Motion in a Plane · Lesson 4 of 13

Adding Vectors with Components

“Calculate resultants exactly and apply vector addition to rain, boats and repeated turns.”

Learning Objectives

• Add and subtract vectors component by component. • Derive resultant magnitude and direction for two vectors at any angle. • Apply vector addition with correct compass directions and units. • Analyse polygonal routes using repeated displacement vectors.

Replace a scale drawing with component arithmetic

A graphical construction explains what the resultant means, but a ruler and protractor limit its accuracy. Components turn the same construction into arithmetic. First choose common axes. Express every vector along those axes, including signs. Add x-components to x-components and y-components to y-components; then recover the resultant magnitude and direction. The same rules apply to subtraction and to more than two vectors.

Component addition and subtractionLaTeX

R = A + B : R_{x} = A_{x} + B_{x}, R_{y} = A_{y} + B_{y}, R_{z} = A_{z} + B_{z}

For A + B − C, subtract each component of C. All terms in a physical sum must have compatible units and meanings.

A three-vector calculation

Problem
Let A = 3 i-hat + 2 j-hat, B = −i-hat + 4 j-hat and C = 2 i-hat − j-hat, all in metres. Find T = A + B − C.

1.T_x = 3 − 1 − 2 = 0 m. T_y = 2 + 4 − (−1) = 7 m.
2.T = 7 j-hat m, so its magnitude is 7 m and it points along positive y.
3.The negative y-component of C becomes positive when C is subtracted. Reversing an entire vector means reversing every component.

Derive a formula for two vectors at an angle

Let A have magnitude A and lie along positive x. Let B have magnitude B and make an angle θ with A, where 0° ≤ θ ≤ 180°. The components of B are B cosθ and B sinθ. The resultant therefore has components A + B cosθ and B sinθ. Square and add these components, expand the square, and use cos²θ + sin²θ = 1.

Magnitude of a two-vector resultantLaTeX

R^{2} = (A + B cos θ)^{2} + (B sin θ)^{2} = A^{2} + B^{2} + 2 A B cos θ

θ is the angle between the original vectors drawn with a common tail, not the supplementary interior angle in a head-to-tail triangle.

Direction relative to ALaTeX

tan α = \frac{B sin θ}{A + B cos θ}

Use the signs of numerator and denominator to select α. If the denominator is zero, the resultant is perpendicular to A; if both components vanish, direction is undefined.

The sine rule offers another geometric route. For a nondegenerate vector triangle, R/sinθ = B/sinα = A/sinβ, where α is the angle between R and A and β is the angle between R and B. Here α + β = θ for 0° < θ < 180°. The triangle’s interior angle opposite R is 180° − θ, whose sine equals sinθ. Components avoid the ambiguous-angle choices that sometimes accompany inverse sine.

Sine-rule formLaTeX

\frac{R}{sin θ} = \frac{B}{sin α} = \frac{A}{sin β}

Use this only for a nondegenerate triangle and interpret the angles geometrically.

Rain and an umbrella

Wind gives falling rain a horizontal component

Problem
Rain has vertical downward speed 35 m/s. A wind gives it a westward component of 12 m/s. How should a stationary person orient an umbrella?

1.Choose east as +x and upward as +y. The rain velocity is −12 i-hat − 35 j-hat m/s.
2.The speed is √(12² + 35²) = 37 m/s. The fall direction makes arctan(12/35) = 18.9° with downward vertical, toward the west.
3.The rain arrives from the opposite direction: above and east. Tilt the umbrella’s upward-facing shaft about 19° toward the east, into the incoming rain.
4.This adds velocity components of the rain in one ground frame. If the person walks, first find rain velocity relative to that person.

Resultant velocity of wind-driven rain— The blue arrow is the rain velocity, directed down and west. A person faces the umbrella upward and east, opposite to this incoming velocity.

A boat and a water current

Northward boat in a southeast current

Problem
A boat moves north at 25 km/h relative to water. The current is 10 km/h at 60° east of south. Find the ground velocity.

1.Choose east as +x and north as +y. Boat velocity relative to water is (0,25) km/h.
2.The current has east component 10 sin60° = 8.660 km/h and north component −10 cos60° = −5 km/h.
3.Add components: ground velocity = (8.660,20) km/h. Its magnitude is √(75 + 400) = √475 = 21.79 km/h.
4.The direction is arctan(8.660/20) = 23.41° east of north. Equivalently, the angle between boat and current is 120°, giving the same magnitude through the cosine formula.

Repeated turns create a polygon

A route that turns left through 60° after each equal straight leg traces a regular hexagon. Choose the first leg along +x and sum displacements with headings 0°, 60°, 120°, 180°, 240° and 300°. After six equal legs the polygon closes. The physical interpretation is that each arrival at a corner counts as the next turn after the starting corner.

A 500 m leg followed by a 60° left turn

Problem
Find displacement after the third, sixth and eighth turns, using the first heading as +x.

1.After three legs, x = 500(1 + cos60° + cos120°) = 500 m and y = 500(sin60° + sin120°) = 500√3 m. Magnitude is 1000 m at 60° to the first heading; path length is 1500 m.
2.After six legs, all components cancel. Displacement is zero while distance is 3000 m.
3.After eight legs, the completed hexagon contributes zero, leaving two legs. Displacement = 750 i-hat + 250√3 j-hat m, with magnitude 500√3 ≈ 866 m at 30° to the first heading. Distance is 4000 m.
4.Saying “eight turns” without defining the initial counting event can be ambiguous. The counting convention used here is explicit.

Quick check

Two vectors of magnitudes 3 and 4 are perpendicular. What is the magnitude of their sum?

Key Takeaways

• Resolve every vector in one coordinate system before adding. • The two-vector magnitude formula uses the angle between common-tail arrows. • Recover direction with its quadrant. • An umbrella points into the incoming rain, opposite to the falling velocity. • Closed polygonal routes can have zero displacement and positive distance.

Previous · Lesson 3

Unit Vectors and Components

Next · Lesson 5

Position, Velocity and Acceleration in a Plane