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This page sets out the method for vector proof questions in the order you would use it, then walks through the full lesson explanation. For three complete worked examples with the slips that cost marks, see the Vector proof questions topic page.
When you see 'find in terms of and ', do this.
When you see 'show that PQ is parallel to AB', do this.
When you see 'prove that M, N and X are collinear' or 'lie on a straight line', do this.
OA = a means the vector from the origin to A is a. It is not the length on its own.
The vector AB is OB - OA, which is b - a.
The midpoint of AB has position 1/2(a+b).
If one vector is k times another, the two vectors are parallel.
If AC is a scalar multiple of AB and they share A, then C lies on the line AB.
Draw the arrows from the origin before you combine them.
Choose an origin O and write overrightarrowOA = mathbfa and overrightarrowOB = mathbfb. Those arrows both start at O. The arrow you usually want is along a side, which does not start at O. To go from A to B, go backwards along mathbfa to the origin and then forwards along mathbfb. In symbols that is -mathbfa + mathbfb, which is written mathbfb - mathbfa. The order matters: mathbfa - mathbfb is the journey from B back to A.
With numbers the same subtraction is done on each component. If mathbfa = (2, 5) and mathbfb = (8, 9), then overrightarrowAB = (8 - 2, 9 - 5) = (6, 4). The midpoint of A and B averages the components: ((2 + 8)/2, (5 + 9)/2) = (5, 7). In letters the midpoint of AB has position 1/2(mathbfa + mathbfb). Halving only one of the vectors, or adding them and forgetting the half, puts the point in the wrong place.
Write the path in words before you simplify. "From O to A, then halfway along AB" becomes mathbfa + 1/2(mathbfb - mathbfa). Expand that and you reach the average again: mathbfa + 1/2mathbfb - 1/2mathbfa = 1/2mathbfa + 1/2mathbfb. Two routes that land on the same expression are a useful check. If they disagree, one of the routes started at the wrong point.
Method: 1. Write overrightarrowOA=mathbfa and overrightarrowOB=mathbfb. 2. The side from A to B is overrightarrowAB=mathbfb-mathbfa. 3. With mathbfa=(2,5) and mathbfb=(8,9), that side is (6,4) and the midpoint is (5,7).
Let M be the midpoint of OA and let N be the midpoint of OB, in triangle OAB. Then overrightarrowOM = 1/2mathbfa and overrightarrowON = 1/2mathbfb. The journey from M to N is the end minus the start: overrightarrowMN = 1/2mathbfb - 1/2mathbfa = 1/2(mathbfb - mathbfa). But overrightarrowAB = mathbfb - mathbfa, so overrightarrowMN = 1/2overrightarrowAB. The scalar is 1/2. That single equation says two things: MN is parallel to AB, and MN is half as long as AB.
A different midline uses the midpoint of a side rather than both arms from the origin. Let M again be the midpoint of OA, so its position is 1/2mathbfa. Let N now be the midpoint of AB, so its position is 1/2(mathbfa + mathbfb). Then overrightarrowMN = 1/2(mathbfa + mathbfb) - 1/2mathbfa = 1/2mathbfa + 1/2mathbfb - 1/2mathbfa = 1/2mathbfb. Since overrightarrowOB = mathbfb, you have overrightarrowMN = 1/2overrightarrowOB. The scalar is again 1/2, so MN is parallel to OB and half as long. The mathbfa terms cancelled. If they do not cancel, the midpoint positions were not both halved.
A numeric check keeps the algebra honest. Take mathbfa = (6, 2) and mathbfb = (4, 8) for this second midline. Then M = (3, 1) and N = (5, 5), so overrightarrowMN = (2, 4). Half of mathbfb is 1/2(4, 8) = (2, 4). The vectors match, so k = 1/2. Equal length on its own would not have been enough: you needed one vector to be a scalar times the other. Here the scalar is positive, so the arrows point the same way. A negative scalar would still mean parallel, but the arrows would point in opposite senses.
Method: 1. Halve the position vectors of the endpoints to place each midpoint. 2. Then subtract: overrightarrowMN=1/2mathbfb-1/2mathbfa=1/2(mathbfb-mathbfa). 3. A scalar of 1/2 shows MN is parallel to AB and half as long.
Suppose C lies on AB with AC:CB = m:n. The fraction of the journey from A to B is m/(m + n), because the whole side is split into m + n equal parts and C is m parts from A. So overrightarrowAC = m/(m + n)(mathbfb - mathbfa), and the position of C is overrightarrowOC = mathbfa + m/(m + n)(mathbfb - mathbfa). The ratio 1:1 gives the midpoint, since 1/2(mathbfb - mathbfa) is half the side. The ratio 1:2 gives a point one third of the way from A to B.
The exam ratio AC:CB = 2:1 means C is two thirds of the way from A to B, and closer to B than to A. Then overrightarrowOC = mathbfa + 2/3(mathbfb - mathbfa). Expand the bracket: mathbfa + 2/3mathbfb - 2/3mathbfa = 1/3mathbfa + 2/3mathbfb. The coefficient of mathbfb is the larger one. Swapping them describes the ratio 1:2, which is a different point. You can remember the weights as "the other part of the ratio": the weight on A is the CB part, and the weight on B is the AC part, then divide by the total.
Collinear points use the same scalar test, plus a shared point. If overrightarrowAC = koverrightarrowAB and C is reached from A, then C lies on the line AB. For the ratio 2:1 the scalar from A along AB is 2/3, which is between 0 and 1, so C is between A and B. A scalar bigger than 1 would put the point beyond B. A negative scalar would put it on the extension beyond A. State the scalar, and state what it tells you about the position, rather than only saying "parallel".
Method: 1. For AC:CB=m:n, travel m/(m+n) of the way from A to B. 2. The position is overrightarrowOC=mathbfa+m/(m+n)(mathbfb-mathbfa). 3. Ratio 2:1 puts C at 1/3mathbfa+2/3mathbfb.
Define every letter before you use it. The first lines are usually overrightarrowOA = mathbfa and overrightarrowOB = mathbfb, then the midpoint or the ratio you were given. Build the unknown vector as a path: position of the end minus position of the start, or position of the start plus a fraction of a side. Simplify by collecting mathbfa and mathbfb. Only then compare with the vector you hope to match.
To show a length ratio, write one vector as k times the other and state k. To show parallel, the same line is enough, because a scalar multiple is the definition you are using. To show three points are collinear, you need both the scalar multiple and a point in common. A diagram helps you choose the path, but the marks sit on the algebra and the sentence that names k.
Method: 1. Define mathbfa and mathbfb before you build a path. 2. Simplify the path, then write the result as k times the vector you want to match. 3. State the value of k, and use a shared point as well when you claim three points are collinear.
A vector question usually gives a triangle or quadrilateral with a few sides labelled, such as overrightarrowOA = mathbfa and overrightarrowOB = mathbfb, and asks for another vector "in terms of mathbfa and mathbfb". The skill is to travel along known sides from the start point to the end point. Going with an arrow adds its vector, going against an arrow subtracts it. So overrightarrowAB = overrightarrowAO + overrightarrowOB = -mathbfa + mathbfb.
Write the route first, for example overrightarrowAB = overrightarrowAO + overrightarrowOB, and then replace each part by its letter. Collect like terms at the end and write them in a tidy order, usually mathbfb - mathbfa rather than -mathbfa + mathbfb. Command phrases such as "give your answer in its simplest form" mean you must simplify any fractions and combine multiples of the same vector.
Method: 1. Choose a route from the start point to the end point using known sides. 2. Write the route as a sum of vectors, using a minus for any side travelled backwards. 3. Replace each part by its vector in terms of mathbfa and mathbfb. 4. Collect like terms and write the answer in simplest form.
Tip: Reversing a vector changes its sign: overrightarrowBA = -overrightarrowAB.
When a point X divides AB so that AX : XB = m : n, the vector overrightarrowAX is m/(m + n) of overrightarrowAB. For example, if AX : XB = 1 : 3 then overrightarrowAX = tfrac14overrightarrowAB, because there are four equal parts in total. Reading the ratio the wrong way, or using tfracmn instead of tfracmm + n, is the most frequent slip.
To find the position vector of X from O, go from O to A and then along part of the line: overrightarrowOX = mathbfa + tfrac14(mathbfb - mathbfa) = tfrac34mathbfa + tfrac14mathbfb. The coefficients add to 1, which is a good quick check for a point on the line AB. Two ratios in one question, on two different sides, simply need the same method twice.
Method: 1. Add the ratio parts to find the total number of equal parts. 2. Write the part of the line as a fraction of the whole vector. 3. Add that part to the vector to the start of the line. 4. Check that the coefficients of mathbfa and mathbfb add up to 1.
Two vectors are parallel when one is a multiple of the other. If overrightarrowMN = 3overrightarrowAB then MN is parallel to AB and three times as long. To prove three points A, B and C lie on a straight line you show that overrightarrowAB and overrightarrowAC are multiples, for example overrightarrowAC = 3overrightarrowAB, and then say they share the point A.
That last sentence is the mark that is often missed. Parallel vectors alone do not make a straight line, because two parallel lines can be a long way apart. Write "overrightarrowAC = 3overrightarrowAB, so AC is parallel to AB. They share the point A, so A, B and C lie on a straight line." Factorise the vectors first if the numbers are large, since 15mathbfa + 6mathbfb = 3(5mathbfa + 2mathbfb).
Method: 1. Write each vector in terms of mathbfa and mathbfb and simplify. 2. Factorise to show one vector is a scalar multiple of the other. 3. State that they are parallel, and for collinear points add that they share a point. 4. Give the conclusion in words, using the letters of the question.
A full vector proof has four parts: the route for each vector, the simplification, the comparison, and the conclusion. For example, to show a line joining two midpoints is parallel to the third side and half its length, find overrightarrowMN by going via the origin, simplify to tfrac12(mathbfb - mathbfa), then compare it with overrightarrowAB = mathbfb - mathbfa and say what the factor tfrac12 means.
A "hence" part may ask for a ratio of lengths, such as OX : XM, or for an area ratio. Read it directly from the multiples. If overrightarrowOM = tfrac32overrightarrowOX then OX : OM = 2 : 3 and OX : XM = 2 : 1. Column vectors can also be used in a proof. For example, vectors beginpmatrix k \ 3 endpmatrix and beginpmatrix 4 \ k + 1 endpmatrix are parallel when k(k + 1) = 4 × 3.
Method: 1. Write the route for each vector you need. 2. Simplify each vector fully and factorise it. 3. Compare the two vectors and state the multiple. 4. Write a conclusion that uses the words in the question, such as parallel, collinear or midpoint.
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