Add vectors by the component method

Two perpendicular displacements: 9 m east and 12 m north. Enter the resultant's magnitude in m.

Two displacements point north of east, 50° apart. They are not perpendicular, so the Pythagorean theorem will not take them as they stand. You could draw them tip to tail and measure. There is a faster way, and it needs no ruler.

A hiker walks 40 m at 20° north of east, then 30 m at 70° north of east. How far from the start does the hiker end up, and in what direction?

The hiker's two legs, tip to tail. They are not perpendicular, so the Pythagorean theorem cannot take them directly.

First, resolve each vector into an east component and a north component, with the cosine and the sine.

First leg: east component 40 cos 20° = 37.59 m; north component 40 sin 20° = 13.68 m.

Second leg: east component 30 cos 70° = 10.26 m; north component 30 sin 70° = 28.19 m.

Next, add the east components together, and add the north components together. Call the two sums the east total and the north total.

East total: 37.59 m + 10.26 m = 47.85 m. North total: 13.68 m + 28.19 m = 41.87 m.

Each leg resolved, then the columns added: one east total and one north total.

The east total and the north total are perpendicular. That is the method's whole trick: any two vectors become one east piece and one north piece, and those two are always perpendicular.

Last, recombine the two totals. The Pythagorean theorem gives the magnitude, and the tangent gives the direction.

magnitude = √(47.85² + 41.87²) = 63.6 m

θ = tan⁻¹(41.87 ÷ 47.85) = 41.2° north of east

So the hiker ends up 63.6 m from the start, at 41.2° north of east.

The two totals are perpendicular, so the Pythagorean theorem and the tangent finish the job: 63.6 m at 41.2° north of east.

Resolve, add, recombine. Resolve each vector into east and north components, add the east components and the north components separately, then recombine the two totals with the Pythagorean theorem and the tangent. The method works for any two vectors, at any angles.

An orienteering runner jogs 60 m at 25° north of east, then 45 m at 65° north of east. Calculate the magnitude and direction of the runner's resultant displacement.

Write down the values in the question:

leg 1 = 60 m at 25° north of east

leg 2 = 45 m at 65° north of east

resultant = ?

Resolve each vector into components:

leg 1 east: 60 cos 25° = 54.38 m; leg 1 north: 60 sin 25° = 25.36 m

leg 2 east: 45 cos 65° = 19.02 m; leg 2 north: 45 sin 65° = 40.78 m

Add the east components, and add the north components:

east total: 54.38 m + 19.02 m = 73.40 m

north total: 25.36 m + 40.78 m = 66.14 m

Recombine with the Pythagorean theorem and the tangent:

magnitude = √(73.40² + 66.14²) = 98.8 m

θ = tan⁻¹(66.14 ÷ 73.40) = 42.0° north of east

The resultant displacement is 98.8 m at 42.0° north of east.

A rover drives 8.0 m at 40° north of east, then 6.0 m at 70° north of east. Enter the east total of its displacement components in m, to one decimal place.

A rover drives 8.0 m at 40° north of east, then 6.0 m at 70° north of east. Enter the north total of its displacement components in m, to one decimal place.

A displacement has an east component of 7.5 m and a north component of 4.0 m. Enter the displacement's magnitude in m, to one decimal place.

Your turn

A kayaker paddles 30 m at 10° north of east, then 20 m at 50° north of east. Enter the magnitude of the resultant displacement in m, to one decimal place.

A sailboat travels 18 m at 20° north of east, then 14 m at 60° north of east. Enter the direction of the resultant displacement, in degrees north of east, to one decimal place.

A trail runner covers 3.0 km at 15° north of east, then 2.0 km at 55° north of east. Enter the magnitude of the resultant displacement in km, to one decimal place.