One movement, then the next
The vector a goes 3 across and 1 up, so a = (3, 1). The vector b goes 2 across and 4 up, so b = (2, 4). Each one is a movement, and each can be made from any starting point.
To add two vectors, make one movement and then the other. Start at the origin and make the movement a: it ends at the point (3, 1). Then make the movement b, starting from there. Drawn this way, the start of the arrow b sits at the head of the arrow a, and the arrows are said to be joined tip to tail.
The arrow a runs from (0, 0) to (3, 1), and the arrow b starts at (3, 1), where a finished.
Where the journey ends
From (3, 1), the movement b goes 2 across and 4 up, so it ends at (3 + 2, 1 + 4) = (5, 5). The single arrow from the start of a to the end of b is the sum a + b. It is also called the resultant: one movement with the same effect as the two.
Read the sum off the components. Across, a gives 3 and b gives 2, and 3 + 2 = 5. Up, a gives 1 and b gives 4, and 1 + 4 = 5. So a + b = (3, 1) + (2, 4) = (5, 5).
This works for any two vectors, because a step across never changes how far up you are, and a step up never changes how far across. Each component is added on its own. In letters, (p, q) + (r, s) = (p + r, q + s).
The sum a + b runs straight from the start of a to the end of b: 3 + 2 = 5 across and 1 + 4 = 5 up.
b = (2, 4)
Make the resultant lie along the y-axis
The first arrow is a = (3, 1), and the second, b = (2, 4), starts at its tip. The gold resultant ends at (5, 5). Drag the head of b to any other point: the resultant always ends at 3 plus the step across of b, and 1 plus its step up.
The order does not matter
Make the movements the other way round. From the origin, b reaches (2, 4), and then a takes you on to (2 + 3, 4 + 1) = (5, 5), the same point. So b + a = a + b, for any two vectors: vector addition is commutative, because each component is an ordinary sum, and 3 + 2 = 2 + 3.
Drawn together, the two routes are the four sides of a parallelogram: a and then b along one pair of sides, b and then a along the other. The sum a + b is the diagonal from the shared start to the shared finish.
The parallelogram matters when two vectors act at the same time rather than one after the other, such as two ropes pulling on one object. Their combined effect is the single vector along the diagonal, which is still their sum.
The route a then b, and the route b then a, are the sides of a parallelogram. Both end at (5, 5), the head of a + b, the diagonal.
Negative components, and more than two vectors
A step to the left or down is negative, and it adds in the same way. (−4, −2) + (−2, 2) = (−4 + (−2), −2 + 2) = (−6, 0). The sum goes 6 to the left and has no step up at all, because 2 down and then 2 up cancel.
Three or more vectors are chained the same way, each starting where the one before it finished. (1, 4) + (3, −2) + (2, 1) = (1 + 3 + 2, 4 − 2 + 1) = (6, 3).
Components add; lengths do not
By Pythagoras, a = (3, 1) has length , which is 3.16 to 2 decimal places, and b = (2, 4) has length , which is 4.47. The sum (5, 5) has length , which is 7.07. That is less than 3.16 + 4.47 = 7.63.
The straight route from start to finish is shorter than the two legs, unless the two vectors point the same way. So to find the length of a sum, add the components first and then use Pythagoras on the result.
The usual mistakes
Counting the second arrow from the origin. In the chain, b starts where a finished, so the journey ends at a + b, not at b.
Stopping where the first arrow ends. The end of a is only halfway; b still has its own steps to make.
Subtracting the components. (3, 1) − (2, 4) = (1, −3) is the difference of the vectors, not their sum.
Multiplying the components. (3 × 2, 1 × 4) = (6, 4) is not a journey of a then b; the steps of a movement after a movement add.
Adding the lengths of the two vectors to get the length of the sum. Add the components, then find the length.
Two legs, and two pulls at once
In the first application below, each leg of a drone’s flight is written as a column vector, kilometers east on top and kilometers north underneath. Adding the columns, top with top and bottom with bottom, gives the drone’s displacement from its base, and its length by Pythagoras is the straight flight back.
In the second, two ropes pull on a crate at the same time, so the resultant is the diagonal of the parallelogram the two forces make. The crate is held still when all the forces on it add to the zero vector, so the force of a third rope must be the negative of that resultant: the same length, pointing the opposite way.
Worked example: A Delivery Drone's Two Legs and the Flight Back to Base
Question A delivery drone leaves its base O. Its first leg is a = 25 and its second leg is b = 43, in kilometers east and north. (a) Find the displacement of the drone from its base as a column vector. (b) The drone has enough charge left for 12 km. Can it fly straight back to base, and how many kilometers of charge will it have to spare?
1.Each leg is a column vector: the top number is kilometers east and the bottom number is kilometers north. The first leg goes 2 km east and 5 km north, and the second goes 4 km east and 3 km north.
Each leg is a column vector, kilometers east on top and north underneath: a = 25 and b = 43. 2.Add the components: a + b = 25 + 43 = 2 + 45 + 3 = 68.
Add the components: a + b = 2 + 45 + 3 = 68. On the grid it is the third side of the triangle, from the start of a to the end of b. 3.(a) The drone is 6 km east and 8 km north of its base, so its displacement is 68.
(a) The drone is 6 km east and 8 km north of its base: its displacement is 68. 4.The straight flight back is the magnitude of this vector: √62 + 82 = √36 + 64 = √100 = 10 km.
The straight flight back is the magnitude: √62 + 82 = √100 = 10 km. 5.(b) 10 km is less than 12 km, so the drone can fly straight back, with 12 − 10 = 2 km of charge to spare. Check: the two legs are √29 ≈ 5.4 km and 5 km, and the straight line back, 10 km, is shorter than the 10.4 km flown, as it must be.
(b) 10 km is less than 12 km, so the drone can fly straight back with 12 − 10 = 2 km of charge to spare.
Answer: (a) 68 km; (b) yes: the flight back is 10 km, leaving 2 km to spare
Common mistakes
- Adding the lengths of the two legs, 5.4 + 5 = 10.4 km, and calling it the distance from base. The legs point in different directions, so their lengths do not add; only their components do.
- Working out √6 + 8 instead of √62 + 82. The magnitude is the hypotenuse of a right-angled triangle with sides 6 and 8, so each component is squared first.
Worked example: Two Ropes Pulling a Crate, and the Third Rope That Holds It Still
Question Two workers pull a crate across a yard with ropes. With components east and north in newtons, the ropes pull with forces F = 4070 and G = 80−20. (a) Find the resultant force R on the crate as a column vector, and its magnitude. (b) A third worker holds the crate still with a single rope. What force must that rope exert?
1.Both forces act on the crate at once, so the resultant is their sum. On the diagram it is the diagonal of the parallelogram whose sides are the two ropes.
Both forces act on the crate at once: F = 4070 N and G = 80−20 N. Their resultant is the diagonal of the parallelogram with the two ropes as sides. 2.Add the components: R = 4070 + 80−20 = 40 + 8070 − 20 = 12050 N.
Add the components: R = 40 + 8070 − 20 = 12050 N. 3.(a) The magnitude is √1202 + 502 = √14400 + 2500 = √16900 = 130 N.
(a) The magnitude is √1202 + 502 = √16900 = 130 N. 4.The crate stays still when the three forces add to zero, so the third force is the negative of the resultant: −R = −120−50 N.
The crate stays still when the forces add to zero, so the third force is −R = −120−50 N. 5.(b) The third rope must pull with −120−50 N, a force of 130 N directly opposite the resultant. Check: 4070 + 80−20 + −120−50 = 00.
(b) The third rope pulls with −120−50 N, 130 N directly opposite the resultant, and the three forces add to 00.
Answer: (a) R = 12050 N, of magnitude 130 N; (b) −120−50 N
Common mistakes
- Adding the sizes of the two forces: √402 + 702 ≈ 80.6 N and √802 + 202 ≈ 82.5 N make about 163 N. The ropes pull in different directions, so part of each pull is canceled by the other.
- Giving the third force as 12050 N, the same as the resultant. That would double the pull; to hold the crate still the force must point the opposite way.