The Unit Vectors i and j

One step across and one step up

Two vectors are given names of their own. The vector i = (1, 0) is one step in the direction of the x-axis, and the vector j = (0, 1) is one step in the direction of the y-axis. Each has length 1, so each is a unit vector. Like other vectors, they are set in bold type in print and underlined by hand.

i is one step along the x-axis and j is one step along the y-axis, both from the origin.

Any vector, written with i and j

The vector (3, 4) is 3 steps across and 4 steps up: 3 steps of i and 4 steps of j. In symbols, 3i + 4j = 3(1, 0) + 4(0, 1) = (3, 0) + (0, 4) = (3, 4). So (3, 4) and 3i + 4j are the same vector, written two ways.

The number in front of i is the x-component and the number in front of j is the y-component. A negative component gives a minus sign: (2, −1) = 2i − j. A component of 0 leaves its letter out: (−5, 0) = −5i, and (0, 7) = 7j. A component of 1 is not written: j means 1j.

3i runs 3 steps along the x-axis, and 4j runs 4 steps up from its head. Together they make 3i + 4j, the vector (3, 4).

Adding, subtracting and scaling

Vectors in i and j are added like terms in algebra: the i parts together and the j parts together. (3i + 4j) + (2i − j) = (3 + 2)i + (4 − 1)j = 5i + 3j. In pairs, that is (3, 4) + (2, −1) = (5, 3), the same sum.

The i parts and the j parts never combine with each other, because i and j point in different directions. 5i + 3j cannot be written as 8 of anything.

Subtraction and scalars work the same way. (3i + 4j) − (2i − j) = (3 − 2)i + (4 − (−1))j = i + 5j, and 2(3i + 4j) = 6i + 8j.

From the head of 3i + 4j, the arrow 2i − j goes 2 across and 1 down. The journey ends at (5, 3), the head of the sum 5i + 3j.

The first arrow is (3, 1), which is 3i + j, and the second is set to 2i + 3j. The resultant ends at (5, 4), which is 5i + 4j: 3 + 2 = 5 steps of i and 1 + 3 = 4 steps of j. Drag the head of the second arrow: its i part always adds to 3, and its j part to 1.

Length and direction in i and j

The magnitude comes from the two numbers in front of i and j, by Pythagoras. $|3i+4j|=\sqrt{3^2+4^2}=\sqrt{25}=5$|3i + 4j| = √(3² + 4²) = √25 = 5, and $|3i-4j|=\sqrt{3^2+{\left(-4\right)}^2}=\sqrt{25}=5$|3i − 4j| = √(3² + (−4)²) = √25 = 5 as well.

Dividing by the length gives the unit vector in the same direction: $\left(3i+4j\right)/5=0.6i+0.8j$(3i + 4j) / 5 = 0.6i + 0.8j.

A different i

The letter i has another meaning in mathematics: in complex numbers, i is the imaginary unit, the number whose square is −1. The unit vector i is a different thing, a step of length 1 along the x-axis. Some books write the unit vector as î, with a hat, to keep the two apart. The context tells you which is meant.

The usual mistakes

Giving i the y-component. i goes with the x-component: (3, 4) is 3i + 4j, not 4i + 3j.

Adding the two parts into one number. 3i + 4j is not 7: the steps across and the steps up point different ways.

Crossing the letters when adding. In (3i + 4j) + (2i − j), the 3i adds to the 2i, never to the −j.

Losing the sign in front of j. 2i − j has a y-component of −1, one step down.

A robot’s moves

In the application below, i is one meter east and j is one meter north. A robot’s two moves, given in i and j, add to its position from the dock. The move still to make is the charger’s position minus the robot’s, and its length by Pythagoras is the distance to travel.