Vector Addition and Subtraction

In short

To add vectors, place them head-to-tail and draw the resultant from the first tail to the last head (triangle law). Equivalently, add their components: R_x = A_x + B_x, R_y = A_y + B_y, then |\vec{R}| = \sqrt{R_x^2 + R_y^2} and \theta = \tan^{-1}(R_y/R_x). Vector subtraction is just adding the negative: \vec{A} - \vec{B} = \vec{A} + (-\vec{B}). The null vector \vec{0} has zero magnitude and no defined direction.

You are standing at India Gate in New Delhi, and a friend gives you directions: walk 3 km east, then turn and walk 4 km north. How far are you from where you started?

If you add the numbers the way you add rupees — 3 + 4 = 7 — you get 7 km. But pull out a map and measure the straight-line distance from India Gate to where you end up: it is 5 km, not 7. The missing 2 km is the price you pay for ignoring direction.

This is why vectors exist. A vector carries both a magnitude and a direction, and adding two vectors is fundamentally different from adding two numbers. The rules for vector addition are what this article is about. These rules show up every time you combine forces, add velocities, or track displacements — which is to say, they show up everywhere in physics.

The triangle law — head-to-tail addition

The simplest way to add two vectors is to draw them head-to-tail.

Take two vectors \vec{A} and \vec{B}. Draw \vec{A} first. Then place the tail of \vec{B} at the head of \vec{A}. The resultant \vec{R} = \vec{A} + \vec{B} is the arrow from the tail of \vec{A} to the head of \vec{B}.

Triangle law: place the tail of $\vec{B}$ at the head of $\vec{A}$. The resultant $\vec{R}$ runs from where you started (O) to where you ended up (Q).

Why does this work? Think of it physically. If you walk displacement \vec{A} and then walk displacement \vec{B}, your net displacement is the straight line from your starting point to your final position. That straight line is the resultant. The triangle that forms — with \vec{A}, \vec{B}, and \vec{R} as sides — gives this method its name.

Why "head-to-tail": the head of the first vector is where the tail of the next one begins. This is the physical sequence of doing one displacement after another.

Notice that the order does not matter: \vec{A} + \vec{B} = \vec{B} + \vec{A}. If you walked 4 km north first and then 3 km east, you would end up at the same point. Vector addition is commutative.

The parallelogram law — same result, different picture

There is another way to see the same addition. Instead of placing \vec{B} at the head of \vec{A}, draw both vectors from the same starting point. Then complete the parallelogram — the diagonal from the common tail is the resultant.

Parallelogram law: both vectors start from O. Complete the parallelogram — the diagonal from O is the resultant. The angle $\alpha$ between $\vec{A}$ and $\vec{B}$ determines the magnitude of $\vec{R}$.

The triangle law and the parallelogram law are not two different methods — they are two different drawings of the same thing. In the parallelogram, the upper-right triangle is exactly the head-to-tail triangle you drew before, just embedded in a parallelogram.

The magnitude formula from the parallelogram law

The parallelogram gives you a formula for the magnitude of the resultant. If |\vec{A}| = A, |\vec{B}| = B, and the angle between them is \alpha, then:

|\vec{R}|^2 = A^2 + B^2 + 2AB\cos\alpha

This is worth deriving, because it is just the cosine rule in disguise.

Step 1. Set up coordinates. Place \vec{A} along the x-axis. Then \vec{B} makes angle \alpha with \vec{A}.

Why: choosing \vec{A} along the x-axis simplifies the components without losing generality.

Step 2. Write the components of \vec{B}.

B_x = B\cos\alpha, \qquad B_y = B\sin\alpha

Why: \vec{B} makes angle \alpha with the x-axis (which is the direction of \vec{A}), so its projection onto x is B\cos\alpha and onto y is B\sin\alpha.

Step 3. The head of the resultant is at coordinates (A + B\cos\alpha,\; B\sin\alpha).

Why: the x-component of the resultant is the sum of the x-components of \vec{A} and \vec{B}: A + B\cos\alpha. The y-component is 0 + B\sin\alpha.

Step 4. Compute |\vec{R}|^2.

|\vec{R}|^2 = (A + B\cos\alpha)^2 + (B\sin\alpha)^2

= A^2 + 2AB\cos\alpha + B^2\cos^2\alpha + B^2\sin^2\alpha

= A^2 + 2AB\cos\alpha + B^2(\cos^2\alpha + \sin^2\alpha)

= A^2 + B^2 + 2AB\cos\alpha

Why: expanding the square and using \cos^2\alpha + \sin^2\alpha = 1 collapses the expression to this clean form. This is the law of cosines applied to the triangle formed by \vec{A}, \vec{B}, and \vec{R}.

\boxed{|\vec{R}| = \sqrt{A^2 + B^2 + 2AB\cos\alpha}}

Check the special cases. When \alpha = 0° (parallel vectors), \cos 0° = 1 and R = A + B — they add directly. When \alpha = 180° (antiparallel), \cos 180° = -1 and R = |A - B| — they oppose each other. When \alpha = 90° (perpendicular), \cos 90° = 0 and R = \sqrt{A^2 + B^2} — the Pythagorean theorem, exactly what you saw with the 3 km east + 4 km north example.

The direction of the resultant, measured from \vec{A}, is:

\tan\beta = \frac{B\sin\alpha}{A + B\cos\alpha}

Why: \beta is the angle the resultant makes with \vec{A}. Its tangent is the y-component divided by the x-component of the resultant.

The component method — the practical workhorse

The triangle and parallelogram laws are beautiful for two vectors, but what if you need to add five forces acting on a kite during Uttarayan — tension in the string, gravity, a side-gust from the west, lift from the wind below, and drag opposing the kite's motion? Drawing five vectors head-to-tail and measuring with a protractor is not practical.

The component method handles any number of vectors with pure arithmetic. The idea is simple: break every vector into its x- and y-components, add all the x-components together, add all the y-components together, then recombine.

The component method: decompose each vector into x and y parts, add the parts separately, then reconstruct the resultant. The x-components add along the x-axis, and the y-components add along the y-axis.

Here is the method, step by step.

Step 1. Choose a coordinate system (x-axis and y-axis). The choice is yours — pick axes that make the geometry simple.

Step 2. For each vector \vec{V} that makes angle \theta with the positive x-axis, compute:

V_x = V\cos\theta, \qquad V_y = V\sin\theta

Why: V_x and V_y are the projections of the vector onto the axes. This decomposes a single two-dimensional problem into two one-dimensional problems.

Step 3. Add all the x-components. Add all the y-components.

R_x = A_x + B_x + C_x + \ldots

R_y = A_y + B_y + C_y + \ldots

Why: components along the same axis are just numbers (positive or negative), and numbers add the ordinary way. This is the power of the component method — it reduces vector addition to scalar addition.

Step 4. Find the magnitude and direction of the resultant.

|\vec{R}| = \sqrt{R_x^2 + R_y^2}

\theta = \tan^{-1}\!\left(\frac{R_y}{R_x}\right)

Why: the resultant's x- and y-components form the two legs of a right triangle, with |\vec{R}| as the hypotenuse. The angle \theta is measured from the positive x-axis — but be careful to check which quadrant \vec{R} falls in, because \tan^{-1} by itself only gives values between -90° and +90°.

The component method works for any number of vectors. Three forces on a carrom coin at different angles? Add their x-components, add their y-components, done. Twenty velocity contributions? Same procedure. This is why the component method is the one you will use most often in physics problems.

Why components are better than geometry for hard problems

You might wonder: if the triangle law and the component method give the same answer, why bother with components? The answer is scalability and precision. The triangle law requires you to draw accurate diagrams and measure angles with a protractor — good for building intuition, terrible for getting exact answers. The component method turns vector addition into ordinary algebra: decompose, add numbers, recombine. No protractor, no drawing, no measurement error. For a problem with three or more forces at arbitrary angles — say, a block on a rough inclined plane with an applied force at some angle to the surface — components are the only practical approach.

Here is a quick recipe you can use for every vector-addition problem:

Draw a rough sketch (even a bad one helps you get signs right).
Choose axes. If one vector points along a natural direction (the incline, the string, the river bank), use that as an axis.
Decompose every vector into components using V_x = V\cos\theta and V_y = V\sin\theta.
Sum all x-components. Sum all y-components.
Compute |\vec{R}| and \theta from R_x and R_y.

This five-step recipe works whether you have two vectors or twenty.

Subtraction — adding the negative

How do you subtract \vec{B} from \vec{A}? You don't need a new rule. Subtracting a vector is the same as adding the negative of that vector:

\vec{A} - \vec{B} = \vec{A} + (-\vec{B})

The negative of a vector \vec{B} is a vector with the same magnitude but the opposite direction. If \vec{B} points northeast, -\vec{B} points southwest. If \vec{B} has components (B_x, B_y), then -\vec{B} has components (-B_x, -B_y).

Left: the original vectors $\vec{A}$ and $\vec{B}$. Right: to compute $\vec{A} - \vec{B}$, reverse $\vec{B}$ to get $-\vec{B}$, then add it to $\vec{A}$ head-to-tail. The dashed arrow is $-\vec{B}$; the solid coloured arrow is the result.

In components, subtraction is effortless:

R_x = A_x - B_x, \qquad R_y = A_y - B_y

Why: replacing \vec{B} with -\vec{B} just flips the sign of each component. Adding a negative number is the same as subtracting it.

Vector subtraction appears constantly in physics. Relative velocity is a subtraction: the velocity of a boat relative to the ground is the velocity of the boat relative to the water plus the velocity of the water relative to the ground. Change in velocity is a subtraction: \Delta\vec{v} = \vec{v}_f - \vec{v}_i. Whenever you see "change in" or "relative to," you are working with vector addition or subtraction.

Here is a concrete example of why subtraction matters. A cricket ball is bowled at 140 km/h due east. After hitting the bat, it leaves at 100 km/h due west. What is the change in velocity?

\Delta\vec{v} = \vec{v}_f - \vec{v}_i = (-100\hat{i}) - (140\hat{i}) = -240\hat{i} \text{ km/h}

The change in velocity is 240 km/h westward — far larger than either the initial or final speed alone. This is because the ball reversed direction, and subtraction captures that reversal. If you had naively subtracted magnitudes (140 - 100 = 40), you would have missed the reversal entirely. This is exactly the kind of mistake that vector subtraction prevents.

The polygon law — adding many vectors at once

What if you have three or more vectors to add? Use the same head-to-tail idea, extended. Place \vec{A}, then place \vec{B} at the head of \vec{A}, then place \vec{C} at the head of \vec{B}, and so on. The resultant is the arrow from the tail of the first vector to the head of the last.

\vec{R} = \vec{A} + \vec{B} + \vec{C} + \vec{D}

The chain of vectors and the resultant form a polygon — hence the name polygon law of vector addition.

If the chain of vectors closes on itself — that is, if the head of the last vector lands exactly on the tail of the first — then the resultant is zero. The vectors cancel out completely. This is a powerful check in equilibrium problems: if an object is in equilibrium, the forces on it sum to zero, so the force vectors form a closed polygon.

Think of walking around a city block in Jaipur — you walk north, then east, then south, then west, each block the same length. You end up exactly where you started. Your net displacement is zero because the four displacement vectors form a closed rectangle.

This closed-polygon condition is exactly the test for equilibrium in mechanics. When a lamp hangs from two strings attached to the ceiling, three forces act on the knot: tension \vec{T}_1 in the left string, tension \vec{T}_2 in the right string, and the weight \vec{W} downward. If the lamp hangs motionless, these three forces must sum to zero — and the three force vectors, drawn head-to-tail, form a closed triangle.

In practice, you almost always use the component method for adding three or more vectors. The polygon law is the geometric picture; components are the arithmetic. But the geometric picture is worth carrying in your head, because it gives you an instant sanity check: if your computed resultant is absurdly large or points in the wrong direction, the polygon you draw from the vectors will immediately look wrong.

Explore the parallelogram law

The interactive figure below lets you change the angle \alpha between two vectors and watch how the resultant changes in real time. Notice that the resultant is longest when the two vectors are parallel (\alpha = 0°) and shortest when they are antiparallel (\alpha = 180°).

Drag the red point to change the angle $\alpha$ between two vectors of magnitudes $A = 5$ and $B = 3$. Watch how the resultant $|\vec{R}| = \sqrt{A^2 + B^2 + 2AB\cos\alpha}$ varies. At $\alpha = 0°$, the resultant peaks at 8 (parallel). At $\alpha = 180°$, it drops to 2 (antiparallel).

The null vector

When \vec{A} + (-\vec{A}) = \vec{0}, the result is the null vector (or zero vector). It has zero magnitude and no defined direction.

Null vector

The null vector \vec{0} is a vector with magnitude zero: |\vec{0}| = 0. Its direction is undefined (or, equivalently, it can be assigned any direction without contradiction). It is the identity element for vector addition: \vec{A} + \vec{0} = \vec{A} for any vector \vec{A}.

The null vector is not "nothing" — it is a vector that represents zero displacement, zero force, or zero velocity. When you say the net force on an object in equilibrium is zero, you mean \vec{F}_{\text{net}} = \vec{0}. The direction is undefined because there is no direction associated with standing still or with forces that perfectly cancel.

Properties of the null vector:

Adding it to any vector leaves that vector unchanged: \vec{A} + \vec{0} = \vec{A}
Its magnitude is zero: |\vec{0}| = 0
It is the result of adding a vector to its own negative: \vec{A} + (-\vec{A}) = \vec{0}
Multiplying any vector by the scalar zero gives the null vector: 0 \cdot \vec{A} = \vec{0}

Worked examples

Example 1: Two forces on a carrom striker

A carrom striker sits at rest on the board. You flick it with a force of 30 N directed along the positive x-axis (0°). At the same instant, a second player's striker (in a doubles game) hits it with 40 N directed along the positive y-axis (90°). Find the magnitude and direction of the resultant force on the striker.

Two perpendicular forces on the striker — 30 N along x and 40 N along y — produce a resultant of 50 N at 53.1° from the x-axis. The 3-4-5 right triangle appears.

Step 1. Identify the components.

The 30 N force is along x, so: F_{1x} = 30 N, F_{1y} = 0 N.

The 40 N force is along y, so: F_{2x} = 0 N, F_{2y} = 40 N.

Why: each force is already aligned with an axis, so the components are just the magnitudes. No trigonometry needed here.

Step 2. Add the components.

R_x = F_{1x} + F_{2x} = 30 + 0 = 30 \text{ N}

R_y = F_{1y} + F_{2y} = 0 + 40 = 40 \text{ N}

Why: this is the component method at its simplest — each component adds independently along its own axis.

Step 3. Find the magnitude.

|\vec{R}| = \sqrt{R_x^2 + R_y^2} = \sqrt{30^2 + 40^2} = \sqrt{900 + 1600} = \sqrt{2500} = 50 \text{ N}

Why: the x- and y-components are perpendicular, so the resultant is the hypotenuse of a right triangle. You might recognise 30-40-50 as the 3-4-5 Pythagorean triple scaled by 10.

Step 4. Find the direction.

\theta = \tan^{-1}\!\left(\frac{R_y}{R_x}\right) = \tan^{-1}\!\left(\frac{40}{30}\right) = \tan^{-1}(1.333) \approx 53.1°

Why: \theta is measured from the positive x-axis. Both R_x and R_y are positive, so the resultant is in the first quadrant — the angle is between 0° and 90°, which checks out.

Result: The resultant force is 50 N at 53.1° from the x-axis (or equivalently, 36.9° from the y-axis). The striker accelerates in this direction.

What this shows: Two perpendicular forces of 30 N and 40 N don't produce a 70 N resultant — they produce a 50 N resultant at an angle. The direction is closer to the larger force, which makes physical sense: the 40 N force has more influence on the direction than the 30 N force.

Example 2: A boat crossing the Ganga

A boat can travel at 4 m/s in still water. It aims straight across the Ganga (perpendicular to the banks), but the river flows at 3 m/s from west to east. Find the boat's actual velocity relative to the ground — its speed and direction.

The boat aims north at 4 m/s, but the river pushes it east at 3 m/s. The actual velocity (dashed) is 5 m/s, directed 36.9° east of north. The boat drifts downstream even though it points straight across.

Step 1. Set up. Let north be the y-direction and east be the x-direction.

The boat's velocity in still water: \vec{v}_b = (0,\; 4) m/s (pointing north).

The river's velocity: \vec{v}_r = (3,\; 0) m/s (pointing east).

Why: the boat aims straight across (north), so its velocity has no east-west component. The river flows east, so its velocity has no north-south component.

Step 2. Add the velocities.

\vec{v}_{\text{actual}} = \vec{v}_b + \vec{v}_r = (0 + 3,\; 4 + 0) = (3,\; 4) \text{ m/s}

Why: the boat is carried by the water. Its actual velocity relative to the ground is the vector sum of its velocity through the water and the water's velocity relative to the ground.

Step 3. Find the speed.

|\vec{v}_{\text{actual}}| = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5 \text{ m/s}

Step 4. Find the direction.

\theta = \tan^{-1}\!\left(\frac{v_x}{v_y}\right) = \tan^{-1}\!\left(\frac{3}{4}\right) = 36.9° \text{ east of north}

Why: here \theta is measured from north (the y-axis) toward east (the x-axis), so the tangent is the east component divided by the north component, not the other way around. Always match the angle definition to the components you divide.

Result: The boat's actual speed is 5 m/s, directed 36.9° east of north. It reaches the opposite bank, but downstream of the point directly across.

What this shows: The 3-4-5 triangle appears again. A boat pointing straight north at 4 m/s in a river flowing east at 3 m/s ends up moving at 5 m/s at an angle. If the river is 200 m wide, the boat will drift 200 \times \frac{3}{4} = 150 m downstream by the time it crosses.

Common confusions

"Just add the magnitudes." This is wrong whenever the vectors are not in the same direction. Two forces of 3 N and 4 N can produce a resultant anywhere between 1 N (antiparallel) and 7 N (parallel), depending on the angle between them. Adding magnitudes directly only works when both vectors point in exactly the same direction.
"The resultant is always bigger than either vector." Not true. If the angle between the two vectors is large enough (greater than 90°), the resultant can be smaller than one of the individual vectors. Two forces of 10 N at 150° apart give a resultant of only \sqrt{100 + 100 + 2(100)(-0.866)} \approx 5.2 N — smaller than either individual force.
"Subtracting vectors means subtracting magnitudes." Only when the vectors point in the same direction. In general, |\vec{A} - \vec{B}| \neq |A - B|. Subtraction is addition of the negative, and you must use the full vector arithmetic.
"Components depend on the vectors, not on the axes." It is the other way around. The components of a vector depend entirely on which axes you choose. Rotate your coordinate system by 45°, and every component changes. The vector itself — its magnitude and direction in space — does not change. Only its description in your particular coordinate system does.
"The angle in \tan^{-1}(R_y/R_x) always gives the right answer." It gives an answer, but not always the right one. The \tan^{-1} function returns values between -90° and +90°, so it cannot distinguish between a vector in the first quadrant and one in the third quadrant (both have positive R_y/R_x). Always check the signs of R_x and R_y to determine the correct quadrant, then adjust the angle.

If you came here to learn how to add vectors and use the component method, you have what you need — the rest of the article is for readers who want the formal proofs and some subtleties.

Proof of the parallelogram law from the cosine rule

The result |\vec{R}|^2 = A^2 + B^2 + 2AB\cos\alpha looks like the law of cosines, and for good reason — it is the law of cosines, applied to the right triangle.

In a triangle with sides a, b, c and the angle C opposite side c, the law of cosines states:

c^2 = a^2 + b^2 - 2ab\cos C

Now look at the triangle formed by \vec{A}, \vec{B}, and \vec{R}. The angle inside the triangle between sides A and B is \pi - \alpha (not \alpha), because \alpha is the angle between the two vectors measured outside the triangle. Since \cos(\pi - \alpha) = -\cos\alpha:

R^2 = A^2 + B^2 - 2AB\cos(\pi - \alpha) = A^2 + B^2 + 2AB\cos\alpha

Why: the angle between the vectors (\alpha) and the angle inside the triangle (\pi - \alpha) are supplementary. Substituting \cos(\pi - \alpha) = -\cos\alpha flips the sign, turning the minus into a plus.

This confirms that the parallelogram law formula is a restatement of the cosine rule. No new physics here — just trigonometry applied to the geometry of vector addition.

The null vector and vector space axioms

The null vector \vec{0} is not a curiosity — it is a foundational object. In the language of mathematics, the set of all vectors in a plane (or in 3D space), together with vector addition and scalar multiplication, forms a vector space. The null vector is the additive identity of this vector space, meaning:

\vec{A} + \vec{0} = \vec{A} \quad \text{for all } \vec{A}

The negative vector -\vec{A} is the additive inverse — the vector you add to \vec{A} to get \vec{0}. These are the same ideas as 0 and -a in ordinary arithmetic, extended to objects with direction.

The reason the direction of \vec{0} is undefined is that every direction is equally valid. If \vec{0} had a specific direction, then adding it to another vector would have to be consistent with that direction, which would conflict with \vec{A} + \vec{0} = \vec{A} for vectors \vec{A} pointing in different directions. So the null vector has magnitude zero and direction "anything you like."

This idea becomes load-bearing in more advanced physics. When you write \vec{F}_{\text{net}} = \vec{0} for equilibrium, you are saying that the system's force vector is the identity element of the vector space — no direction, no magnitude, nothing to push the object anywhere.

Generalising to three dimensions

Everything in this article extends seamlessly to three dimensions. A vector in 3D has three components: \vec{A} = A_x\hat{i} + A_y\hat{j} + A_z\hat{k}. To add two 3D vectors, add the components along each axis separately:

R_x = A_x + B_x, \quad R_y = A_y + B_y, \quad R_z = A_z + B_z

|\vec{R}| = \sqrt{R_x^2 + R_y^2 + R_z^2}

The direction now requires two angles (typically \theta and \phi in spherical coordinates), but the principle is identical. The component method works in any number of dimensions — it is the universal workhorse for adding vectors.

Where this leads next

Scalar Product — multiplying two vectors to get a number. This is how you compute work done by a force.
Vector Product — multiplying two vectors to get a third vector perpendicular to both. This is how you compute torque.
Position, Distance, and Displacement — displacement is a vector, distance is not. Adding displacement vectors is how you track motion.
Vectors Introduction and Notation — if you need a refresher on what vectors are, how to notate them, and why direction matters.