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This page sets out the method for vectors 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 Vectors topic page.
When you see vectors being added, subtracted or multiplied by a number.
When a shape has moved on a grid, or you must move a point by a given vector.
Add the vectors (3, 1) and (2, 5). The sum is (5, 6).
Multiply the vector (3, 4) by 2. Both components double, so the result is (6, 8).
Subtract the vector (1, 5) from (7, 2). Take 7 − 1 and 2 − 5, which is (6, -3).
The vector from (1, 2) to (4, 6) is (3, 4). Subtract the start from the end.
Subtraction is not symmetric. From A to B is the reverse of from B to A.
Never add an x-component to a y-component.
Draw the arrow from its start. For beginpmatrix 3 \ -2 endpmatrix, go 3 squares right and 2 squares down. The order matters. beginpmatrix -2 \ 3 endpmatrix is a different move: 2 left and 3 up. The zero vector beginpmatrix 0 \ 0 endpmatrix stays where it is. Two vectors are equal only when both components match.
On a grid, the arrow can start anywhere. The vector records the move, not the starting point. An arrow from (1, 2) to (4, 0) is 3 right and 2 down, so it is still beginpmatrix 3 \ -2 endpmatrix. Parallel arrows of the same length and direction are the same vector, even if they sit in different places on the page.
It helps to say the vector out loud before you calculate. "3 right, 2 down" is harder to muddle than a pair of numbers on their own. If a diagram gives the arrow the other way, both signs flip. The return journey of beginpmatrix 3 \ -2 endpmatrix is beginpmatrix -3 \ 2 endpmatrix. Same length, opposite direction. That return journey is what you get when you multiply the vector by -1.
Method: 1. Read the top number as across and the bottom number as up. 2. Vector beginpmatrix3\\-2endpmatrix is 3 right and 2 down. 3. The return journey is beginpmatrix-3\\2endpmatrix.
To add, place the second arrow on the end of the first, or just add the components. beginpmatrix 3 \ -2 endpmatrix + beginpmatrix 1 \ 5 endpmatrix = beginpmatrix 4 \ 3 endpmatrix. Check the arithmetic in each row on its own: 3 + 1 = 4 and -2 + 5 = 3. The result is 4 right and 3 up.
Subtracting mathbfq from mathbfp means adding the reverse of mathbfq. If mathbfp = beginpmatrix 4 \ 1 endpmatrix and mathbfq = beginpmatrix -2 \ 5 endpmatrix, then mathbfp - mathbfq = beginpmatrix 4 - (-2) \ 1 - 5 endpmatrix = beginpmatrix 6 \ -4 endpmatrix. Subtracting a negative is the same as adding. The y-component 1 - 5 is -4, so the move is 6 right and 4 down. Addition can be done in either order. Subtraction cannot.
You can check a sum by drawing. Start at the origin, follow the first vector, then follow the second from where you landed. The single arrow from the origin to the finish is the sum. For beginpmatrix 3 \ -2 endpmatrix + beginpmatrix 1 \ 5 endpmatrix, that is 3 right and 2 down, then 1 right and 5 up, which finishes 4 right and 3 up. If the drawing and the column disagree, the components were combined in the wrong order or a sign was dropped.
Method: 1. Add the top numbers together, then add the bottom numbers together. 2. beginpmatrix3\\-2endpmatrix+beginpmatrix1\\5endpmatrix=beginpmatrix4\\3endpmatrix. 3. Subtract by reversing the second vector: beginpmatrix4\\1endpmatrix-beginpmatrix-2\\5endpmatrix=beginpmatrix6\\-4endpmatrix.
A scalar changes the length and may change the sense of the arrow. 3 beginpmatrix 2 \ -4 endpmatrix = beginpmatrix 6 \ -12 endpmatrix. Both components are trebled. The arrow is three times as long and still points down to the right, because the x-component stays positive and the y-component stays negative.
Multiplying by -1 reverses the arrow: -beginpmatrix 6 \ -4 endpmatrix = beginpmatrix -6 \ 4 endpmatrix. Multiplying by 1/2 halves both components. Do not multiply the top number and add the bottom, or square them. A scalar is one multiplication, applied twice, once to each component. If the scalar is 0, both components become 0.
Method: 1. Multiply both components by the same number. 2. 3beginpmatrix2\\-4endpmatrix=beginpmatrix6\\-12endpmatrix. 3. Multiplying by -1 reverses the arrow, and multiplying by 0 gives beginpmatrix0\\0endpmatrix.
The magnitude is the straight-line length. For beginpmatrix 6 \ 8 endpmatrix, work out sqrt6^2 + 8^2 = √(36 + 64) = √(100) = 10. A 3, 4, 5 triangle scaled by 2 is a 6, 8, 10 triangle, so this length is exact. Leave a surd when the squares do not make a square number (Higher). The magnitude of beginpmatrix 2 \ 6 endpmatrix is √(4 + 36) = √(40) = 2√(10).
If A is (1, 2) and B is (7, 10), the position vector of A is mathbfa = beginpmatrix 1 \ 2 endpmatrix and the position vector of B is mathbfb = beginpmatrix 7 \ 10 endpmatrix. Then overrightarrowAB = mathbfb - mathbfa = beginpmatrix 6 \ 8 endpmatrix, and overrightarrowBA is the reverse, beginpmatrix -6 \ -8 endpmatrix. The midpoint averages the coordinates: ((1 + 7)/2, (2 + 10)/2) = (4, 6). The vector from A to the midpoint is half of overrightarrowAB.
Method: 1. Find the length with sqrtx^2+y^2. 2. beginpmatrix6\\8endpmatrix has magnitude √(36+64)=10. 3. From (1,2) to (7,10), the vector is beginpmatrix6\\8endpmatrix and the midpoint is (4,6).
A column vector beginpmatrix x \ y endpmatrix tells you how far to move right (or left if negative) and how far to move up (or down if negative). In exam questions it appears as a translation of a shape, or as a journey in a context such as a drone that flies 6 km east and 2 km north. The top number is always the horizontal move.
To find the vector from a point P to a point Q, subtract: take the coordinates of Q and subtract the coordinates of P. The vector in the opposite direction, from Q back to P, has every sign changed. Applying the same translation twice means adding the vector to itself, which is the same as multiplying it by 2.
Method: 1. Write the coordinates of the start point and the end point. 2. Subtract: end - start, one coordinate at a time, to get the column vector. 3. For the reverse journey, change the sign of both numbers. 4. To translate a point, add the vector to its coordinates.
Tip: The vector from B to A is the negative of the vector from A to B.
Exam questions give column vectors mathbfa and mathbfb and ask for combinations such as 2mathbfa + mathbfb or 3mathbfp - 2mathbfq. Multiply each vector by its scalar first, then add or subtract the top numbers and the bottom numbers separately. Write the column of each multiple on its own line so you can see each step.
Some questions work backwards. If mathbfa + mathbfc = beginpmatrix 6 \ 1 endpmatrix, then mathbfc = beginpmatrix 6 \ 1 endpmatrix - mathbfa. Take care with subtracting a negative number, which becomes an addition. Always finish by putting the answer back in the original equation to check that it works. Taking a minute to check the signs often catches the single slip that costs both marks.
Method: 1. Multiply each vector by its scalar: multiply both the top and the bottom number. 2. Add or subtract the tops, then add or subtract the bottoms. 3. Watch the signs, especially subtracting a negative. 4. Check by substituting back or by working the top and bottom numbers a second time.
In a Higher question the vectors are given as letters, not numbers. For example overrightarrowOA = mathbfa and overrightarrowOB = mathbfb, and you must write another vector in terms of mathbfa and mathbfb. Find a route along the lines of the diagram from the start to the end of the vector you want. Practise on a few different diagrams, because the choice of route is the main skill.
Travelling along an arrow adds that vector. Travelling against an arrow subtracts it. So overrightarrowAB = -mathbfa + mathbfb = mathbfb - mathbfa. A midpoint is half-way along, so it is half of that vector. A point that divides a line in the ratio 2 : 1 is 2/3 of the way along. Simplify your answer by collecting the mathbfa terms and the mathbfb terms.
Method: 1. Choose a route from the start point to the end point that uses known vectors. 2. Add a vector when you travel with its arrow and subtract it when you travel against it. 3. For a midpoint or a ratio, use the right fraction of the whole vector. 4. Collect like terms and write the answer in its simplest form.
Two vectors are parallel when one is a multiple of the other. For example beginpmatrix 6 \ 3 endpmatrix and beginpmatrix 4 \ 2 endpmatrix are parallel because the first is 1.5 times the second. A negative multiple means the vectors point in opposite directions, but they are still parallel. Parallel vectors can also be written as a common factor times the same vector, for example 2mathbfa + 4mathbfb and mathbfa + 2mathbfb.
To show that three points A, B and C lie on a straight line, find overrightarrowAB and overrightarrowBC, show that one is a multiple of the other, and say that they share the point B. Parallel alone is not enough, because parallel lines can be apart. The common point is what makes the line a single straight line.
Method: 1. Find the two vectors, such as overrightarrowAB and overrightarrowBC. 2. Show that one is a multiple of the other by dividing the tops and the bottoms. 3. Note the common point they share. 4. Write a conclusion: the vectors are parallel and share a point, so the points are on a straight line.
Tip: In a "show that" question, write the conclusion in words as well as the working.
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