Position Vector - Definition, Formula, and Examples

What Is A Position Vector?

A position vector is a vector that indicates the location of a point relative to a fixed reference point, the origin $O$. Its tail always sits at the origin and its head sits at the point; its length is the distance from the origin to the point, and its direction is the orientation of the point as seen from the origin.

A vector is a quantity with both magnitude (size) and direction, drawn as an arrow. What makes a position vector special is that its starting point is fixed at the origin - it is not free to float. That anchoring is what lets it name a location.

For a point $P$ with coordinates $(x, y)$ in the plane, the position vector of $P$ is written:

$$\vec{OP} = x\hat{i} + y\hat{j}$$

Here $\hat{i}$ is the unit vector (a vector of length 1) pointing along the positive x-axis, and $\hat{j}$ is the unit vector along the positive y-axis. So $x\hat{i} + y\hat{j}$ means "go $x$ units along x, then $y$ units along y." The position vector and the coordinates carry the same information in two notations.

What Is The Formula For A Position Vector In 3D?

In three dimensions, a point $P(x, y, z)$ has position vector:

$$\vec{OP} = x\hat{i} + y\hat{j} + z\hat{k}$$

where $\hat{k}$ is the unit vector along the positive z-axis. Each component tells you how far to travel along one axis. The 2D case is just this formula with $z = 0$.

Magnitude. The length of the position vector — the straight-line distance from the origin to the point — comes from the Pythagorean theorem extended to vectors:

$$|\vec{OP}| = \sqrt{x^2 + y^2 + z^2}$$

In 2D this is $|\vec{OP}| = \sqrt{x^2 + y^2}$. Each term under the root is a coordinate; the square root turns the components into one distance.

How Do You Find the Vector Between Two Points?

This is where position vectors do their most useful work. To find the vector that points from point $A$ to point $B$, subtract the position vector of the start from the position vector of the end:

$$\vec{AB} = \vec{OB} - \vec{OA}$$

If $A = (x_1, y_1)$ and $B = (x_2, y_2)$, then in components:

$$\vec{AB} = (x_2 - x_1)\hat{i} + (y_2 - y_1)\hat{j}$$

The rule is always "head minus tail" - the destination's position vector minus the starting point's. Get the order backwards and you get the vector pointing the wrong way.

Examples of the Position Vector

Each example is worked step by step. The problem statement is in bold; the working is not.

Example 1

Write the position vector of the point P(2, 5).

The point has $x = 2$, $y = 5$. Substitute into $\vec{OP} = x\hat{i} + y\hat{j}$:

$$\vec{OP} = 2\hat{i} + 5\hat{j}$$

Final answer: $\vec{OP} = 2\hat{i} + 5\hat{j}$.

Example 2

Find the vector from A(2, 3) to B(7, 1). A student writes AB = OA − OB. Is that right?

The tempting move is to subtract in reading order: $A$ first, so $\vec{OA} - \vec{OB}$.

Wrong path: $$\vec{OA} - \vec{OB} = (2 - 7)\hat{i} + (3 - 1)\hat{j} = -5\hat{i} + 2\hat{j}$$

Check the direction. Going from $A(2,3)$ to $B(7,1)$ moves right (x increases from 2 to 7) and down (y drops from 3 to 1). So the x-component should be positive and the y-component negative. The wrong answer has them exactly reversed - it describes the trip from $B$ to $A$, not $A$ to $B$.

The rule is head minus tail: the head is $B$, so $\vec{AB} = \vec{OB} - \vec{OA}$: $$\vec{AB} = (7 - 2)\hat{i} + (1 - 3)\hat{j} = 5\hat{i} - 2\hat{j}$$

Now the signs match the picture: right and down.

Final answer: $\vec{AB} = 5\hat{i} - 2\hat{j}$. Always subtract the tail from the head.

Example 3

Find the magnitude of the position vector of P(3, 4).

The position vector is $3\hat{i} + 4\hat{j}$. Its magnitude:

$$|\vec{OP}| = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5$$

Final answer: $|\vec{OP}| = 5$ units.

Example 4

Write the position vector of the point Q(1, −2, 4) in three dimensions, and find its magnitude.

Position vector, using all three components: $$\vec{OQ} = \hat{i} - 2\hat{j} + 4\hat{k}$$

Magnitude: $$|\vec{OQ}| = \sqrt{1^2 + (-2)^2 + 4^2} = \sqrt{1 + 4 + 16} = \sqrt{21} \approx 4.58$$

Final answer: $\vec{OQ} = \hat{i} - 2\hat{j} + 4\hat{k}$, magnitude $\sqrt{21}$ units.

Example 5

The position vectors of A and B are OA = 2î + 3ĵ and OB = 6î + 6ĵ. Find the length of AB.

First the vector $\vec{AB} = \vec{OB} - \vec{OA}$: $$\vec{AB} = (6 - 2)\hat{i} + (6 - 3)\hat{j} = 4\hat{i} + 3\hat{j}$$

Then its magnitude: $$|\vec{AB}| = \sqrt{4^2 + 3^2} = \sqrt{16 + 9} = \sqrt{25} = 5$$

Final answer: $|\vec{AB}| = 5$ units.

Example 6

The position vectors of points A and B are a = 2î + ĵ and b = 4î + 5ĵ. Find the position vector of the midpoint M of AB.

The midpoint's position vector is the average of the two endpoint position vectors: $$\vec{OM} = \frac{\vec{a} + \vec{b}}{2}$$ $$= \frac{(2\hat{i} + \hat{j}) + (4\hat{i} + 5\hat{j})}{2}$$ $$= \frac{6\hat{i} + 6\hat{j}}{2} = 3\hat{i} + 3\hat{j}$$

Final answer: $\vec{OM} = 3\hat{i} + 3\hat{j}$, so $M = (3, 3)$.

Why Position Vectors Matter - "Fixing where things are"

A position vector solves a problem so basic it is easy to overlook: how do you say exactly where something is? Coordinates alone are just a pair of numbers. Turning them into a vector - an arrow from a fixed origin - lets you do arithmetic with locations: subtract two positions to get the path between them, average two positions to find a midpoint, add a change to update a position.

That is why position vectors sit under so much applied work:

• Navigation and GPS - a satellite reports your location as a vector from Earth's centre; the difference between two such vectors is the route between two places.

• Physics - the position vector $\vec{r}(t)$ of a moving object, tracked over time, is the starting point for velocity and acceleration. Change in position over change in time is velocity; the whole of motion analysis begins here.

• Computer graphics and robotics - every vertex of a 3D model and every joint of a robot arm is stored as a position vector, then moved by adding other vectors.

The idea worth carrying forward: a position vector is tied to the origin, but the vector between two points is free — it depends only on the difference, not on where the origin sits. That distinction between "anchored" and "free" vectors is the seed of the whole subject of vectors.

What Are The Most Common Mistakes With Position Vectors?

Mistake 1: Subtracting in the wrong order

Where it slips in: Finding the vector from one point to another.

Don't do this: Writing $\vec{AB} = \vec{OA} - \vec{OB}$ (tail minus head).

The correct way: Always head minus tail: $\vec{AB} = \vec{OB} - \vec{OA}$. The first-instinct error is to subtract in the order the letters are read ($A$ then $B$), which produces the vector pointing from $B$ back to $A$ - correct magnitude, reversed direction. Match the sign of each component against the picture before trusting the answer.

Mistake 2: Confusing a position vector with a displacement vector

Where it slips in: Any problem that mixes "where a point is" with "how far something moved."

Don't do this: Treating the vector between two points as a position vector anchored at the origin.

The correct way: A position vector is anchored at the origin and names where a point is; a displacement vector is free and names how much the position changed. The point of confusion between these two close ideas is exactly this - both are arrows with the same components in some problems, but only the position vector must start at the origin. Ask which the question wants: a location, or a change.

Mistake 3: Forgetting the square root in the magnitude

Where it slips in: Computing the length of a position vector.

Don't do this: Reporting $x^2 + y^2$ as the magnitude.

The correct way: The magnitude is $\sqrt{x^2 + y^2}$ - the sum of squares is only halfway there. The rusher who stops at $9 + 16 = 25$ and calls the length $25$ has skipped the root; the real length is $\sqrt{25} = 5$. This is the vector version of a mistake with real stakes: an aircraft computing straight-line distance to a runway from its east–north offsets must take the square root of the sum of squares, and stopping early would put the reported distance wildly off - the same failure that unit and formula slips have caused in aviation navigation incidents.

Conclusion

• A position vector points from the origin to a point, written $\vec{OP} = x\hat{i} + y\hat{j}$ (or with $z\hat{k}$ in 3D).

• The vector from $A$ to $B$ is $\vec{AB} = \vec{OB} - \vec{OA}$ - head minus tail.

• Its magnitude is $\sqrt{x^2 + y^2}$ (2D) or $\sqrt{x^2 + y^2 + z^2}$ (3D).

• A position vector is anchored at the origin; a displacement vector is free.

• Position vectors let you do arithmetic with locations - subtract for paths, average for midpoints.

To take position vectors and vector geometry further with a teacher, explore Bhanzu's geometry tutor or high school math tutor programs, or browse math classes online.

A Practical Next Step

Practice these problems to solidify your understanding. Write the position vector of $P(-2, 6)$ and find its magnitude. Then, given $\vec{OA} = 3\hat{i} + \hat{j}$ and $\vec{OB} = -\hat{i} + 4\hat{j}$, find $\vec{AB}$ and its length. If the "head minus tail" order trips you up, return to the two-points section and sketch the arrow. Want a live Bhanzu trainer to walk through more vector problems? Book a free demo class.

Read More

• Addition of vectors

• Collinear vectors

• Multiplication of vectors

• Dot product of vectors

• What is a vector