Subtracting Vectors

Subtracting is adding the negative

Take a = (5, 4) and b = (2, 1). The negative of b is −b = (−2, −1): the same arrow turned to point the opposite way, so both of its components change sign.

Subtracting b means adding its negative: a − b = a + (−b). Make the movement a, and then the movement b in reverse. In components, (5, 4) − (2, 1) = (5, 4) + (−2, −1) = (5 − 2, 4 − 1) = (3, 3).

So each component subtracts on its own, the step across from the step across and the step up from the step up. In letters, (p, q) − (r, s) = (p − r, q − s).

The movement a = (5, 4), and then −b = (−2, −1) from its head: 2 to the left and 1 down. The journey ends at (3, 3), the head of a − b.

The first arrow is (3, 1), and the second is set to (−1, −2), the negative of (1, 2). The gold resultant (3, 1) + (−1, −2) ends at (2, −1), which is (3, 1) − (1, 2). Drag the head of the second arrow: the resultant is always (3, 1) plus that arrow, so setting it to the negative of a vector subtracts that vector.

Drawn from one start

Now draw a = (5, 4) and b = (2, 1) both from the origin. The arrow a − b = (3, 3) runs from the head of b, at (2, 1), to the head of a, at (5, 4): from (2, 1), 3 across and 3 up reaches (5, 4).

The reason is the sum b + (a − b) = a. Going along b and then along a − b ends exactly where a ends. In components, (2, 1) + (3, 3) = (5, 4).

That sum is also the way to check a subtraction: add the answer back to the vector that was taken away, and the first vector should return. For (3, 1) − (1, 2) = (2, −1), the check is (2, −1) + (1, 2) = (3, 1).

a and b drawn from the origin. The arrow from the head of b to the head of a is a − b = (3, 3).

The other way round

b − a = (2 − 5, 1 − 4) = (−3, −3). That is the arrow from the head of a back to the head of b, and it is the negative of a − b = (3, 3).

So the order matters: a − b and b − a have the same length and point in opposite directions. Vector subtraction, like the subtraction of numbers, is not commutative.

Negative components

Subtracting a negative component adds. (−1, 3) − (2, −4) = (−1 − 2, 3 − (−4)) = (−3, 7). Up, taking away 4 down is the same as going 4 up, so 3 becomes 7.

The vector between two points

The arrow from the origin to a point has the coordinates of the point as its components, so the subtraction above also gives the vector between two points. Take B at (2, 1) and A at (5, 4). The vector from B to A is the end minus the start, A minus B: (5 − 2, 4 − 1) = (3, 3).

That is destination minus start, one coordinate at a time. The vector from A back to B is start minus destination, (−3, −3), which points the other way.

The vector from B(2, 1) to A(5, 4) is (5 − 2, 4 − 1) = (3, 3): destination minus start.

The usual mistakes

Adding when the question subtracts. (5, 4) + (2, 1) = (7, 5) chains the two movements; a − b is the gap between them.

Subtracting the wrong way round. b − a = (−3, −3) is the arrow from the head of a back to the head of b, the negative of a − b.

Changing the sign of only one component of b. The negative of (2, 1) is (−2, −1); both signs change.

Losing a sign when a component is negative. 3 − (−4) = 3 + 4 = 7, not −1.

A robot and its charger

In the application below, a robot’s moves are written with i and j: i stands for one meter east and j for one meter north, so 4i + j means 4 m east and 1 m north, the vector (4, 1). Moves written this way add and subtract like any vectors, the i parts with the i parts and the j parts with the j parts.

Once the robot’s position is known, the move that takes it to the charger is the charger’s position minus the robot’s: destination minus start. The length of that move by Pythagoras is the distance it travels.