In short

A vector is a quantity that has both magnitude (size) and direction. Displacement, velocity, and force are vectors. A scalar has only magnitude — temperature, mass, speed. Any vector in a plane can be written as \vec{A} = A_x\hat{i} + A_y\hat{j}, where A_x and A_y are its components along the x and y axes. The magnitude is |\vec{A}| = \sqrt{A_x^2 + A_y^2}.

You are standing at India Gate in the centre of Delhi. A friend asks you: "How far is Connaught Place from here?" You answer: "About 3 kilometres." That is useful — it tells your friend the distance. But now suppose your friend is a delivery driver who needs to actually get there. "3 kilometres" is no longer enough. They need to know: 3 kilometres in which direction? North-west, as it happens. The distance alone is a number. The distance plus the direction is something fundamentally different — it is a vector.

This distinction — between quantities that need only a number and quantities that need a number and a direction — runs through all of physics. Temperature is 38°C. That is a complete statement; temperature has no direction. But the wind is blowing at 20 km/h from the south. Drop the "from the south" and you have lost half the information. The wind speed alone does not tell you whether your kite will fly north or get pushed into your neighbour's terrace.

Physics is full of both kinds. This article is about learning to tell them apart, and about the notation that makes vectors precise enough to compute with.

Scalars — when a number is enough

Some physical quantities are completely described by a single number (with a unit). These are called scalars.

Scalar

A scalar is a physical quantity that has magnitude only — no direction. It is fully specified by a number and a unit.

Examples you already know:

Quantity Example value Direction?
Temperature 42°C in Delhi, May No — temperature has no direction
Mass 160 g (a cricket ball) No — a cricket ball's mass is the same in every direction
Time 3.5 seconds No
Speed 140 km/h No — speed is how fast, not which way
Energy 500 joules No
Distance 3 km No — distance is how far, not which way

The key word is magnitude — the "how much" of a quantity. A scalar is all magnitude, no direction. When you add two scalars, ordinary arithmetic works: 2 kg + 3 kg = 5 kg. Simple.

Vectors — when direction changes everything

Now consider force. You push a cricket ball with 5 N of force. Is that a complete description? Not at all. A 5 N push to the right sends the ball rolling right. A 5 N push upward sends it into the air. The same magnitude, but a different direction, produces a completely different physical outcome. Force is not a scalar — it is a vector.

Vector

A vector is a physical quantity that has both magnitude (how much) and direction (which way). It is not fully described by a number alone.

Here are the quantities that need direction:

Quantity Example Why direction matters
Displacement 3 km north-west Tells you where you end up, not just how far
Velocity 140 km/h due east A fast bowler's delivery: speed and direction
Acceleration 9.8 m/s² downward Gravity pulls down, not sideways
Force 50 N at 37° above horizontal Push direction changes the motion
Momentum 6.2 kg·m/s north Carries the direction of the velocity

Notice the pair: distance is a scalar, but displacement is a vector. Speed is a scalar, but velocity is a vector. The scalar version answers "how much?" The vector version answers "how much, and in which direction?"

Here is a concrete example of why this matters. Suppose you walk 4 km east through a Delhi market, then turn and walk 3 km north. The total distance you walked is 7 km — scalar addition, simple arithmetic. But your displacement — how far you actually are from where you started, and in which direction — is not 7 km. It is 5 km, pointing north-east. The displacement depends on direction, and adding directed quantities is not the same as adding plain numbers.

Representing vectors — arrows with purpose

How do you draw a vector? With an arrow. The arrow has two features that carry the two pieces of information:

  1. The length of the arrow represents the magnitude.
  2. The direction the arrow points represents the direction.
A vector represented as an arrow An arrow pointing to the upper right, labeled with its magnitude and direction. The tail is marked as the starting point and the head (arrowhead) as the tip. Tail Head (tip) Magnitude = length of arrow θ (direction)
A vector is drawn as an arrow. The length encodes how big the quantity is; the direction the arrow points encodes which way it acts. The angle θ specifies the direction relative to a reference line (here, the horizontal).

Notation — how to write a vector on paper

When you write a vector, you need to distinguish it from a scalar. Physics uses several conventions:

So \vec{v} is a velocity vector (has direction), and v = |\vec{v}| is the speed (just the magnitude, a scalar).

Unit vectors — direction without magnitude

A unit vector is a vector whose magnitude is exactly 1. It carries pure direction information and no magnitude information. You write it with a "hat" (a small caret on top):

\hat{A} = \frac{\vec{A}}{|\vec{A}|}

Why: dividing any vector by its own magnitude strips out the "how big" part and leaves only the "which way" part. The result has magnitude 1.

If \vec{A} points north-east with a magnitude of 50 N, then \hat{A} still points north-east but has magnitude 1 (dimensionless). It is a pure direction marker.

Position vectors, displacement vectors, and free vectors

Not all vectors behave the same way. There are three important categories to understand early.

A position vector points from the origin to a specific point. If you set up a coordinate system with your school as the origin, then the position vector of your home tells you exactly where your home is — both distance and direction from the school. A position vector is tied to the origin; move the origin, and the vector changes.

A displacement vector tells you how to get from one point to another. If you walk from your school gate to the chai stall across the road — 50 m due south — that is a displacement of 50 m south. The displacement does not depend on where the origin is. It depends only on where you started and where you ended.

A free vector is one that can be moved anywhere in space without changing its meaning, as long as you keep its magnitude and direction the same. Force and velocity are free vectors — a 10 N force pushing east is the same physical force whether it acts at point A or point B (assuming the same body).

The key insight: two arrows that have the same length and point in the same direction represent the same free vector, regardless of where they are drawn.

Free vector: same vector drawn at three positions Three arrows of equal length, all pointing to the upper right at the same angle, drawn at different positions in the plane. All three represent the same displacement vector. →A →A →A Same length, same direction → same vector
All three arrows have the same length and the same direction. As free vectors, they are identical — the starting position does not matter. A displacement of "120 m north-east" is the same displacement whether you start from India Gate, Lodhi Garden, or Hauz Khas.

The unit vectors \hat{i}, \hat{j}, \hat{k} — a coordinate language for vectors

Saying "50 N at 37° above the horizontal" is fine for everyday speech, but physics needs something more systematic. You need a way to break any vector into a standard set of directions that you can compute with. That is where the standard unit vectors come in.

Set up a coordinate system — a pair of perpendicular axes. The x-axis points to the right, the y-axis points upward. Now define:

Each has magnitude 1. Each points along one axis and no other.

Unit vectors i-hat, j-hat, and k-hat along the coordinate axes A right-handed coordinate system showing three unit vectors: i-hat along the positive x-axis, j-hat along the positive y-axis, and k-hat along the positive z-axis (drawn at an angle to suggest depth). î x ĵ y k̂ z O magnitude = 1
The three standard unit vectors: $\hat{i}$ along the x-axis (red), $\hat{j}$ along the y-axis (dark), and $\hat{k}$ along the z-axis (grey, shown at an angle to suggest depth). Each has magnitude 1. Together they form a right-handed coordinate system.

Components — breaking a vector into \hat{i} and \hat{j} pieces

Here is the most powerful idea in this article. Any vector in a 2D plane can be written as a sum of two pieces — one along the x-axis and one along the y-axis.

Take a vector \vec{A} that makes an angle \theta with the positive x-axis. Drop a perpendicular from the tip of \vec{A} to the x-axis. You get a right triangle. The horizontal side is the x-component of \vec{A}, and the vertical side is the y-component.

Vector components: decomposing a vector into x and y parts A vector A makes angle theta with the x-axis. A right triangle is formed by the vector, its x-component along the horizontal axis, and its y-component along the vertical axis. The components are labeled Ax and Ay. x y →A Ax = |→A| cos θ Ay = |→A| sin θ θ
The vector $\vec{A}$ decomposes into a horizontal component $A_x$ along the x-axis and a vertical component $A_y$ along the y-axis. The angle $\theta$ is measured from the positive x-axis. The three form a right triangle — the vector is the hypotenuse.

From the right triangle, basic trigonometry gives you the components:

Step 1. The adjacent side of the triangle is the x-component.

A_x = |\vec{A}|\cos\theta \tag{1}

Why: in a right triangle, \cos\theta = \text{adjacent}/\text{hypotenuse}. The adjacent side is A_x and the hypotenuse is |\vec{A}|. Multiply both sides by |\vec{A}|.

Step 2. The opposite side is the y-component.

A_y = |\vec{A}|\sin\theta \tag{2}

Why: \sin\theta = \text{opposite}/\text{hypotenuse}. The opposite side is A_y.

Step 3. Write the vector using the unit vectors.

\vec{A} = A_x\hat{i} + A_y\hat{j} \tag{3}

Why: the x-component is a number times the unit vector in the x-direction, and the y-component is a number times the unit vector in the y-direction. Adding them reconstructs the original vector. This is the component form — the standard way to write any 2D vector.

This is the central result. Once you have the components, the vector is just two numbers — one for each axis. All the geometry of arrows and angles gets converted into algebra with numbers, and algebra is something you can compute with.

Magnitude from components — Pythagoras does the work

You now know how to go from magnitude and angle to components. What about the reverse? If someone hands you A_x = 4 and A_y = 3, what is |\vec{A}|?

Look at the right triangle again. The vector \vec{A} is the hypotenuse. The two components are the legs. The Pythagorean theorem gives:

Step 1. Apply Pythagoras to the component triangle.

|\vec{A}|^2 = A_x^2 + A_y^2 \tag{4}

Why: in any right triangle, the square of the hypotenuse equals the sum of the squares of the other two sides. The hypotenuse here is |\vec{A}|, and the sides are A_x and A_y.

Step 2. Take the positive square root.

|\vec{A}| = \sqrt{A_x^2 + A_y^2} \tag{5}

Why: magnitude is always a positive number (or zero). You take the positive root because a length cannot be negative.

Step 3. Find the direction angle.

\theta = \tan^{-1}\!\left(\frac{A_y}{A_x}\right) \tag{6}

Why: from the right triangle, \tan\theta = \text{opposite}/\text{adjacent} = A_y/A_x. Inverting gives the angle. Be careful with the quadrant — \tan^{-1} by itself gives angles only in the first and fourth quadrants, so check the signs of A_x and A_y to determine the correct quadrant.

These three equations — (1), (2), and (5) — let you convert freely between the two representations of a vector:

Both carry the same information. The component form is almost always easier to compute with.

Worked examples

Example 1: Walking through a Delhi neighbourhood

A student walks 400 m due east from their house to a bookshop, then turns and walks 300 m due north to a chai stall. Find the magnitude and direction of the student's displacement from their house to the chai stall.

Displacement vector from walking 400 m east then 300 m north A right-angled path: 400 m east (horizontal arrow) then 300 m north (vertical arrow). The displacement vector is the diagonal from start to finish, forming a 3-4-5 right triangle. The angle theta is shown at the starting point. East (x) North (y) 400 m east 300 m north Displacement = ? θ House Chai stall
The student's path forms a right triangle. The two legs are the 400 m east walk and the 300 m north walk. The displacement — the straight-line distance from start to finish — is the hypotenuse.

Step 1. Identify the components.

The student walked 400 m east and 300 m north. Taking east as the positive x-direction and north as the positive y-direction:

A_x = 400 m, \quad A_y = 300 m

Why: the eastward walk is entirely along the x-axis, so it contributes only to A_x. The northward walk is entirely along the y-axis, so it contributes only to A_y. No trigonometry needed here — the walks are already aligned with the axes.

Step 2. Compute the magnitude using Pythagoras.

|\vec{A}| = \sqrt{A_x^2 + A_y^2} = \sqrt{400^2 + 300^2}
= \sqrt{160000 + 90000} = \sqrt{250000} = 500 \text{ m}

Why: this is a 3-4-5 right triangle scaled by 100. The sides 300 and 400 give a hypotenuse of 500. The student is 500 m from their starting point in a straight line — even though they walked 700 m in total.

Step 3. Compute the direction.

\theta = \tan^{-1}\!\left(\frac{A_y}{A_x}\right) = \tan^{-1}\!\left(\frac{300}{400}\right) = \tan^{-1}(0.75) \approx 36.87° \approx 37°

Why: the angle is measured from the east (x) axis toward north (y). Both components are positive, so the angle is in the first quadrant — north of east.

Result: The displacement is 500 m at 37° north of east. In component form: \vec{A} = 400\hat{i} + 300\hat{j} m.

What this shows: The distance walked (700 m) and the displacement (500 m) are very different numbers. Distance is a scalar — it just adds up. Displacement is a vector — it cares about direction, and the 400 m east and 300 m north partially "cancel" each other's contribution to the total straight-line distance.

Example 2: Force on a cricket ball at an angle

A fielder throws a cricket ball with a force of 50 N at an angle of 37° above the horizontal. Find the horizontal and vertical components of this force.

Decomposing a 50 N force into horizontal and vertical components A force vector of 50 N at 37 degrees above horizontal, decomposed into a horizontal component of 40 N and a vertical component of 30 N, forming a right triangle. x (horizontal) y (vertical) →F = 50 N Fx = 50 cos 37° = 40 N Fy = 50 sin 37° = 30 N 37° ball
The 50 N force at 37° decomposes into a 40 N horizontal push and a 30 N vertical lift. The angle 37° is a favourite in physics problems because $\cos 37° \approx 0.8$ and $\sin 37° \approx 0.6$, giving clean numbers.

Step 1. Write down the knowns.

|\vec{F}| = 50 N, \quad \theta = 37° above the horizontal.

Step 2. Compute the horizontal component.

F_x = |\vec{F}|\cos\theta = 50 \times \cos 37°

Using \cos 37° \approx 0.8:

F_x = 50 \times 0.8 = 40 \text{ N}

Why: the horizontal component is the "shadow" of the force vector on the x-axis. The cosine projects the full magnitude onto the adjacent (horizontal) direction.

Step 3. Compute the vertical component.

F_y = |\vec{F}|\sin\theta = 50 \times \sin 37°

Using \sin 37° \approx 0.6:

F_y = 50 \times 0.6 = 30 \text{ N}

Why: the vertical component is the "shadow" on the y-axis. The sine projects the magnitude onto the opposite (vertical) direction.

Step 4. Write the force in component form.

\vec{F} = 40\hat{i} + 30\hat{j} \text{ N}

Step 5. Verify: does the magnitude check out?

\sqrt{F_x^2 + F_y^2} = \sqrt{40^2 + 30^2} = \sqrt{1600 + 900} = \sqrt{2500} = 50 \text{ N} \checkmark

Why: the components must reconstruct the original magnitude via Pythagoras. This is a quick sanity check — if it fails, you made an error in the decomposition.

Result: F_x = 40 N (horizontal), F_y = 30 N (vertical). The force in component form is \vec{F} = 40\hat{i} + 30\hat{j} N.

What this shows: Decomposition converts a single force at an angle into two perpendicular forces along the axes. This is the first step in almost every force problem in physics — break every force into components, then work with the components separately. The 37° angle appears constantly in Indian physics problems because it produces the clean ratio 3:4:5.

Common confusions

Going deeper: three dimensions and direction cosines

If you are comfortable with 2D vectors and want to see the full 3D picture, read on. Otherwise, you have everything you need for most of class 11 physics.

Vectors in three dimensions

Everything you learned in 2D extends naturally. A vector in 3D has three components:

\vec{A} = A_x\hat{i} + A_y\hat{j} + A_z\hat{k}

The magnitude uses all three:

|\vec{A}| = \sqrt{A_x^2 + A_y^2 + A_z^2}

This is the 3D version of Pythagoras. First, A_x and A_y give a diagonal in the x-y plane: \sqrt{A_x^2 + A_y^2}. Then that diagonal and A_z form another right triangle, giving the full magnitude as \sqrt{(\sqrt{A_x^2 + A_y^2})^2 + A_z^2} = \sqrt{A_x^2 + A_y^2 + A_z^2}.

For example, consider an ISRO rocket that, at a particular instant, has a velocity of \vec{v} = 200\hat{i} + 300\hat{j} + 600\hat{k} m/s (where \hat{k} points vertically upward). The speed is:

|\vec{v}| = \sqrt{200^2 + 300^2 + 600^2} = \sqrt{40000 + 90000 + 360000} = \sqrt{490000} = 700 \text{ m/s}

Direction cosines

In 2D, one angle \theta specifies the direction of a vector. In 3D, you need more information. The standard approach is direction cosines — the cosines of the angles the vector makes with each axis.

Define three angles:

  • \alpha = angle between \vec{A} and the x-axis
  • \beta = angle between \vec{A} and the y-axis
  • \gamma = angle between \vec{A} and the z-axis

The direction cosines are:

\cos\alpha = \frac{A_x}{|\vec{A}|}, \qquad \cos\beta = \frac{A_y}{|\vec{A}|}, \qquad \cos\gamma = \frac{A_z}{|\vec{A}|}

These three numbers satisfy a remarkable identity:

\cos^2\alpha + \cos^2\beta + \cos^2\gamma = 1

Why: substitute the definitions. You get \frac{A_x^2 + A_y^2 + A_z^2}{|\vec{A}|^2}. But the numerator is |\vec{A}|^2 by the magnitude formula. So the fraction equals 1.

This identity is not just a curiosity — it is a constraint. You cannot choose the three direction cosines independently. If two of them are known, the third is determined (up to a sign). A vector's direction in 3D has two degrees of freedom, not three.

For the ISRO rocket example above, the direction cosines are:

\cos\alpha = \frac{200}{700} = \frac{2}{7}, \qquad \cos\beta = \frac{300}{700} = \frac{3}{7}, \qquad \cos\gamma = \frac{600}{700} = \frac{6}{7}

Check: \left(\frac{2}{7}\right)^2 + \left(\frac{3}{7}\right)^2 + \left(\frac{6}{7}\right)^2 = \frac{4 + 9 + 36}{49} = \frac{49}{49} = 1. It works.

The unit vector in the direction of \vec{A} can be written using direction cosines:

\hat{A} = \cos\alpha\,\hat{i} + \cos\beta\,\hat{j} + \cos\gamma\,\hat{k}

This is just the definition of \hat{A} = \vec{A}/|\vec{A}| with each component divided by the magnitude. Direction cosines appear frequently in JEE Advanced problems on 3D geometry and electrostatics.

Where this leads next