The triangle law of vector addition
NCERT §3.4 begins with the operational definition: a vector is anything that obeys the triangle law (equivalently the parallelogram law) of addition. So the addition rule is not a derived theorem — it is part of what makes a vector a vector. The companion piece on scalars and vectors argues why electric current, despite having direction along the wire, is a scalar — it fails this very test.
The rule is concrete. To add $\mathbf{A}$ and $\mathbf{B}$, draw $\mathbf{A}$ to scale; without changing its direction, place $\mathbf{B}$'s tail at $\mathbf{A}$'s head; draw a straight arrow from the tail of $\mathbf{A}$ to the head of $\mathbf{B}$. That arrow is the resultant $\mathbf{R} = \mathbf{A} + \mathbf{B}$. Because the three vectors form a triangle, the construction is also called the head-to-tail or triangle method. The order does not matter — NCERT Fig. 3.4 verifies that $\mathbf{B} + \mathbf{A}$ produces the same arrow.
The parallelogram law
NCERT §3.4 also gives a second picture which yields the same resultant. Bring the tails of $\mathbf{A}$ and $\mathbf{B}$ to a common origin $O$. From the head of $\mathbf{A}$ draw a line parallel to $\mathbf{B}$, and from the head of $\mathbf{B}$ draw a line parallel to $\mathbf{A}$. The two lines intersect at a point $S$, and $OAS B$ is a parallelogram. The diagonal $OS$, drawn from the common origin, is the resultant $\mathbf{R}$.
To see why this is the triangle law in disguise, slide $\mathbf{B}$ from $O$ to the head of $\mathbf{A}$. Since opposite sides of a parallelogram are equal and parallel, the side from the head of $\mathbf{A}$ to $S$ is a faithful copy of $\mathbf{B}$, and you recover the head-to-tail picture with resultant $OS$. The triangle law is cleaner when many vectors must be summed; the parallelogram law is the one most NEET problems are drawn with, because the angle $\theta$ between the two vectors is shown directly with both tails meeting at $O$.
The polygon law and closure
The triangle law extends naturally to more than two vectors. To add $\mathbf{A}$, $\mathbf{B}$, $\mathbf{C}$, $\mathbf{D}$, $\ldots$, place each tail at the previous head in turn. The arrow from the tail of the first vector to the head of the last is the resultant. The construction is called the polygon law.
If the polygon closes — the head of the last vector lands exactly on the tail of the first — then the resultant is zero. This is the geometric statement of equilibrium and the basis of NCERT Exercise 3.7: given $\mathbf{a} + \mathbf{b} + \mathbf{c} + \mathbf{d} = \mathbf{0}$, the four vectors form a closed quadrilateral, and $|\mathbf{a}| \le |\mathbf{b}| + |\mathbf{c}| + |\mathbf{d}|$ because $\mathbf{a} = -(\mathbf{b} + \mathbf{c} + \mathbf{d})$.
Commutative, associative and the role of the null vector
Vector addition shares two algebraic properties with ordinary scalar addition, both proved graphically in NCERT §3.4. The commutative property says
$$\mathbf{A} + \mathbf{B} = \mathbf{B} + \mathbf{A}$$which is just the observation that walking $\mathbf{A}$ first and $\mathbf{B}$ second lands you at the same point as walking $\mathbf{B}$ first and $\mathbf{A}$ second — the parallelogram diagonal is the same regardless of which side you traverse first. The associative property says
$$(\mathbf{A} + \mathbf{B}) + \mathbf{C} = \mathbf{A} + (\mathbf{B} + \mathbf{C})$$and is illustrated in Fig. 3.4(d) — bracketing makes no difference to the polygon resultant.
The null vector $\mathbf{0}$ has zero magnitude (and hence no well-defined direction). It plays the role of the additive identity:
$$\mathbf{A} + \mathbf{0} = \mathbf{A}, \qquad \lambda \mathbf{0} = \mathbf{0}, \qquad 0\,\mathbf{A} = \mathbf{0}$$NCERT illustrates its physical meaning with the displacement example — an object that leaves $P$, moves around, and returns to $P$ has a null displacement vector. The same idea makes $\mathbf{A} + (-\mathbf{A}) = \mathbf{0}$ — two equal-and-opposite vectors cancel.
Subtracting vectors: $\mathbf{A} - \mathbf{B}$
NCERT defines subtraction in terms of addition. The difference $\mathbf{A} - \mathbf{B}$ is the sum of $\mathbf{A}$ and the reversed vector $-\mathbf{B}$:
$$\mathbf{A} - \mathbf{B} = \mathbf{A} + (-\mathbf{B})$$Graphically, reverse $\mathbf{B}$ — rotate it through 180° while keeping its tail fixed — and then apply the triangle law to $\mathbf{A}$ and $-\mathbf{B}$. The resultant arrow is $\mathbf{A} - \mathbf{B}$. NCERT Fig. 3.5 draws both $\mathbf{R}_1 = \mathbf{A} + \mathbf{B}$ and $\mathbf{R}_2 = \mathbf{A} - \mathbf{B}$ on the same diagram so you can compare.
Equivalently, in the parallelogram with tails of $\mathbf{A}$ and $\mathbf{B}$ at $O$, the outward diagonal is $\mathbf{A} + \mathbf{B}$ and the diagonal from the head of $\mathbf{B}$ to the head of $\mathbf{A}$ represents $\mathbf{A} - \mathbf{B}$ — the two diagonals describe the sum and the difference simultaneously.
The analytical formula for the resultant
Graphical addition is intuitive but unsuited to exam timing. The analytical method derives a closed-form expression that you can evaluate in seconds. NCERT §3.6 (Example 3.2) does the derivation using the parallelogram law.
Place the tails of $\mathbf{A}$ and $\mathbf{B}$ at $O$, with $\mathbf{A}$ along $OP$. Let the angle between them be $\theta$. Complete the parallelogram and draw the diagonal $OS = \mathbf{R}$. Drop a perpendicular from $S$ onto the extended line of $OP$, hitting it at $N$. Then triangle $PSN$ is right-angled at $N$, with
$$PN = B\cos\theta, \qquad SN = B\sin\theta$$The right triangle $OSN$ now gives
$$OS^2 = ON^2 + SN^2 = (A + B\cos\theta)^2 + (B\sin\theta)^2$$Expanding and using $\sin^2\theta + \cos^2\theta = 1$,
$$\boxed{R = \sqrt{A^2 + B^2 + 2AB\cos\theta}}$$This is NCERT Eq. (3.24a), the law of cosines for vector addition. It is the single most-tested formula in the chapter — every problem that asks "find the magnitude of the resultant of two vectors" is solved by plugging into this expression.
Direction of the resultant
The same diagram gives the direction of $\mathbf{R}$. Let $\alpha$ be the angle that $\mathbf{R}$ makes with $\mathbf{A}$. From triangle $OSN$, $\tan\alpha = SN/ON$, so
$$\tan\alpha = \frac{B\sin\theta}{A + B\cos\theta}$$NCERT Eq. (3.24f). Two derived statements follow from the geometry — the law of sines (NCERT Eq. 3.24d):
$$\frac{R}{\sin\theta} = \frac{A}{\sin\beta} = \frac{B}{\sin\alpha}$$where $\beta$ is the angle $\mathbf{R}$ makes with $\mathbf{B}$. The law of sines is the right tool when you know the angles in the triangle but not all three side magnitudes — typical for boat-and-current and aircraft-and-wind problems, which is exactly how NCERT Example 3.3 uses it.
NCERT Example 3.1. Rain falls vertically at $35$ m s$^{-1}$. A wind starts blowing at $12$ m s$^{-1}$ from east to west. In which direction should a boy at a bus stop hold his umbrella?
The two velocities are perpendicular, so $\theta = 90°$ and $\cos\theta = 0$. The resultant has magnitude
$$R = \sqrt{35^2 + 12^2} = \sqrt{1225 + 144} = \sqrt{1369} = 37 \text{ m s}^{-1}$$
Its angle with the vertical is $\tan^{-1}(12/35) = \tan^{-1}(0.343) \approx 19°$. So the umbrella should tilt $19°$ from the vertical towards the east (into the apparent rain direction).
Three special cases NEET examiners love
Three angles between $\mathbf{A}$ and $\mathbf{B}$ collapse the analytical formula into one-line answers. Burn these into memory.
| Angle $\theta$ | Configuration | Resultant magnitude | Comment |
|---|---|---|---|
| $0°$ (parallel) | Same direction | $R = A + B$ | Maximum — vectors reinforce |
| $90°$ (perpendicular) | At right angles | $R = \sqrt{A^2 + B^2}$ | Pythagorean — projectile-style |
| $180°$ (antiparallel) | Opposite directions | $R = |A - B|$ | Minimum — vectors cancel partially |
From these three rows the general bound follows: for any angle, the resultant magnitude lies between $|A - B|$ and $A + B$. This is precisely the triangle inequality of NCERT Exercise 3.6:
$$|A - B| \;\le\; |\mathbf{A} + \mathbf{B}| \;\le\; A + B$$The left inequality is equality iff the vectors are antiparallel; the right inequality is equality iff they are parallel.
The equal-magnitude shortcut: $R = 2A\cos(\theta/2)$
If $|\mathbf{A}| = |\mathbf{B}| = A$, the analytical formula simplifies dramatically. Substituting $B = A$,
$$R = \sqrt{A^2 + A^2 + 2A^2\cos\theta} = \sqrt{2A^2(1 + \cos\theta)} = \sqrt{2A^2 \cdot 2\cos^2(\theta/2)}$$ $$\boxed{R = 2A\cos(\theta/2)}$$using the half-angle identity $1 + \cos\theta = 2\cos^2(\theta/2)$. Three consequences follow.
First, the direction of $\mathbf{R}$ is the angle bisector of $\mathbf{A}$ and $\mathbf{B}$. Substituting $A = B$ in $\tan\alpha = B\sin\theta / (A + B\cos\theta)$ and applying half-angle identities gives $\alpha = \theta/2$. So the resultant of two equal-magnitude vectors makes angle $\theta/2$ with each of them.
Second, the angle for which the resultant equals the magnitude of each vector. Set $R = A$:
$$A = 2A\cos(\theta/2) \;\Rightarrow\; \cos(\theta/2) = \tfrac{1}{2} \;\Rightarrow\; \theta = 120°$$This is the classic NEET statement: if two vectors of equal magnitude have a resultant of the same magnitude, the angle between them is $120°$. Three equal-magnitude vectors at $120°$ to each other close their polygon — they are in equilibrium.
Third, the angle for which the resultant has magnitude $A\sqrt{2}$. Set $R = A\sqrt{2}$:
$$A\sqrt{2} = 2A\cos(\theta/2) \;\Rightarrow\; \cos(\theta/2) = \tfrac{1}{\sqrt{2}} \;\Rightarrow\; \theta = 90°$$which is just the Pythagorean special case in equal-magnitude disguise. Two equal-magnitude perpendicular vectors give a resultant $\sqrt{2}$ times either of them.
The component method — bridge to coordinates
Graphical addition gives pictures; the analytical formula gives a single magnitude. The component method gives both at once and scales to any number of vectors. The mechanics is in the sibling piece on resolution of vectors; here we only state the operational rule for adding.
If $\mathbf{A} = A_x\hat{\mathbf{i}} + A_y\hat{\mathbf{j}}$ and $\mathbf{B} = B_x\hat{\mathbf{i}} + B_y\hat{\mathbf{j}}$, then NCERT Eq. (3.21) says
$$\mathbf{R} = \mathbf{A} + \mathbf{B} = (A_x + B_x)\hat{\mathbf{i}} + (A_y + B_y)\hat{\mathbf{j}}$$That is, each component of the resultant is the sum of the corresponding components of the addends. Once $R_x$ and $R_y$ are known, the magnitude and direction follow from the standard reconstruction:
$$R = \sqrt{R_x^2 + R_y^2}, \qquad \tan\theta_R = \frac{R_y}{R_x}$$This is the workhorse for NEET problems. It scales to many vectors, it removes the need to identify pairwise angles, and it is signed — a $B_x$ pointing in the $-\hat{\mathbf{i}}$ direction is automatically subtracted from $A_x$, with no separate "subtraction case".
Worked examples — three NEET-style problems
Two forces of magnitude $5$ N and $12$ N act at right angles to each other on a point. Find the magnitude and direction of the resultant.
With $\theta = 90°$, $\cos\theta = 0$ and the formula collapses to the Pythagorean case:
$$R = \sqrt{5^2 + 12^2} = \sqrt{25 + 144} = \sqrt{169} = 13 \text{ N}$$
The angle $\alpha$ the resultant makes with the $5$ N force is
$$\tan\alpha = \frac{12\sin 90°}{5 + 12\cos 90°} = \frac{12}{5} \;\Rightarrow\; \alpha = \tan^{-1}(12/5) \approx 67.4°$$
This is the classic $5$-$12$-$13$ triangle. The same arithmetic appears every year disguised as "displacements of $30$ m east and $40$ m north" ($R = 50$ m) or "velocities of $6$ m s$^{-1}$ horizontal and $8$ m s$^{-1}$ vertical" ($R = 10$ m s$^{-1}$).
Two velocities each of magnitude $10$ m s$^{-1}$ have a resultant of magnitude $10$ m s$^{-1}$. What is the angle between them?
Use the equal-magnitude shortcut $R = 2A\cos(\theta/2)$. Substituting $R = A = 10$:
$$10 = 2(10)\cos(\theta/2) \;\Rightarrow\; \cos(\theta/2) = \tfrac{1}{2} \;\Rightarrow\; \theta/2 = 60° \;\Rightarrow\; \theta = 120°$$
If a student blindly applies $R^2 = A^2 + B^2 + 2AB\cos\theta$ with $R = A = B = 10$, the algebra is more painful but lands on the same place: $100 = 200(1 + \cos\theta)$, giving $\cos\theta = -1/2$ and $\theta = 120°$. The shortcut saves three lines of working under exam pressure.
Three coplanar forces of magnitudes $F$, $F$ and $F$ act on a point and the body remains in equilibrium. What must be the angle between any two of these forces?
For equilibrium the polygon closes — the three forces drawn head-to-tail form a triangle. Three equal sides force an equilateral triangle, in which every interior angle is $60°$. But the interior angle at a vertex is $180° - \theta$ (where $\theta$ is the angle between the original concurrent vectors), so $180° - \theta = 60°$ and
$$\theta = 120°$$
between any two of the three concurrent forces. This is the same $120°$ that emerged in the equal-magnitude shortcut: three equal forces in equilibrium must be at $120°$ to each other. NCERT Exercise 3.7 is the four-vector cousin of this question — replace "triangle" with "quadrilateral".
A motorboat heads north at $25$ km h$^{-1}$ in a current of $10$ km h$^{-1}$ flowing $60°$ east of south. Find the resultant velocity.
The angle between the two vectors is $180° - 60° = 120°$ (boat heading north, current $60°$ east of south). Using the law of cosines:
$$R = \sqrt{25^2 + 10^2 + 2(25)(10)\cos 120°} = \sqrt{625 + 100 - 250} = \sqrt{475} \approx 21.8 \text{ km h}^{-1}$$
For the direction, apply the law of sines: $R/\sin 120° = 10/\sin\alpha$, giving $\sin\alpha \approx 0.397$ and $\alpha \approx 23.4°$ east of the boat's heading. NCERT works this problem in full at §3.6.
What this subtopic locked in
- Triangle law. Tail of $\mathbf{B}$ at head of $\mathbf{A}$; resultant from tail of $\mathbf{A}$ to head of $\mathbf{B}$.
- Parallelogram law. Tails coincide; diagonal from the common origin is the resultant. Equivalent to the triangle law.
- Polygon law. Place all vectors head-to-tail. Closure ⇒ resultant zero ⇒ equilibrium.
- Commutative, associative, null. $\mathbf{A} + \mathbf{B} = \mathbf{B} + \mathbf{A}$; $(\mathbf{A} + \mathbf{B}) + \mathbf{C} = \mathbf{A} + (\mathbf{B} + \mathbf{C})$; $\mathbf{A} + \mathbf{0} = \mathbf{A}$.
- Analytical formula. $R = \sqrt{A^2 + B^2 + 2AB\cos\theta}$ with $\theta$ measured tail-to-tail.
- Direction. $\tan\alpha = B\sin\theta / (A + B\cos\theta)$, with $\alpha$ measured from $\mathbf{A}$.
- Special cases. $\theta = 0° \Rightarrow R = A + B$; $\theta = 90° \Rightarrow R = \sqrt{A^2 + B^2}$; $\theta = 180° \Rightarrow R = |A - B|$.
- Equal magnitudes. $R = 2A\cos(\theta/2)$. $R = A$ at $\theta = 120°$; $R = A\sqrt{2}$ at $\theta = 90°$.
- $|\mathbf{A} + \mathbf{B}| = |\mathbf{A} - \mathbf{B}|$ iff $\mathbf{A} \perp \mathbf{B}$.
- Component method. $R_x = A_x + B_x$, $R_y = A_y + B_y$; then $R = \sqrt{R_x^2 + R_y^2}$.