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

Concept: Vector Addition (Triangle Law) — displacements add as vectors; the resultant is the vector from the start to the final point.

Step 1: Represent each displacement as a vector.

Take east as $+x$, north as $+y$.

First displacement: $\vec{A} = 4$ km west $= (-4, 0)$ km.

Second displacement: $3$ km at $30^\circ$ east of north means $30^\circ$ from the north toward east.

Components:

$x$-component: $3 \sin 30^\circ = 3 \times 0.5 = 1.5$ km (east, so $+1.5$)

$y$-component: $3 \cos 30^\circ = 3 \times \frac{\sqrt{3}}{2} = \frac{3\sqrt{3}}{2}$ km (north, so $+ \frac{3\sqrt{3}}{2}$)

Thus $\vec{B} = \left(1.5,\ \frac{3\sqrt{3}}{2}\right)$ km.

Step 2: Add the vectors.

Resultant $\vec{R} = \vec{A} + \vec{B} = \left(-4 + 1.5,\ 0 + \frac{3\sqrt{3}}{2}\right) = \left(-2.5,\ \frac{3\sqrt{3}}{2}\right)$ km.

Step 3: Find magnitude and direction.

Magnitude: $|\vec{R}| = \sqrt{(-2.5)^2 + \left(\frac{3\sqrt{3}}{2}\right)^2} = \sqrt{6.25 + \frac{27}{4}} = \sqrt{6.25 + 6.75} = \sqrt{13} \approx 3.606$ km.

Direction: angle $\theta$ measured from the positive $x$-axis (east).

$\tan \theta = \frac{\frac{3\sqrt{3}}{2}}{-2.5} = -\frac{3\sqrt{3}}{5} \approx -1.0392$.

Since $x$ is negative and $y$ positive, the vector lies in the second quadrant.

$\theta = 180^\circ - \tan^{-1}(1.0392) \approx 180^\circ - 46.1^\circ = 133.9^\circ$ from east, i.e., $43.9^\circ$ west of north.

Take $\hat i$ pointing east and $\hat j$ pointing north.

Walk 1 (4 km west): $\vec{OP} = -4\,\hat i$.

Walk 2 (3 km, $30^\circ$ east of north): the unit direction is $\sin 30^\circ\,\hat i + \cos 30^\circ\,\hat j = \tfrac12\hat i + \tfrac{\sqrt3}{2}\hat j$, so

Displacement from the start:

Magnitude:

Method: Resultant Displacement by Resolving into Components

Use this for 'walks one way, then another' problems: represent each leg as a vector, add component-wise, then take the magnitude.

Steps

Step 1: Fix axes and resolve each leg

Choose $\hat i$ = east, $\hat j$ = north. Resolve each displacement into east and north parts. Mind the compass phrasing: '$30^\circ$ east of north' is measured from north towards east, so the north part uses $\cos30^\circ$ and the east part uses $\sin30^\circ$.

Step 2: Add the legs (triangle law)

The net displacement is the vector sum — the single arrow from start to finish:

adding the $\hat i$ parts together and the $\hat j$ parts together.

Step 3: Find magnitude (and direction if asked)

If a direction is needed, use $\tan\phi=R_y/R_x$ and fix the quadrant from the signs of $R_x,R_y$.

Common Mistakes

Mistake 1: Swapping sine and cosine for '$30^\circ$ east of north'

Why it's wrong: the angle is measured from the north axis, so north $=3\cos30^\circ$ and east $=3\sin30^\circ$; swapping mislabels the components. Correct approach: draw the direction first — the perpendicular (east) part gets $\sin$ of the given angle.

Mistake 2: Getting the sign of 'west' wrong

Why it's wrong: west is the negative $x$-direction, so 4 km west is $-4\hat i$, not $+4\hat i$. Correct approach: assign signs from your chosen axes before adding.

Mistake 3: Adding the distances ($4+3=7$) instead of the vectors

Why it's wrong: the legs are not collinear, so their magnitudes do not simply add. Correct approach: add as vectors and use $\sqrt{R_x^2+R_y^2}$, giving $\sqrt{13}$, not $7$.