Position and displacement vectors

Stand in a field and try to tell a friend exactly where you are. "Near the tree" will not do — you need a fixed starting point and a rule for measuring from it. Physics solves this with the position vector: pick an origin $O$ , draw an arrow from $O$ straight to the particle, and that arrow — with its length and its direction — pins the particle down completely. Move the particle, and a second arrow, the displacement vector, tells you exactly how the motion changed its location, no matter how crooked the path in between. This sounds almost too simple to be a chapter of physics, and that is precisely its power. Once position is written as $\vec r = x\,\hat i + y\,\hat j$ , all of two-dimensional kinematics — projectile motion, circular motion, everything in "Motion in a Plane" — becomes bookkeeping with these two arrows and their derivatives. Everything else in the chapter is built on top of this scaffolding, so getting it exactly right now saves confusion for weeks. The one idea worth carrying out of this article is a distinction that trips up almost every student on their first test: displacement is not the same as distance travelled. A hiker who walks 8 m east and then 6 m north has covered 14 m of ground, yet a bird flying straight from her start to her end point would only need to cross 10 m. Both numbers are "correct" — they just answer different questions. This article builds the vector machinery to compute both, precisely, every time, and shows why the origin you choose changes where you say a particle is but never changes how far it has moved.

Key takeaways

The position vector $\vec r = x\,\hat i + y\,\hat j$ is the arrow from a chosen origin $O$ to the particle; it fixes both distance and direction from $O$ in one object.
The displacement vector $\Delta\vec r = \vec r_2 - \vec r_1$ is the arrow from the initial point to the final point — it depends only on start and end, never on the path taken.
Magnitude and direction come from the components: $|\Delta\vec r| = \sqrt{(\Delta x)^2+(\Delta y)^2}$ and $\theta = \tan^{-1}(\Delta y/\Delta x)$ .
Distance travelled is always greater than or equal to $|\Delta\vec r|$ ; equality holds only for straight-line motion that never reverses.
Shifting the origin changes every position vector in a problem but leaves the displacement vector between two fixed points completely unchanged, because the shift cancels in the subtraction $\vec r_2 - \vec r_1$ .
Position and displacement vectors are the entry point to the whole "Motion in a Plane" chapter — velocity and acceleration vectors are just their rates of change with time.
Key idea: a position vector locates a particle relative to a chosen origin as $\vec r = x\hat i + y\hat j$ , while the displacement vector $\Delta\vec r = \vec r_2 - \vec r_1$ measures only the net change between two points, independent of origin and of the actual path, and is never larger in magnitude than the distance actually travelled.

The core insight

A position vector fixes a particle relative to a chosen origin, but a displacement vector — the difference of two position vectors — depends only on the start and end points, not on the origin or the path, and its magnitude never exceeds the distance actually travelled.

Where it fits in the syllabus

This article sits at the start of the vector-foundations section of CBSE Class 11 Physics, chapter Motion in a Plane, right after scalars and vectors are introduced and just before equality of vectors and vector addition. The chapter's prescribed scope keeps this piece narrow and concrete: set up an origin and axes in a plane, write a particle's position as $\vec r = x\hat i + y\hat j$ , and build displacement as the difference of two position vectors — the natural two-dimensional extension of the one-dimensional "final minus initial" idea from the previous chapter, Motion in a Straight Line. Nothing here reaches beyond components in a flat plane; three-dimensional vector geometry and anything resembling relative-velocity river-boat problem chains are kept for later, lighter treatment.

The physical picture

Every position vector needs two ingredients fixed in advance: an origin $O$ , chosen once and held fixed for the whole problem, and a set of perpendicular axes ( $x$ and $y$ ) through it. With those fixed, the position vector of a particle sitting at the point $P(x, y)$ is simply the arrow drawn from $O$ to $P$ :

\vec r = x\,\hat i + y\,\hat j

where: $x$ and $y$ are the coordinates of the particle measured along the two axes, and $\hat i, \hat j$ are unit vectors — arrows of length exactly 1, with no units of their own — pointing along the positive $x$ - and $y$ -directions respectively. Multiplying $\hat i$ by the number $x$ stretches (or, if $x$ is negative, reverses) it into a vector of length $|x|$ along the $x$ -axis; adding that to $y\,\hat j$ gives the full arrow to $P$ . The length of this arrow is its magnitude,

r = |\vec r| = \sqrt{x^2+y^2}

read straight off the right triangle formed by the arrow and its two components. Because the position vector is anchored to $O$ , moving the origin to a new point instantly changes every $x$ and every $y$ in the problem, and therefore changes $\vec r$ for every particle — position is only ever "position relative to this origin," never absolute. Displacement measures motion, not location. If a particle moves from an initial point $P_1$ , with position vector $\vec r_1$ , to a final point $P_2$ , with position vector $\vec r_2$ , its displacement is defined as the vector difference

\Delta\vec r = \vec r_2 - \vec r_1 = (x_2-x_1)\hat i + (y_2-y_1)\hat j

where: the subtraction is done component by component — $x$ -components subtract from $x$ -components, $y$ from $y$ — and the result is itself a vector, with a magnitude $|\Delta\vec r| = \sqrt{(\Delta x)^2+(\Delta y)^2}$ and a direction angle $\theta = \tan^{-1}(\Delta y/\Delta x)$ measured from the $x$ -axis. The physical reading is that $\Delta\vec r$ is the single straight arrow you would need to draw to jump directly from where the particle started to where it ended up — regardless of how twisted, looping, or backtracking the actual journey was. This is exactly why displacement is origin-independent: if the origin is shifted by some fixed vector $\vec s$ , both $\vec r_1$ and $\vec r_2$ shift by that same $\vec s$ , and in the subtraction $\vec r_2-\vec r_1$ the shift cancels perfectly, leaving $\Delta\vec r$ untouched.

Figure

Two dots on a grid labelled start and end joined by a straight dashed displacement arrow, with a curved solid path between them representing the actual distance travelled, and position vectors drawn from a marked origin to each dot

Derivation ABuilding the displacement vector from two position vectors

The goal here is to show, step by step, exactly why the displacement between two points can be written as the simple component-wise subtraction $\Delta\vec r = (x_2-x_1)\hat i + (y_2-y_1)\hat j$ , and, in the same argument, why that result cannot depend on where the origin was placed in the first place — a fact that is easy to state but genuinely worth proving rather than just accepting, because it explains why physicists are comfortable choosing the origin wherever is most convenient for a given problem without worrying that the answer for "how far did it move" will change.

Step 1. Fix an origin $O$ and axes, and write the position vectors of the two points using components: $\vec r_1 = x_1\hat i + y_1\hat j$ for the initial point $P_1$ , and $\vec r_2 = x_2\hat i + y_2\hat j$ for the final point $P_2$ . Both are arrows starting at the same origin $O$ , so they can be legitimately combined by ordinary vector algebra.

Step 2. Displacement is defined as the vector that, added to $\vec r_1$ , gives $\vec r_2$ — the arrow that carries the particle from where it was to where it now is. Algebraically this means $\vec r_1 + \Delta\vec r = \vec r_2$ , so rearranging gives $\Delta\vec r = \vec r_2 - \vec r_1$ .

Step 3. Substitute the components from Step 1 and subtract like components (unit vectors of the same axis combine, different axes never mix): $\Delta\vec r = (x_2\hat i+y_2\hat j) - (x_1\hat i+y_1\hat j) = (x_2-x_1)\hat i + (y_2-y_1)\hat j$ . Writing $\Delta x = x_2-x_1$ and $\Delta y = y_2-y_1$ gives the compact form $\Delta\vec r = \Delta x\,\hat i + \Delta y\,\hat j$ , with magnitude $|\Delta\vec r| = \sqrt{(\Delta x)^2+(\Delta y)^2}$ .

Step 4. Now shift the origin to a new point $O'$ , displaced from $O$ by a fixed vector $\vec s$ . Every position vector measured from $O'$ loses that same $\vec s$ : the new position vectors are $\vec r_1' = \vec r_1 - \vec s$ and $\vec r_2' = \vec r_2 - \vec s$ . Recomputing the displacement in the new frame, $\vec r_2' - \vec r_1' = (\vec r_2-\vec s) - (\vec r_1-\vec s) = \vec r_2 - \vec r_1 = \Delta\vec r$ — the $\vec s$ terms cancel exactly, so the displacement vector is identical no matter which origin was used. This is the formal reason displacement is treated as a more physically fundamental quantity than position whenever only the motion, not the absolute location, matters.

Worked examples

Board level - locating and displacing a particle

A particle is at point $A(2,3)$ m at time $t=0$ and moves to point $B(6,6)$ m at time $t=4$ s. Find (a) the position vectors $\vec r_A$ and $\vec r_B$ , (b) the displacement vector, and (c) its magnitude and direction.

Step 1. Write the position vectors directly from the given coordinates: $\vec r_A = 2\hat i+3\hat j$ m and $\vec r_B = 6\hat i+6\hat j$ m.

Step 2. Subtract component-wise to get the displacement: $\Delta\vec r = \vec r_B-\vec r_A = (6-2)\hat i+(6-3)\hat j = 4\hat i+3\hat j$ m.

Step 3. Find the magnitude using the Pythagorean theorem on the components: $|\Delta\vec r| = \sqrt{4^2+3^2} = \sqrt{16+9} = \sqrt{25} = 5$ m.

Step 4. Find the direction from $\tan\theta = \Delta y/\Delta x = 3/4$ , so $\theta = \tan^{-1}(3/4) \approx 36.9^\circ$ above the $x$ -axis.

The displacement is 5 m, directed at about 36.9° above the x-axis, regardless of any curve the particle's actual path might have taken between A and B.

Competitive level - distance versus displacement on a two-leg journey

A cyclist starts at the origin, rides 8 m east, then 6 m north, arriving at point $C$ . Find the distance travelled and the magnitude of the displacement, compare the two, and find the direction of the displacement from the starting direction of travel (east).

Step 1. Take east as $+\hat i$ and north as $+\hat j$ . After the first leg the cyclist is at $(8,0)$ ; after the second leg, at $(8,6)$ , so $\vec r_C = 8\hat i+6\hat j$ m measured from the origin.

Step 2. Distance travelled is the sum of the actual path lengths covered, not a vector calculation: $8\ \text{m} + 6\ \text{m} = 14$ m.

Step 3. Displacement is the position vector of $C$ minus the position vector of the start (the origin, $0\hat i+0\hat j$ ): $\Delta\vec r = 8\hat i+6\hat j$ m, so $|\Delta\vec r| = \sqrt{8^2+6^2} = \sqrt{64+36} = \sqrt{100} = 10$ m.

Step 4. Direction from east: $\tan\theta = 6/8 = 0.75$ , so $\theta = \tan^{-1}(0.75) \approx 36.9^\circ$ north of east — the cyclist could have reached $C$ directly with a single straight ride of 10 m at that bearing instead of the two-leg 14 m journey.

Distance travelled is 14 m; displacement magnitude is 10 m, directed about 36.9° north of east — distance exceeds displacement because the path bends through a right angle instead of running straight. Timing note: in an exam, spot immediately whether a question asks for "distance" (add the leg lengths as plain numbers) or "displacement" (subtract position vectors, then use Pythagoras) — this single read saves a wrong-formula error and costs under 20 seconds.

Common traps

Confusing the position vector (anchored to a chosen origin, changes if the origin changes) with the displacement vector (anchored to nothing but the two points themselves, unaffected by origin choice).
Writing displacement as simply the final position vector $\vec r_2$ and forgetting to subtract the initial position vector $\vec r_1$ — a very common one-mark loss.
Assuming displacement magnitude equals distance travelled just because the described path "sounds short" — the two are equal only for straight-line motion that never reverses direction; any turn or curve makes distance strictly greater.
Adding the $x$ - and $y$ -leg lengths of a journey as ordinary numbers to get the displacement magnitude, instead of combining perpendicular components with the Pythagorean theorem.
Forgetting units on components — a position vector like $2\hat i+3\hat j$ is meaningless without stating whether it is in metres, kilometres, or something else.

You should be able to…

Write the position vector of a point in terms of $\hat i,\hat j$ given its coordinates, and find its magnitude.
Compute the displacement vector between two positions as $\vec r_2-\vec r_1$ , working component by component.
Find the magnitude and direction of a displacement vector from its $\Delta x,\Delta y$ components.
Explain why a shift of origin changes position vectors but leaves the displacement vector between two fixed points unchanged.
Distinguish distance travelled from displacement magnitude in a multi-leg journey, and state the inequality relating them.
Decide quickly, from the wording of a problem, whether it is asking for distance or for displacement.

Ask the tutor: Want to practice setting up position vectors for a multi-leg journey with a moved origin, or check whether your displacement calculation for a curved path is right — ask the AI tutor to walk through it with you.

Why do we even need an origin? Can't a particle's position just be a number?

In one dimension a single signed number is enough because there is only one line to move along. In a plane, one number cannot capture both how far a particle is and in which direction, so we need an arrow — the position vector — built from two components. That arrow only makes sense once you have fixed a starting point (the origin) to draw it from, because "3 m from here" means nothing without knowing where "here" is. Different students solving the same problem can choose different origins and still get the same physics, as long as they are each consistent throughout their own solution.

If I choose a different origin, do I get a different displacement for the same motion?

No, and this is one of the most useful facts in the whole topic. Position vectors do change when you move the origin, because every coordinate shifts by the same amount. But displacement is a difference of two position vectors, and that shift cancels out exactly in the subtraction, as shown in the derivation above. So you are always free to place the origin wherever makes the arithmetic easiest — often at the particle's starting point, which makes $\vec r_1$ zero — without changing any displacement answer.

Is displacement the same thing as velocity?

No. Displacement is $\Delta\vec r$ , a change in position, measured in metres. Velocity is how fast that change happens, $\vec v = \Delta\vec r/\Delta t$ for average velocity over a time interval $\Delta t$ , measured in metres per second. Displacement only needs a start and an end point; velocity additionally needs to know how much time elapsed between them. You will meet average and instantaneous velocity vectors properly later in this chapter, but they are always built on top of the displacement vector defined here.

My answer for displacement magnitude came out larger than the distance I calculated. Did I make a mistake?

Yes, and it is worth re-checking the components. Distance travelled along a path is always greater than or equal to the straight-line displacement magnitude between its endpoints — this is essentially the triangle inequality applied to motion, since a straight line is the shortest path between two points. If your displacement magnitude comes out bigger than the total distance, look for an arithmetic slip in the subtraction of coordinates or in applying the Pythagorean theorem, because that situation cannot happen physically.