how far across (right is +) over how far up (up is +). A scalar is an ordinary number.
Method. Find a route from D to E using vectors you know. Moving against an arrow gives the negative of that vector.
- Choose a route. Go from D to E along paths whose vectors we know or can work out. Add the vectors on the way.→DE = →DO + →OE
- Work out DO. D is on OA with OD = ⅓a, so DO = −⅓a.→DO = −13a
- Work out OE. E is on OB with OE = ⅓b.→OE = 13b
- Add the parts and simplify. Collect the a terms together and the b terms together.→DE = (−13a) + (13b)
= 13b − 13a - Compare with AB. AB = b − a.→DE = 13b − 13a = ⅓(b − a) = ⅓→AB
- Write the conclusion. DE is a multiple of AB.DE is parallel to AB and one third of its length.
Check. Try a different route from D to E, or use numbers (say a = (2, 0) and b = (0, 3)): both routes give the same vector. ✓
Watch out. Going the opposite way along a vector reverses its sign: AO = −OA. Always say which way each arrow points.
Key words. A vector has a size (its length) and a direction. A column vector writes it as two numbers: how far across (right is +) over how far up (up is +). A scalar is an ordinary number.
Method. Points are collinear if the vectors between them are parallel and share a point. Show that QR is a multiple of PQ.
- Write the vector PQ. A vector between two points is the end position vector minus the start position vector.→PQ = →OQ − →OP = (3a + 2b) − (a + b) = 2a + b
- Write the vector QR. Do the same for the second pair of points.→QR = →OR − →OQ = (5a + 3b) − (3a + 2b) = 2a + b
- Compare the two vectors. If one is a scalar multiple of the other, they are parallel.→QR = 1(2a + b) = 1 →PQ
- Write the conclusion. The vectors are parallel, and they share the point Q. Parallel vectors through a common point lie on the same straight line.P, Q and R are collinear (on one straight line).
Check. With a = (1, 0) and b = (0, 1): P = (1, 1), Q = (3, 2), R = (5, 3). The gradients PQ and QR are both 0.5. ✓
Watch out. Parallel is not enough: the two vectors must also share a point (here Q), otherwise they could be two separate parallel lines.
Key words. A vector has a size (its length) and a direction. A column vector writes it as two numbers: how far across (right is +) over how far up (up is +). A scalar is an ordinary number.
Method. Points are collinear if the vectors between them are parallel and share a point. Show that QR is a multiple of PQ.
- Write the vector PQ. A vector between two points is the end position vector minus the start position vector.→PQ = →OQ − →OP = (2b − a) − (2a + b) = b − 3a
- Write the vector QR. Do the same for the second pair of points.→QR = →OR − →OQ = (3b − 4a) − (2b − a) = b − 3a
- Compare the two vectors. If one is a scalar multiple of the other, they are parallel.→QR = 1(b − 3a) = 1 →PQ
- Write the conclusion. The vectors are parallel, and they share the point Q. Parallel vectors through a common point lie on the same straight line.P, Q and R are collinear (on one straight line).
Check. With a = (1, 0) and b = (0, 1): P = (2, 1), Q = (−1, 2), R = (−4, 3). The gradients PQ and QR are both −0.333333333333333333333333333333333333333333333333333333333333. ✓
Watch out. Parallel is not enough: the two vectors must also share a point (here Q), otherwise they could be two separate parallel lines.
Key words. A vector has a size (its length) and a direction. A column vector writes it as two numbers: how far across (right is +) over how far up (up is +). A scalar is an ordinary number.
Method. Points are collinear only if the vectors between them are parallel. Show that QR is not a multiple of PQ.
- Write the vector PQ. End position vector minus start position vector.→PQ = (2a + 5b) − (a + 2b) = a + 3b
- Write the vector QR. Do the same for the next pair.→QR = (4a + 9b) − (2a + 5b) = 2a + 4b
- Compare the two vectors. If the points were on one line, one vector would be a scalar multiple of the other. Compare the coefficients of a and b.→PQ: a coefficient 1, b coefficient 3
→QR: a coefficient 2, b coefficient 4
No single number multiplies both coefficients of →PQ to give those of →QR. - Write the conclusion. The vectors are not parallel.P, Q and R are not collinear.
Check. Put a = (1, 0) and b = (0, 1) and plot the three points: they do not lie on a straight line. ✓
Watch out. Compare the ratios of the coefficients (a to b), not just one coefficient.
Key words. A vector has a size (its length) and a direction. A column vector writes it as two numbers: how far across (right is +) over how far up (up is +). A scalar is an ordinary number.
Method. Write OX in two different ways (once along each line), then equate the coefficients of a and b to find the two unknown fractions.
- Write AN and BM. Use the route A → O → N for AN, and B → O → M for BM.→AN = −a + 12b = 12b − a
→BM = −b + 13a = 13a − b - Describe X using AN. X is on AN, so AX is some multiple λ of AN. Go from O to A and then along AN.→OX = →OA + λ→AN = a + λ(12b − a) = (1 − λ)a + 12λb
- Describe X using BM. X is also on BM, so BX is some multiple μ of BM. Go from O to B and then along BM.→OX = →OB + μ→BM = b + μ(13a − b) = 13μa + (1 − μ)b
- Compare the coefficients. a and b are not parallel, so the a parts must match and the b parts must match.1 − λ = 13μ (1)
12λ = 1 − μ (2) - Solve the simultaneous equations. Rearrange (1) to get λ = 1 − 13μ, then substitute into (2) and collect the μ terms.12(1 − 13μ) = 1 − μ
12 − 16μ = 1 − μ
μ − 16μ = 1 − 12
56μ = 12
μ = 35
λ = 1 − 13 × 35 = 45 - Find OX. Substitute λ = 45 back into the expression from AN.→OX = (1 − 45)a + 12 × 45b = 15a + 25b
Check. Use the BM route: 13 × 35 = 15 for a and 1 − 35 = 25 for b. This agrees with 15a + 25b. ✓
Watch out. Both routes must give the same vector. Equating coefficients only works because a and b are not parallel.
Key words. A vector has a size (its length) and a direction. A column vector writes it as two numbers: how far across (right is +) over how far up (up is +). A scalar is an ordinary number.
Method. This is the same set-up as the worked example with M and N the midpoints. Write OX in two ways and compare coefficients.
- Use the general working. With OM = ½a and ON = ½b, find AN and BM and write OX as OA + λAN and as OB + μBM.AN = −a + ½b and BM = ½a − b
- Equate the coefficients. (1 − λ)a + ½λb = ½μa + (1 − μ)b.1 − λ = ½μ and ½λ = 1 − μ
- Solve. From the first equation λ = 1 − ½μ. Substitute into the second: ½(1 − ½μ) = 1 − μ, so ½ − ¼μ = 1 − μ, so ¾μ = ½.μ = 23 and λ = 23
- Find OX. Substitute λ = 2/3 into OX = (1 − λ)a + ½λb.OX = ⅓a + ⅓b = ⅓(a + b)
- Interpret. AX = λAN with λ = 2/3, so X is two thirds of the way along AN.AX : XN = 2 : 1
Check. The medians of a triangle meet at the centroid, which is two thirds of the way along each median from the vertex. ✓
Watch out. Both routes describe the same point X, so the coefficients must match.
Exercise 22M Vector proof
Non-calculatorKey words. A vector has a size (its length) and a direction. A column vector writes it as two numbers: how far across (right is +) over how far up (up is +). A scalar is an ordinary number.
Method. Find a route from M to F using vectors you know. Moving against an arrow gives the negative of that vector.
- Choose a route. Go from M to F along paths whose vectors we know or can work out. Add the vectors on the way.→MF = →MO + →OB + →BF
- Work out MO. MO = −½a.→MO = −12a
- Work out OB. OB = b.→OB = b
- Work out BF. F is the midpoint of AB. BF is half of BA = −(b − a), so BF = ½(a − b).→BF = 12a − 12b
- Add the parts and simplify. Collect the a terms together and the b terms together.→MF = (−12a) + (b) + (12a − 12b)
= 12b - Compare. Compare with OB = b.→MF = ½b = ½→OB
- Write the conclusion. MF is parallel to OB and half its length.
Check. Try a different route from M to F, or use numbers (say a = (2, 0) and b = (0, 3)): both routes give the same vector. ✓
Watch out. Going the opposite way along a vector reverses its sign: AO = −OA. Always say which way each arrow points.
Key words. A vector has a size (its length) and a direction. A column vector writes it as two numbers: how far across (right is +) over how far up (up is +). A scalar is an ordinary number.
Method. Find a route from N to F using vectors you know. Moving against an arrow gives the negative of that vector.
- Choose a route. Go from N to F along paths whose vectors we know or can work out. Add the vectors on the way.→NF = →NO + →OA + →AF
- Work out NO. NO = −½b.→NO = −12b
- Work out OA. OA = a.→OA = a
- Work out AF. AF is half of AB = ½(b − a).→AF = 12b − 12a
- Add the parts and simplify. Collect the a terms together and the b terms together.→NF = (−12b) + (a) + (12b − 12a)
= 12a - Compare. Compare with OA = a.→NF = ½a = ½→OA
- Write the conclusion. NF is parallel to OA and half its length.
Check. Try a different route from N to F, or use numbers (say a = (2, 0) and b = (0, 3)): both routes give the same vector. ✓
Watch out. Going the opposite way along a vector reverses its sign: AO = −OA. Always say which way each arrow points.
Key words. A vector has a size (its length) and a direction. A column vector writes it as two numbers: how far across (right is +) over how far up (up is +). A scalar is an ordinary number.
Method. Find a route from A to B using vectors you know. Moving against an arrow gives the negative of that vector.
- Choose a route. Go from A to B along paths whose vectors we know or can work out. Add the vectors on the way.→AB = →AO + →OC + →CB
- Work out AO. Against OA: −a.→AO = −a
- Work out OC. OC = ½a + b.→OC = 12a + b
- Work out CB. CB = 2a (given).→CB = 2a
- Add the parts and simplify. Collect the a terms together and the b terms together.→AB = (−a) + (12a + b) + (2a)
= 32a + b - Compare the parallel sides. OA = a and CB = 2a.→CB = 2→OA, so CB is parallel to OA and twice as long.
- Check the other pair of sides. A trapezium has exactly one pair of parallel sides, so OC and AB must not be parallel. Compare the coefficients of a and b in each.→OC = 12a + b and →AB = 32a + b
The a:b coefficients are 1/2 : 1 and 3/2 : 1, which are different, so OC and AB are not parallel. - Write the conclusion. One pair of sides is parallel and the other pair is not.OABC is a trapezium: OA is parallel to CB, and CB is twice as long as OA.
Check. Try a different route from A to B, or use numbers (say a = (2, 0) and b = (0, 3)): both routes give the same vector. ✓
Watch out. Going the opposite way along a vector reverses its sign: AO = −OA. Always say which way each arrow points.
Key words. A vector has a size (its length) and a direction. A column vector writes it as two numbers: how far across (right is +) over how far up (up is +). A scalar is an ordinary number.
Method. If Q is the midpoint of PR then PQ = QR. Work out both vectors.
- Find PQ. End position vector minus start position vector.→PQ = (3a + 4b) − (a + 2b) = 2a + 2b
- Find QR. The same way.→QR = (5a + 6b) − (3a + 4b) = 2a + 2b
- Compare. The two vectors are equal, so P, Q, R are on a straight line and PQ = QR.PQ = QR
- Write the conclusion. Q is the midpoint of PR.
Check. Midpoint formula: ½(OP + OR) = ½(6a + 8b) = 3a + 4b = OQ. ✓
Watch out. Equal vectors share direction and length, so Q is exactly half way.
Key words. A vector has a size (its length) and a direction. A column vector writes it as two numbers: how far across (right is +) over how far up (up is +). A scalar is an ordinary number.
Method. Now OA = 4a and OB = 4b, so use these in the route method.
- Find AB. Go from A to O to B.→AB = −4a + 4b = 4(b − a)
- Find AP. AP : PB = 1 : 3, so AP is 1 of the 4 equal parts of AB.→AP = ¼ × 4(b − a) = b − a
- Find OP. Go from O to A, then along AP.→OP = 4a + (b − a) = 3a + b
Check. With a = (1, 0) and b = (0, 1): A = (4, 0), B = (0, 4), P = (3, 1). P is on the line x + y = 4 through A and B. ✓
Watch out. A scalar in front of a or b (like 4a) is part of the vector. Do not change it.
Key words. A vector has a size (its length) and a direction. A column vector writes it as two numbers: how far across (right is +) over how far up (up is +). A scalar is an ordinary number.
Method. Find the position vectors of the four midpoints, then find the vectors PQ and SR. If PQ = SR then two sides are equal and parallel, which makes a parallelogram.
- Find the midpoints. The position vector of a midpoint is half the sum of the position vectors of the ends.P = ½(a + b), Q = ½(b + c), R = ½(c + d), S = ½(d + a)
- Find PQ. End minus start.→PQ = ½(b + c) − ½(a + b) = ½(c − a)
- Find SR. End minus start.→SR = ½(c + d) − ½(d + a) = ½(c − a)
- Compare. The two vectors are equal.PQ = SR
- Write the conclusion. PQ and SR are equal in length and parallel, so PQRS has a pair of opposite sides that are equal and parallel.PQRS is a parallelogram.
Check. Test with A(0, 0), B(4, 0), C(6, 4), D(0, 6): P(2, 0), Q(5, 2), R(3, 5), S(0, 3). PQ = (3, 2) and SR = (3, 2). ✓
Watch out. This works for every quadrilateral, even one that is not a parallelogram.
Key words. A vector has a size (its length) and a direction. A column vector writes it as two numbers: how far across (right is +) over how far up (up is +). A scalar is an ordinary number.
Method. Find the vectors BX and XM, and compare them.
- Find BX. End position vector minus start.→BX = ⅓(a + b) − b = ⅓a − ⅔b
- Find XM. M is the midpoint of OA so OM = ½a.→XM = ½a − ⅓(a + b) = ⅙a − ⅓b
- Compare. Is one a multiple of the other?→BX = ⅓a − ⅔b = 2(⅙a − ⅓b) = 2→XM
- Write the conclusion. They are parallel, share X, and BX is twice XM.B, X, M are collinear and BX : XM = 2 : 1.
Check. X is the centroid, which is 2/3 of the way from B to M. ✓
Watch out. Show the multiple explicitly, then say what it means.