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Chapter 10 · Vector Algebra

Miscellaneous · Q3: 4 km west, then 3 km 30° east of north: the girl's displacement

Miscellaneous ExerciseTriangle law, components, bearings · Sections 10.4 and 10.5.1
The question

A girl walks 4 km towards west, then she walks 3 km in a direction 30° east of north and stops. Determine the girl's displacement from her initial point of departure.

Exam answer

Let O be the starting point, and take i^\hat i and j^\hat j as unit vectors towards east and north. Let the girl walk from O to A and then from A to B.

OA is 4 km towards west, so

OA→=−4i^\overrightarrow{OA} = -4\hat i

AB is 3 km at 30° east of north, so its east part is 3sin⁡30∘3\sin 30^\circ and its north part is 3cos⁡30∘3\cos 30^\circ⁠:

AB→=3sin⁡30∘ i^+3cos⁡30∘ j^=32i^+332j^\begin{aligned} \overrightarrow{AB} &= 3\sin 30^\circ\,\hat i + 3\cos 30^\circ\,\hat j \\ &= \frac32\hat i + \frac{3\sqrt3}{2}\hat j \end{aligned}

By the triangle law of vector addition,

OB→=OA→+AB→=−4i^+32i^+332j^=−52i^+332j^\begin{aligned} \overrightarrow{OB} &= \overrightarrow{OA} + \overrightarrow{AB} \\ &= -4\hat i + \frac32\hat i + \frac{3\sqrt3}{2}\hat j \\ &= -\frac52\hat i + \frac{3\sqrt3}{2}\hat j \end{aligned} ∣OB→∣=254+274=13 km\begin{aligned} |\overrightarrow{OB}| &= \sqrt{\frac{25}{4} + \frac{27}{4}} \\ &= \sqrt{13}\ \text{km} \end{aligned}

∴\therefore the displacement is −52i^+332j^-\dfrac52\hat i + \dfrac{3\sqrt3}{2}\hat j⁠, of magnitude 13\sqrt{13} km.

Want the reason behind every step, the basics and the checks?

The one idea this problem is built on

Displacement is the arrow from where you start to where you stop, whatever path you took. Write each walk as a vector in components (east and north), and the triangle law says the displacement is simply their sum.

Given, to find, and the plan
Given
Walk 1: 4 km towards west. Walk 2: 3 km in the direction 30° east of north.
To find
The displacement from the starting point O to the stopping point B: as a vector, with its length and direction.
Plan
(1) Fix axes: i^\hat i east, j^\hat j north. (2) Write walk 1 as a vector. (3) Split walk 2 into east and north parts. (4) Add by the triangle law. (5) Find the length and the direction.

Basics you need

1Reading a bearing

"30° east of north" means: face north, then turn 30° towards east, and walk. Basics: "30° east of north": start, turn, go

2Splitting a length into two perpendicular parts

A walk of length LL at angle θ\theta from a direction has a part Lcos⁡θL\cos\theta along that direction and Lsin⁡θL\sin\theta across it. Basics: cos θ is a shadow, sin θ is a height

3Triangle law and adding by components

OA→+AB→=OB→\overrightarrow{OA} + \overrightarrow{AB} = \overrightarrow{OB}⁠; vectors add component by component (Sections 10.4 and 10.5.1). Lesson: adding arrows, head to tail

Solution

  1. Set up the axes

    Let O be the starting point. Take i^\hat i as the unit vector towards east and j^\hat j as the unit vector towards north, and measure in km. The girl walks from O to A, then from A to B.

    Any fixed pair of perpendicular directions would do; east and north match the way the question is worded, so both walks are easy to write down. West is then −i^-\hat i and south is −j^-\hat j⁠.
    Unit vectors along perpendicular axes (Section 10.5.1). Lesson: components

  2. The first walk

    OA→=−4i^.\overrightarrow{OA} = -4\hat i.
    Towards west is the direction of −i^-\hat i⁠, and the length is 4 km, so the vector is 4 times −i^-\hat i⁠. It has no north part.
    Multiplying a vector by a scalar (Section 10.5). Lesson: a vector is a length with a direction

  3. The second walk, split into east and north parts

    At A, face north and turn 30° towards east. The walk AB of 3 km makes 30° with north, so

    AB→=3sin⁡30∘ i^+3cos⁡30∘ j^=32i^+332j^.\begin{aligned} \overrightarrow{AB} &= 3\sin 30^\circ\,\hat i + 3\cos 30^\circ\,\hat j \\ &= \tfrac32\hat i + \tfrac{3\sqrt3}{2}\hat j. \end{aligned}
    The angle is measured from north, so the north part is the side next to the angle, 3cos⁡30∘=3323\cos 30^\circ = \frac{3\sqrt3}{2}⁠, and the east part is the side opposite, 3sin⁡30∘=323\sin 30^\circ = \frac32⁠. Both parts are positive because the direction lies between north and east.
    Sides of a right triangle from the hypotenuse and an angle. Basics: cos θ is a shadow, sin θ is a height

  4. Add the two walks

    OB→=OA→+AB→=(−4+32)i^+332j^=−52i^+332j^.\begin{aligned} \overrightarrow{OB} &= \overrightarrow{OA} + \overrightarrow{AB} \\ &= \left(-4 + \tfrac32\right)\hat i + \tfrac{3\sqrt3}{2}\hat j \\ &= -\tfrac52\hat i + \tfrac{3\sqrt3}{2}\hat j. \end{aligned}

    So OB→=−52i^+332j^\overrightarrow{OB} = -\frac52\hat i + \frac{3\sqrt3}{2}\hat j.

    N E O (start) A B (stop) 4 km west 3 km 30° √13 ≈ 3.61 km
    The walk drawn to scale (one scale for east and north): OA = 4 km west, AB = 3 km at 30° east of north, and the displacement OB→\overrightarrow{OB} of length 13≈3.61\sqrt{13} \approx 3.61 km.
    The triangle law: walking O to A and then A to B ends at the same place as walking straight from O to B. Adding by components: east parts −4+32=−52-4 + \frac32 = -\frac52⁠, north parts 0+3320 + \frac{3\sqrt3}{2}⁠.
    Triangle law of vector addition (Section 10.4); sum by components (Section 10.5.1, (i)). Lesson: adding arrows, head to tail

  5. How far, and in which direction

    ∣OB→∣=(−52)2+(332)2=254+274=13.\begin{aligned} |\overrightarrow{OB}| &= \sqrt{\left(-\tfrac52\right)^2 + \left(\tfrac{3\sqrt3}{2}\right)^2} \\ &= \sqrt{\tfrac{25}{4} + \tfrac{27}{4}} = \sqrt{13}. \end{aligned}

    The east part is negative and the north part is positive, so B is north-west of O. If ϕ\phi is the angle of OB from north towards west,

    tan⁡ϕ=5/233/2=533.\tan\phi = \frac{5/2}{3\sqrt3/2} = \frac{5}{3\sqrt3}.

    So the girl ends 13\sqrt{13} km from O, in the direction tan⁡−1533\tan^{-1}\frac{5}{3\sqrt3} west of north.

    (332)2=9×34=274\left(\frac{3\sqrt3}{2}\right)^2 = \frac{9 \times 3}{4} = \frac{27}{4}⁠, and 25+274=13\frac{25 + 27}{4} = 13⁠. The direction comes from the right triangle with legs 52\frac52 (west) and 332\frac{3\sqrt3}{2} (north): the leg opposite ϕ\phi over the leg next to it.
    ∣r⃗∣=x2+y2+z2|\vec r| = \sqrt{x^2 + y^2 + z^2} (Section 10.5.1). Basics: use Pythagoras twice Lesson: where did she end up?
Result

The displacement is OB→=−52i^+332j^\overrightarrow{OB} = -\frac52\hat i + \frac{3\sqrt3}{2}\hat j (with i^\hat i east and j^\hat j north): 13\sqrt{13} km, in the direction tan⁡−1533\tan^{-1}\frac{5}{3\sqrt3} west of north.

(Textbook answer: −52i^+332j^\frac{-5}{2}\hat i + \frac{3\sqrt3}{2}\hat j⁠.)

Check it with numbers

332≈2.598076\frac{3\sqrt3}{2} \approx 2.598076⁠, so B is about 2.5 km west and 2.598076 km north of O.

13≈3.605551\sqrt{13} \approx 3.605551 km, shorter than the 7 km walked, as a straight line must be.

ϕ=tan⁡−1533≈43.90∘\phi = \tan^{-1}\frac{5}{3\sqrt3} \approx 43.90^\circ west of north.

Law of cosines in triangle OAB as a second check: the angle at A between AO (east) and AB (30° east of north) is 60°, so OB2=42+32−2⋅4⋅3cos⁡60∘=16+9−12=13\text{OB}^2 = 4^2 + 3^2 - 2 \cdot 4 \cdot 3\cos 60^\circ = 16 + 9 - 12 = 13⁠.

Common mistakes
  1. Swapping sine and cosine. The angle is measured from north, so the north part uses cos⁡30∘\cos 30^\circ⁠. Using cos⁡\cos for the east part gives 332i^+32j^\frac{3\sqrt3}{2}\hat i + \frac32\hat j⁠, which is 30° north of east.
  2. Wrong sign for west. West is −i^-\hat i⁠; writing +4i^+4\hat i puts the girl on the wrong side.
  3. Adding the lengths. 4+3=74 + 3 = 7 km is the distance walked, not the displacement.
What to take away
  1. Displacement = sum of the separate displacements (triangle law), whatever the path.
  2. For a bearing "θ\theta east of north", the north part is Lcos⁡θL\cos\theta and the east part is Lsin⁡θL\sin\theta⁠.
  3. Give the answer as a vector, or as a length with a direction.

Every fact used, and where it comes from

FactUsed inSource
Unit vectors i^\hat i⁠, j^\hat j along perpendicular axes; component formSteps 1, 2Section 10.5.1 (Lesson: components)
Scalar multiple of a vectorStep 2Section 10.5
Reading a bearing; sin⁡30∘=12\sin 30^\circ = \frac12⁠, cos⁡30∘=32\cos 30^\circ = \frac{\sqrt3}{2}Step 3Trigonometry (Basics: "30° east of north": start, turn, go)
Triangle law of vector additionStep 4Section 10.4, Fig 10.7
Sum by components; ∣r⃗∣=x2+y2+z2|\vec r| = \sqrt{x^2 + y^2 + z^2}Steps 4, 5Section 10.5.1
The answerResultNCERT answer key, Miscellaneous Exercise on Chapter 10, Q3 (Lesson: put it all together)
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