Q.On an open ground, a motorist follows a track that turns to his left by an angle of $60^\circ$ after every $500\ \text{m}$. Starting from a given turn, specify the displacement of the motorist at the third, sixth and eighth turn. Compare the magnitude of the displacement with the total path length covered by the motorist in each case.

The key idea is Displacement Magnitude. Displacement is the shortest straight-line distance from the initial to the final position, a vector quantity. Path length is the total distance covered along the actual trajectory, a scalar quantity.

Essential Reasoning:

1. The motorist's path consists of segments of length $L = 500\ \text{m}$. A $60^\circ$ left turn after each segment means the internal angle between consecutive path segments is $180^\circ - 60^\circ = 120^\circ$. This forms a regular hexagonal path.
2. At the third turn: The motorist covers $3L = 1500\ \text{m}$. The displacement from the starting point to the end of the third segment is the main diagonal of the hexagon, which is $2L$. Thus, displacement magnitude is $2 \times 500\ \text{m} = 1000\ \text{m}$. This is less than the path length.
3. At the sixth turn: The motorist covers $6L = 3000\ \text{m}$. After six $60^\circ$ turns, the motorist completes a full $360^\circ$ rotation and returns to the starting point. Thus, displacement magnitude is $0\ \text{m}$. This is less than the path length. …

The problem asks us to determine the displacement of a motorist at specific turns and compare its magnitude with the total path length covered. Displacement is a vector quantity representing the shortest distance from the starting point to the final point, while path length is a scalar quantity representing the total distance traveled along the actual path.

The key to solving this problem lies in understanding the geometry of the motorist's path.

Concept and Intuition: The Hexagonal Path

1. Understanding the Turns: The motorist travels $500\ \text{m}$ and then turns left by $60^\circ$. This sequence repeats. If we consider the path segments as sides of a polygon, a $60^\circ$ left turn means the exterior angle of the polygon is $60^\circ$.
2. Identifying the Polygon: For a regular polygon, the sum of exterior angles is $360^\circ$. If each exterior angle is $60^\circ$, the number of sides ($n$) is $360^\circ / 60^\circ = 6$. This means the motorist is tracing the sides of a regular hexagon.
3. Side Length: Each segment of $500\ \text{m}$ is a side of this hexagon. Let $a = 500\ \text{m}$.
4. Displacement vs. Path Length:
   • Path Length: This is straightforward. If the motorist completes $N$ segments, the total path length is $N \times a$.
   • Displacement: This requires vector addition. We need to find the resultant vector from the starting point to the position after $N$ segments.

Let's denote the starting point as $P_0$. The first segment takes the motorist to $P_1$, the second to $P_2$, and so on. The "third turn" means the motorist has completed three segments and is at point $P_3$.

We will use a coordinate system for clarity. Let the starting point $P_0$ be the origin $(0,0)$. Let the first segment be along the positive x-axis.

1. Define Segment Vectors:

   Let $a = 500\ \text{m}$.

   The first segment, $\vec{s_1}$, is along the x-axis:

   $\vec{s_1} = (a, 0)$

   After the first segment, the motorist turns $60^\circ$ to the left. So, the second segment, $\vec{s_2}$, makes an angle of $60^\circ$ with the positive x-axis:

   $\vec{s_2} = (a \cos 60^\circ, a \sin 60^\circ) = (a/2, a\sqrt{3}/2)$

   After the second segment, the motorist turns another $60^\circ$ to the left. So, the third segment, $\vec{s_3}$, makes an angle of $60^\circ + 60^\circ = 120^\circ$ with the positive x-axis:

   $\vec{s_3} = (a \cos 120^\circ, a \sin 120^\circ) = (-a/2, a\sqrt{3}/2)$

   Continuing this pattern:

   $\vec{s_4} = (a \cos 180^\circ, a \sin 180^\circ) = (-a, 0)$

   $\vec{s_5} = (a \cos 240^\circ, a \sin 240^\circ) = (-a/2, -a\sqrt{3}/2)$

   $\vec{s_6} = (a \cos 300^\circ, a \sin 300^\circ) = (a/2, -a\sqrt{3}/2)$

   Notice that the sum of these six vectors is $\vec{s_1} + \vec{s_2} + \vec{s_3} + \vec{s_4} + \vec{s_5} + \vec{s_6} = (0,0)$, which confirms that after 6 segments, the motorist returns to the starting point.

2. Displacement and Path Length at the Third Turn:

   The motorist has completed three segments. The displacement $\vec{D_3}$ is the vector sum of the first three segments:

   $\vec{D_3} = \vec{s_1} + \vec{s_2} + \vec{s_3}$

   $\vec{D_3} = (a, 0) + (a/2, a\sqrt{3}/2) + (-a/2, a\sqrt{3}/2)$

   $\vec{D_3} = (a + a/2 - a/2, 0 + a\sqrt{3}/2 + a\sqrt{3}/2)$

   $\vec{D_3} = (a, a\sqrt{3})$

   The magnitude of the displacement is:

   $|\vec{D_3}| = \sqrt{a^2 + (a\sqrt{3})^2} = \sqrt{a^2 + 3a^2} = \sqrt{4a^2} = 2a$

   Substituting $a = 500\ \text{m}$:

   $|\vec{D_3}| = 2 \times 500\ \text{m} = 1000\ \text{m}$

   The direction of $\vec{D_3}$ is given by $\theta_3 = \arctan\left(\frac{a\sqrt{3}}{a}\right) = \arctan(\sqrt{3}) = 60^\circ$ with respect to the initial segment.

   The total path length covered is the sum of the lengths of the three segments:

   Path length$_3 = 3 \times a = 3 \times 500\ \text{m} = 1500\ \text{m}$

   Comparison: The magnitude of displacement ($1000\ \text{m}$) is less than the total path length ($1500\ \text{m}$).

3. Displacement and Path Length at the Sixth Turn:

   The motorist has completed six segments, which means one full cycle of the regular hexagon.

   The displacement $\vec{D_6}$ is the vector sum of all six segments:

   $\vec{D_6} = \vec{s_1} + \vec{s_2} + \vec{s_3} + \vec{s_4} + \vec{s_5} + \vec{s_6}$

   As established earlier, this sum is $(0,0)$.

   $|\vec{D_6}| = 0\ \text{m}$

   The total path length covered is:

   Path length$_6 = 6 \times a = 6 \times 500\ \text{m} = 3000\ \text{m}$

   Comparison: The magnitude of displacement ($0\ \text{m}$) is significantly less than the total path length ($3000\ \text{m}$).

4. Displacement and Path Length at the Eighth Turn:

   The motorist has completed eight segments. This is equivalent to completing one full hexagon (6 segments) and then two more segments.

   So, $\vec{D_8} = \vec{D_6} + \vec{s_7} + \vec{s_8}$.

   Since $\vec{D_6} = (0,0)$, the displacement is simply the sum of the 7th and 8th segments.

   The 7th segment, $\vec{s_7}$, will have the same direction as $\vec{s_1}$ (angle $0^\circ$). …

Concept: Segments as Complex-Number Rotations — a Geometric Series

Method: Represent Each Leg as $ae^{ik\pi/3}$ and Sum a Finite Geometric Series (No Explicit $\hat i,\hat j$ Components, No Law of Cosines)

Both existing solutions work with the segments individually — one lists all six $(\hat i,\hat j)$ components by hand, the other uses the Law of Cosines on a triangle for the diagonal. This method instead represents every $500\ \text{m}$ leg as a single complex number rotated by a fixed angle from the previous one, and sums the whole run of legs at once using the closed-form formula for a finite geometric series — turning "add up several vectors" into "evaluate one algebraic expression."

Setting up the complex representation

Let $a = 500\ \text{m}$ and let the $k$-th segment (for $k=0,1,2,\dots$) be the complex number

since each successive segment is rotated $60^\circ = \pi/3\ \text{rad}$ to the left of the one before it. The displacement after $n$ segments is the complex sum

(the standard finite geometric-series formula, with common ratio $e^{i\pi/3}$).

Displacement at the 3rd turn ($n=3$)

The denominator's magnitude is a standard chord-length identity, $\left|e^{i\theta}-1\right| = 2\sin(\theta/2)$, so $\left|e^{i\pi/3}-1\right| = 2\sin(30^\circ) = 1$. Hence:

Displacement at the 6th turn ($n=6$)

Here $e^{i\pi/3}$ is a primitive 6th root of unity, and it is a standard algebraic fact that the sum of all $n$-th roots of unity is zero:

So the displacement vanishes exactly because six $60^\circ$ rotations complete a full $360^\circ$ turn — the six unit vectors are symmetrically spaced around the circle and cancel.

Displacement at the 8th turn ($n=8$)

Since $D_6=0$, the sum restarts fresh from the 7th segment:

$D_8 = D_6 + s_6 + s_7 = 0 + a\,e^{i2\pi} + a\,e^{i7\pi/3} = a(1) + a\,e^{i\pi/3}$ …