Scalars and vectors
Every physical quantity belongs to one of two families. A scalar has only a magnitude — a number with a unit. Mass, length, time, energy, temperature, electric charge and density are scalars, and they combine by ordinary arithmetic. A vector has both a magnitude and a direction in space. Displacement, velocity, acceleration, force, momentum, torque, electric field and magnetic field are vectors. They cannot be added arithmetically because the direction changes the result.
The defining test is not "does it point somewhere" but "does it add by the parallelogram law". A quantity qualifies as a vector only if it has magnitude and direction and obeys that addition rule. This distinction separates true vectors from quantities that merely have a sense of direction, and NEET examiners exploit the gap regularly.
| Scalars | Vectors |
|---|---|
| Mass, time, energy, temperature | Displacement, velocity, acceleration |
| Electric charge, density, speed | Force, momentum, torque |
| Electric current, pressure, work | Electric field, magnetic field |
| Added algebraically (\(+,-,\times,\div\)) | Added by triangle / parallelogram law |
Representation and notation
A vector is drawn as a directed line segment: the length of the arrow encodes the magnitude and the arrowhead encodes the direction. In print a vector is written in bold, \(\mathbf{A}\), or with an arrow overhead, \(\vec{A}\); its magnitude is \(|\vec{A}|\) or simply \(A\), always a non-negative number. The tail of the arrow is the point of application or origin, and the head marks the terminus.
A vector is unchanged by parallel translation — sliding it to a new position without rotating or stretching it leaves it the same vector. Rotating it, however, produces a different vector even if the length is preserved, because direction is part of the vector's identity. This is why two equal-looking arrows in different orientations are not the same vector.
Types of vectors
A handful of named vectors recur throughout mechanics. Knowing them by name removes ambiguity when a question hinges on a definition.
- Equal vectors — same magnitude and same direction, regardless of where they are drawn. A parallel shift never breaks equality.
- Negative vector — \(-\vec{B}\) has the same magnitude as \(\vec{B}\) but the exactly opposite direction. It is the key to subtraction.
- Unit vector — magnitude exactly one, dimensionless, carrying direction only. Written with a hat, e.g. \(\hat{A}=\vec{A}/|\vec{A}|\).
- Null (zero) vector — magnitude zero, direction undefined. It is the result of \(\vec{A}-\vec{A}\) and of the displacement around any closed path.
- Collinear / parallel vectors — lie along the same line or parallel lines; the angle between them is \(0^\circ\) or \(180^\circ\).
| Type | Defining property | Typical role |
|---|---|---|
| Equal | Same magnitude and direction | Translating a vector in a free-body sketch |
| Negative \(-\vec{B}\) | Same magnitude, opposite direction | Vector subtraction \(\vec{A}-\vec{B}=\vec{A}+(-\vec{B})\) |
| Unit \(\hat{A}\) | Magnitude \(=1\), no units | Specifying direction; \(\vec{A}=A\,\hat{A}\) |
| Null \(\vec{0}\) | Magnitude \(=0\), direction undefined | Closed-loop displacement; \(\vec{A}-\vec{A}\) |
Triangle and parallelogram laws
To add two vectors \(\vec{A}\) and \(\vec{B}\) by the triangle law, place the tail of \(\vec{B}\) at the tip of \(\vec{A}\); the resultant \(\vec{R}\) is drawn from the tail of \(\vec{A}\) to the tip of \(\vec{B}\). Equivalently, by the parallelogram law, draw both vectors from a common point and complete the parallelogram — the diagonal through that point is the resultant. The two constructions are the same rule seen from two angles and always yield the same \(\vec{R}\).
Vector addition is commutative, \(\vec{A}+\vec{B}=\vec{B}+\vec{A}\), and associative, \(\vec{A}+(\vec{B}+\vec{C})=(\vec{A}+\vec{B})+\vec{C}\). The triangle law extends naturally: to add several vectors, lay them head-to-tail in any order and the resultant runs from the very first tail to the very last head. If the vectors form a closed polygon, the resultant is the null vector.
Magnitude and direction of the resultant
When \(\vec{A}\) and \(\vec{B}\) make an angle \(\theta\) with each other, the parallelogram-law geometry gives the magnitude of the resultant directly.
\(R=\sqrt{A^2+B^2+2AB\cos\theta}\)
Magnitude of the resultant of two vectors at angle \(\theta\).
The direction of \(\vec{R}\), measured as the angle \(\alpha\) it makes with \(\vec{A}\), follows from the same triangle:
\(\tan\alpha=\dfrac{B\sin\theta}{A+B\cos\theta}\)
Direction of the resultant relative to \(\vec{A}\).
The magnitude formula collapses cleanly at the limits. When \(\theta=0^\circ\) (parallel), \(\cos\theta=1\) and \(R=A+B\) — the maximum. When \(\theta=180^\circ\) (antiparallel), \(\cos\theta=-1\) and \(R=|A-B|\) — the minimum. When \(\theta=90^\circ\) the cross term vanishes and \(R=\sqrt{A^2+B^2}\), the Pythagorean case NEET reaches for in projectile and field-superposition problems.
| Angle \(\theta\) | \(\cos\theta\) | Resultant \(R\) | Meaning |
|---|---|---|---|
| \(0^\circ\) | \(+1\) | \(A+B\) | Maximum (parallel) |
| \(90^\circ\) | \(0\) | \(\sqrt{A^2+B^2}\) | Perpendicular (Pythagoras) |
| \(120^\circ\) | \(-\tfrac12\) | \(\sqrt{A^2+B^2-AB}\) | Equal forces \(\Rightarrow R=A\) |
| \(180^\circ\) | \(-1\) | \(|A-B|\) | Minimum (antiparallel) |
Subtraction of vectors
Subtraction is not a new operation — it is addition of a reversed vector. To form \(\vec{A}-\vec{B}\), reverse \(\vec{B}\) to get \(-\vec{B}\) (same length, opposite direction) and add it to \(\vec{A}\) by the triangle law: \(\vec{A}-\vec{B}=\vec{A}+(-\vec{B})\).
Because the angle between \(\vec{A}\) and \(-\vec{B}\) is \(180^\circ-\theta\), the magnitude of the difference follows from the resultant formula with the cross-term sign flipped: \(|\vec{A}-\vec{B}|=\sqrt{A^2+B^2-2AB\cos\theta}\). This is the relative-velocity construction in disguise — the velocity of one body as seen from another is \(\vec{v}_{1}-\vec{v}_{2}\).
Resolution into rectangular components
Graphical addition handles two vectors but becomes unwieldy for several and impossible to write into an exam answer. The standard tool is resolution: replace a vector by its perpendicular projections onto the coordinate axes — its rectangular components. A vector in the xy-plane that makes angle \(\theta\) with the +x axis has components
\(A_x=A\cos\theta,\qquad A_y=A\sin\theta\)
Rectangular components of a vector at angle \(\theta\) from the +x axis.
The original vector is rebuilt from its components as \(\vec{A}=A_x\hat{i}+A_y\hat{j}\), and the inverse relations recover magnitude and direction: \(A=\sqrt{A_x^2+A_y^2}\) and \(\tan\theta=A_y/A_x\). The scheme extends to three dimensions simply by appending a z-component, \(\vec{A}=A_x\hat{i}+A_y\hat{j}+A_z\hat{k}\), with \(A=\sqrt{A_x^2+A_y^2+A_z^2}\).
Unit vectors î, ĵ, k̂
A unit vector has magnitude exactly one and carries direction only; it is dimensionless. The Cartesian unit vectors \(\hat{i}\), \(\hat{j}\), \(\hat{k}\) point along the +x, +y and +z axes and are mutually perpendicular. They are the scaffolding of component notation: a vector with components \((A_x,A_y,A_z)\) is written compactly as \(\vec{A}=A_x\hat{i}+A_y\hat{j}+A_z\hat{k}\).
Any non-zero vector defines a unit vector along its own direction, \(\hat{A}=\vec{A}/|\vec{A}|\), so that \(\vec{A}=A\,\hat{A}\) cleanly separates "how much" from "which way". This separation is exactly what makes the component method work: each axis is handled independently because \(\hat{i}\), \(\hat{j}\) and \(\hat{k}\) do not mix.
The analytical (component) method
Once every vector is resolved into components, addition reduces to ordinary arithmetic on each axis. There are no parallelograms to draw and no triangles to solve. The recipe is short.
- Resolve each vector: \(\vec{A}=A_x\hat{i}+A_y\hat{j}\), \(\vec{B}=B_x\hat{i}+B_y\hat{j}\), and so on.
- Add components axis by axis: \(R_x=A_x+B_x+\dots\), \(R_y=A_y+B_y+\dots\)
- Reassemble: \(\vec{R}=R_x\hat{i}+R_y\hat{j}\), with \(R=\sqrt{R_x^2+R_y^2}\) and \(\tan\alpha=R_y/R_x\).
This is the workhorse for inclined-plane dynamics, projectile motion and equilibrium of concurrent forces. The parallelogram formula is its two-vector special case; the component method is what scales to three forces, four forces, or a force in three dimensions.
Worked examples
Two forces of magnitude \(3\ \text{N}\) and \(4\ \text{N}\) act at a point with an angle of \(90^\circ\) between them. Find the magnitude and direction of the resultant.
Magnitude. With \(\theta=90^\circ\), \(\cos\theta=0\), so \(R=\sqrt{A^2+B^2}=\sqrt{3^2+4^2}=\sqrt{25}=5\ \text{N}\).
Direction. Taking \(A=3\ \text{N}\) along the reference and \(B=4\ \text{N}\) perpendicular, \(\tan\alpha=\dfrac{B\sin\theta}{A+B\cos\theta}=\dfrac{4(1)}{3+0}=\dfrac{4}{3}\), so \(\alpha=\tan^{-1}(4/3)\approx 53^\circ\) measured from the 3 N force.
Sanity check. \(5\ \text{N}\) lies between \(|4-3|=1\ \text{N}\) and \(4+3=7\ \text{N}\), as required.
Two vectors of equal magnitude \(F\) make an angle of \(120^\circ\). Find the magnitude of their resultant.
Apply the formula. With \(A=B=F\) and \(\cos 120^\circ=-\tfrac12\):
\(R=\sqrt{F^2+F^2+2F\cdot F\left(-\tfrac12\right)}=\sqrt{2F^2-F^2}=\sqrt{F^2}=F.\)
Reading. Two equal vectors at \(120^\circ\) produce a resultant equal in magnitude to either one. This is the geometry behind three equal coplanar forces at \(120^\circ\) being in equilibrium — the resultant of any two exactly cancels the third.
Add the displacements \(\vec{A}=3\hat{i}+4\hat{j}\) and \(\vec{B}=1\hat{i}-2\hat{j}\) (in metres) by the analytical method, and find the magnitude of the resultant.
Add components. \(R_x=3+1=4\), \(R_y=4+(-2)=2\). So \(\vec{R}=4\hat{i}+2\hat{j}\ \text{m}\).
Magnitude and direction. \(R=\sqrt{4^2+2^2}=\sqrt{20}=2\sqrt{5}\approx 4.47\ \text{m}\), with \(\tan\alpha=R_y/R_x=2/4=0.5\), giving \(\alpha=\tan^{-1}(0.5)\approx 26.6^\circ\) above the +x axis.
Note. No parallelogram was drawn — once vectors are in component form, addition is pure arithmetic, and the same routine handles a third or fourth vector unchanged.
The vector toolkit in one breath
- A vector needs magnitude, direction and obedience to the parallelogram law. Current and pressure point but are scalars.
- Triangle and parallelogram laws are one rule; the resultant is \(R=\sqrt{A^2+B^2+2AB\cos\theta}\), bounded by \(|A-B|\le R\le A+B\).
- Direction of the resultant: \(\tan\alpha=\dfrac{B\sin\theta}{A+B\cos\theta}\).
- Subtract by reversing: \(\vec{A}-\vec{B}=\vec{A}+(-\vec{B})\); magnitude \(\sqrt{A^2+B^2-2AB\cos\theta}\). Order matters.
- Resolve into rectangular components: \(A_x=A\cos\theta\), \(A_y=A\sin\theta\); rebuild as \(\vec{A}=A_x\hat{i}+A_y\hat{j}\).
- The analytical method adds components axis by axis — the scalable replacement for graphical addition.