Physics Lab

Unit 1: Units, Dimensions & Vectors

This is the foundational unit of JEE Physics. Although seemingly elementary, it appears every year in JEE Main (1–2 questions, ~8 marks) and frequently in JEE Advanced as part of a multi-concept question or a Match-the-Column. Mastery of dimensions, errors and vectors is a prerequisite for every later unit — kinematics, electrodynamics and modern physics all rely on vector calculus and the ability to detect dimensionally inconsistent options under exam pressure.

Typical question types:

  • JEE Main: "Find the dimensional formula of …", "If a=bc/da=bc/d and the percentage errors in b,c,db, c, d are …, find the maximum % error in aa", "Resultant of two vectors", "Angle between A⃗\vec A and B⃗\vec B given A⃗⋅B⃗\vec A\cdot\vec B and ∣A⃗×B⃗∣\vert \vec A\times\vec B\vert ".
  • JEE Advanced: Dimensional formulas of unfamiliar constants (e.g. coefficient in a given empirical equation), trap problems combining dimensions with error analysis, vector geometry in 3D, scalar triple product interpretation.

Concept Map

  • Units & Measurement
    • SI base units (7), supplementary units (radian, steradian), derived units
    • Practical units (light-year, parsec, AU, atomic mass unit, electron-volt, bar, torr)
    • Conversion of units across systems
  • Dimensional Analysis
    • Dimensional formula of mechanical, thermal, electromagnetic quantities
    • Principle of homogeneity
    • Deducing formulae (pendulum, viscous force, drag, Bohr radius)
    • Limitations of dimensional analysis
  • Significant Figures & Error Analysis
    • Rules for significant figures and rounding
    • Absolute, relative and percentage error
    • Propagation in +,−,×,÷,xn+, -, \times, \div, x^n
    • Combination of independent errors (max & probable)
  • Vectors
    • Scalars vs vectors; equal, negative, null, unit, position, displacement
    • Triangle, parallelogram, polygon laws; addition, subtraction
    • Resolution into rectangular components in 2D and 3D
    • Direction cosines, direction ratios
  • Vector Products
    • Dot (scalar) product: definition, properties, geometric meaning
    • Cross (vector) product: definition, right-hand rule, area interpretation
    • Scalar triple product (volume), vector triple product (mention)

Topic 1: Units and Measurement

Sub-topic A: SI System

The SI system has 7 base units:

QuantityUnitSymbol
Lengthmetrem
Masskilogramkg
Timeseconds
Electric currentampereA
Thermodynamic temperaturekelvinK
Amount of substancemolemol
Luminous intensitycandelacd

Two supplementary units (now subsumed under derived in modern SI):

  • Plane angle: radian (rad)
  • Solid angle: steradian (sr)

A solid angle subtended at the centre of a sphere of radius rr by area AA on its surface is Ω=A/r2\Omega = A/r^2. The full sphere subtends 4π4\pi sr.

Sub-topic B: Practical units

  • 1 AU=1.496×10111\,\text{AU} = 1.496\times10^{11} m (mean Earth–Sun distance)
  • 1 light-year=9.46×10151\,\text{light-year} = 9.46\times10^{15} m
  • 1 parsec=3.08×10161\,\text{parsec} = 3.08\times10^{16} m =3.26= 3.26 ly
  • 1 u=1.66×10−271\,\text{u} = 1.66\times10^{-27} kg (atomic mass unit)
  • 1 eV=1.602×10−191\,\text{eV} = 1.602\times10^{-19} J
  • 1 bar=1051\,\text{bar} = 10^5 Pa, 1 atm=1.013×1051\,\text{atm} = 1.013\times10^5 Pa, 1 torr=133.31\,\text{torr} = 133.3 Pa

Sub-topic C: Conversion of units

A physical quantity has the same magnitude in all unit systems when expressed dimensionally; i.e.

n1[M1aL1bT1c]=n2[M2aL2bT2c]n_1 [M_1^a L_1^b T_1^c] = n_2 [M_2^a L_2^b T_2^c]

so

n2=n1(M1M2)a(L1L2)b(T1T2)c.n_2 = n_1 \left(\frac{M_1}{M_2}\right)^a \left(\frac{L_1}{L_2}\right)^b \left(\frac{T_1}{T_2}\right)^c.

Worked Examples (JEE Main level)

Example 1.1. Convert 11 joule into ergs using dimensional analysis.

[J]=ML2T−2[J] = M L^2 T^{-2}. In SI, M1=1M_1=1 kg, L1=1L_1=1 m, T1=1T_1=1 s. In CGS, M2=1M_2=1 g, L2=1L_2=1 cm, T2=1T_2=1 s.

n2=1⋅(1000 g1 g)1(100 cm1 cm)2(11)−2=103⋅104=107.n_2 = 1 \cdot \left(\frac{1000\,\text{g}}{1\,\text{g}}\right)^1 \left(\frac{100\,\text{cm}}{1\,\text{cm}}\right)^2 \left(\frac{1}{1}\right)^{-2} = 10^3 \cdot 10^4 = 10^7.

1 J=107 erg1\,\text{J} = 10^7\,\text{erg}.

Example 1.2. The density of mercury is 13.6 g/cm313.6\,\text{g/cm}^3. Convert it into kg/m³.

[ρ]=ML−3[\rho] = ML^{-3}. With M1M_1=g, L1L_1=cm and target M2M_2=kg, L2L_2=m:

n2=13.6⋅10−3 kg1 kg⋅(10−2 m1 m)−3=13.6⋅10−3⋅106=1.36×104 kg/m3.n_2 = 13.6 \cdot \frac{10^{-3}\,\text{kg}}{1\,\text{kg}} \cdot \left(\frac{10^{-2}\,\text{m}}{1\,\text{m}}\right)^{-3} = 13.6\cdot 10^{-3}\cdot 10^{6} = 1.36\times10^{4}\,\text{kg/m}^3.


Topic 2: Dimensional Analysis

Sub-topic A: Dimensional formulae of common quantities

QuantityDim. Formula
VelocityLT−1LT^{-1}
AccelerationLT−2LT^{-2}
ForceMLT−2MLT^{-2}
Energy / Work / TorqueML2T−2ML^2T^{-2}
PowerML2T−3ML^2T^{-3}
Pressure / Stress / Young's modulusML−1T−2ML^{-1}T^{-2}
Surface tensionMT−2MT^{-2}
Coefficient of viscosityML−1T−1ML^{-1}T^{-1}
Momentum / ImpulseMLT−1MLT^{-1}
Angular momentum / Planck's constant hhML2T−1ML^2T^{-1}
Frequency / Angular freq.T−1T^{-1}
Gravitational constant GGM−1L3T−2M^{-1}L^3T^{-2}
Permittivity ε0\varepsilon_0M−1L−3T4A2M^{-1}L^{-3}T^{4}A^{2}
Permeability μ0\mu_0MLT−2A−2MLT^{-2}A^{-2}
Electric fieldMLT−3A−1MLT^{-3}A^{-1}
Magnetic field BBMT−2A−1MT^{-2}A^{-1}
ResistanceML2T−3A−2ML^2T^{-3}A^{-2}

Sub-topic B: Principle of homogeneity

Every additive term and every side of a physical equation must have the same dimensions. This is used to:

  1. Check correctness of an equation.
  2. Convert units.
  3. Deduce relations among physical quantities (up to a dimensionless constant).

Sub-topic C: Deduction of formulae

Example: Period of a simple pendulum. Suppose TT depends on length ℓ\ell, mass mm, gravity gg:

T=k ℓambgc.T = k\,\ell^a m^b g^c.

Dimensionally: T1=LaMb(LT−2)c=MbLa+cT−2cT^1 = L^a M^b (LT^{-2})^c = M^b L^{a+c} T^{-2c}.

Equating exponents: b=0b=0, a+c=0a+c=0, −2c=1⇒c=−1/2-2c=1 \Rightarrow c=-1/2, a=1/2a=1/2. So T=kℓ/gT=k\sqrt{\ell/g}. (Experiment gives k=2πk=2\pi.)

Example: Stokes' viscous force. Assume FF on a sphere of radius rr moving with velocity vv in a fluid of viscosity η\eta is

F=k ηarbvc.F=k\,\eta^a r^b v^c.

[F]=MLT−2[F]=MLT^{-2}, [η]=ML−1T−1[\eta]=ML^{-1}T^{-1}, [r]=L[r]=L, [v]=LT−1[v]=LT^{-1}.

MLT−2=MaL−a+b+cT−a−c⇒a=1,  −a−c=−2⇒c=1,  −a+b+c=1⇒b=1.MLT^{-2}=M^a L^{-a+b+c}T^{-a-c}\Rightarrow a=1,\;-a-c=-2\Rightarrow c=1,\;-a+b+c=1\Rightarrow b=1. So F=kηrvF=k\eta r v, with k=6πk=6\pi from experiment.

Example: Bohr radius. Estimate the size of a hydrogen atom from hh, mem_e, ee, ε0\varepsilon_0:

a0=k ε0ahbmeced.a_0 = k\,\frac{\varepsilon_0^a h^b}{m_e^c e^d}.

A dimensional check yields a0=ε0h2πmee2≈0.529a_0 = \frac{\varepsilon_0 h^2}{\pi m_e e^2}\approx 0.529 Å.

Sub-topic D: Limitations

Dimensional analysis cannot:

  • determine dimensionless constants (π,e\pi, e, numbers like 2, 1/2),
  • distinguish quantities with the same dimensions (work vs torque, angular momentum vs Planck's constant),
  • handle equations containing functions like sin⁡x\sin x, exe^x, log⁡x\log x (their arguments must be dimensionless),
  • handle quantities depending on more than three mechanical quantities (M, L, T) — over-determined.

Worked Examples (JEE Main level)

Example 2.1. Find [a][a] and [b][b] in (P+aV2)(V−b)=RT\left(P+\dfrac{a}{V^2}\right)(V-b)=RT (van der Waals).

Since a/V2a/V^2 must have the dimensions of pressure: [a]=[P][V2]=ML−1T−2⋅L6=ML5T−2[a]=[P][V^2]= ML^{-1}T^{-2}\cdot L^6 = ML^{5}T^{-2}. And [b]=[V]=L3[b]=[V]=L^3.

Example 2.2. The dimensions of 1μ0ε0\dfrac{1}{\mu_0\varepsilon_0}.

[μ0ε0]=[1/c2]=T2L−2⇒[1/μ0ε0]=L2T−2[\mu_0\varepsilon_0]=[1/c^2]=T^2L^{-2}\Rightarrow[1/\mu_0\varepsilon_0]=L^2T^{-2} — square of speed. Confirms c=1/μ0ε0c=1/\sqrt{\mu_0\varepsilon_0}.

Example 2.3. The position of a particle is x=at+bt2−ct3x=at+bt^2-ct^3. Find dimensions of a,b,ca, b, c.

Each term has units of length: [a]=LT−1[a]=LT^{-1}, [b]=LT−2[b]=LT^{-2}, [c]=LT−3[c]=LT^{-3}.

Worked Examples (JEE Advanced level)

Example 2.A1. In a given system, mass is measured in units α\alpha kg, length in β\beta m and time in γ\gamma s. The unit of force in this system, expressed in newtons, is:

A force of 11 N has dimensions MLT−2MLT^{-2}. Number of new units of force in 11 N is n2=1⋅(1/α)⋅(1/β)⋅(1/γ)−2=γ2/(αβ)n_2 = 1\cdot(1/\alpha)\cdot(1/\beta)\cdot(1/\gamma)^{-2} = \gamma^2/(\alpha\beta). Hence 11 unit of the new force =αβ/γ2= \alpha\beta/\gamma^2 N.

Example 2.A2. The frequency of vibration of a stretched string depends on its length LL, tension TT and mass per unit length μ\mu. Derive the formula.

ν=kLaTbμc\nu = k L^a T^b \mu^c. [ν]=T−1[\nu]=T^{-1}, [L]=L[L]=L, [T]=MLT−2[T]=MLT^{-2}, [μ]=ML−1[\mu]=ML^{-1}.

T−1=La(MLT−2)b(ML−1)c=Mb+cLa+b−cT−2b.T^{-1}= L^a (MLT^{-2})^b (ML^{-1})^c = M^{b+c}L^{a+b-c}T^{-2b}.

So b+c=0b+c=0, a+b−c=0a+b-c=0, −2b=−1⇒b=1/2-2b=-1\Rightarrow b=1/2, c=−1/2c=-1/2, a=−1a=-1.

ν=kLTμ.\boxed{\nu = \frac{k}{L}\sqrt{\frac{T}{\mu}}}.

Experimentally k=1/2k=1/2 for the fundamental.


Topic 3: Significant Figures, Rounding and Error Analysis

Sub-topic A: Rules for significant figures

  1. All non-zero digits are significant.
  2. All zeros between non-zero digits are significant (1002→41002 \to 4 s.f.).
  3. Leading zeros are NOT significant (0.0023→20.0023\to 2 s.f.).
  4. Trailing zeros after a decimal point are significant (2.300→42.300\to 4 s.f.).
  5. Trailing zeros in an integer without a decimal point are ambiguous. Use scientific notation: 45004500 could be 2,32, 3 or 44 s.f.; write 4.500×1034.500\times10^3 for 44 s.f.
  6. In addition/subtraction, the result has the same number of decimal places as the operand with the fewest.
  7. In multiplication/division, the result has the same number of significant figures as the operand with the fewest.

Sub-topic B: Rounding rules

Round half to even (banker's rounding) is recommended; in NCERT-style: round-half-up is acceptable. 2.345→2.342.345 \to 2.34 or 2.352.35 depending on convention — JEE typically accepts either.

Sub-topic C: Errors — definitions

  • Absolute error Δx=∣xmeasured−xtrue∣\Delta x = \vert x_{\text{measured}} - x_{\text{true}}\vert . When the true value is unknown, use xˉ=1n∑xi\bar x = \tfrac1n\sum x_i and Δxi=∣xi−xˉ∣\Delta x_i = \vert x_i - \bar x\vert .
  • Mean absolute error Δx‾=1n∑∣Δxi∣\overline{\Delta x} = \tfrac1n\sum \vert \Delta x_i\vert .
  • Relative error =Δx‾/xˉ= \overline{\Delta x}/\bar x.
  • Percentage error =(Δx‾/xˉ)×100%= (\overline{\Delta x}/\bar x)\times 100\%.

Sub-topic D: Propagation of errors

Suppose A,BA, B have absolute errors ΔA,ΔB\Delta A, \Delta B.

Sum / Difference. Z=A±B⇒ΔZ=ΔA+ΔBZ = A\pm B \Rightarrow \Delta Z = \Delta A + \Delta B (errors add). Relative error in a difference can blow up if A≈BA\approx B — exam favourite trap.

Product / Quotient. Z=ABZ = A B or A/BA/B:

ΔZZ=ΔAA+ΔBB.\frac{\Delta Z}{Z} = \frac{\Delta A}{A} + \frac{\Delta B}{B}.

Power. Z=ApBq/CrZ = A^p B^q / C^r:

ΔZZ=∣p∣ΔAA+∣q∣ΔBB+∣r∣ΔCC.\frac{\Delta Z}{Z} = |p|\frac{\Delta A}{A} + |q|\frac{\Delta B}{B} + |r|\frac{\Delta C}{C}.

Derivation of the power rule. Take logs: ln⁡Z=pln⁡A+qln⁡B−rln⁡C\ln Z = p\ln A + q\ln B - r\ln C. Differentiate: dZ/Z=p dA/A+q dB/B−r dC/CdZ/Z = p\,dA/A + q\,dB/B - r\,dC/C. The maximum (worst-case) error replaces d→Δd\to \Delta and signs by ∣⋅∣\vert \cdot\vert , giving the formula above.

Worked Examples (JEE Main level)

Example 3.1. A physical quantity X=a2b3cdX = \dfrac{a^2 b^3}{c\sqrt{d}}. Given % errors Δa/a=1%\Delta a/a = 1\%, Δb/b=2%\Delta b/b = 2\%, Δc/c=3%\Delta c/c = 3\%, Δd/d=4%\Delta d/d = 4\%. Find the maximum % error in XX.

ΔXX=2(1%)+3(2%)+1(3%)+12(4%)=2+6+3+2=13%.\frac{\Delta X}{X} = 2(1\%) + 3(2\%) + 1(3\%) + \tfrac12(4\%) = 2+6+3+2 = 13\%.

Example 3.2. Two resistors of 100±3 Ω100\pm 3\,\Omega and 200±4 Ω200\pm 4\,\Omega are connected in series. Find equivalent resistance with absolute uncertainty.

R=300 ΩR = 300\,\Omega, ΔR=3+4=7 Ω\Delta R = 3+4 = 7\,\Omega.

Example 3.3. Same resistors in parallel.

Rp=(100)(200)/300=66.6‾ ΩR_p = (100)(200)/300 = 66.\overline 6\,\Omega. Using 1/Rp=1/R1+1/R21/R_p = 1/R_1+1/R_2, ΔRp/Rp2=ΔR1/R12+ΔR2/R22\Delta R_p/R_p^2 = \Delta R_1/R_1^2 + \Delta R_2/R_2^2 so

ΔRp=Rp2(31002+42002)=(66.7)2(3⋅10−4+10−4)≈1.78 Ω.\Delta R_p = R_p^2 \left(\frac{3}{100^2}+\frac{4}{200^2}\right) = (66.7)^2 (3\cdot10^{-4} + 10^{-4}) \approx 1.78\,\Omega.

Worked Examples (JEE Advanced level)

Example 3.A1. A screw gauge has pitch 0.50.5 mm and 5050 divisions on the circular scale. A wire reading is main scale 33 mm, circular scale 3535. Find the diameter and its uncertainty assuming a half-division least count.

Least count =0.5/50=0.01= 0.5/50 = 0.01 mm. Diameter =3+35×0.01=3.35= 3 + 35\times0.01 = 3.35 mm. Uncertainty ±0.005\pm 0.005 mm.

Example 3.A2. Young's modulus is determined from Y=FLAΔL=(Mg)L(πd2/4)ΔLY = \dfrac{FL}{A\Delta L} = \dfrac{(Mg) L}{(\pi d^2/4)\Delta L}. Given M=2.00±0.01M=2.00\pm0.01 kg, L=1.000±0.001L=1.000\pm0.001 m, d=0.500±0.005d=0.500\pm0.005 mm, ΔL=0.250±0.005\Delta L = 0.250\pm0.005 mm. Maximum % error in YY?

ΔYY=ΔMM+ΔLL+2Δdd+Δ(ΔL)ΔL\frac{\Delta Y}{Y} = \frac{\Delta M}{M} + \frac{\Delta L}{L} + 2\frac{\Delta d}{d} + \frac{\Delta(\Delta L)}{\Delta L}

=0.5%+0.1%+2(1%)+2%=4.6%= 0.5\% + 0.1\% + 2(1\%) + 2\% = 4.6\%.


Topic 4: Vectors — Algebra

Sub-topic A: Definitions

  • Scalar: magnitude only (mass, time, temperature, work, energy, power, pressure, charge, frequency, current).
  • Vector: magnitude + direction + obeys parallelogram law (displacement, velocity, acceleration, force, momentum, torque, angular velocity, electric field, magnetic field).
  • Trap: current is a scalar in spite of having a direction in a wire (it does not satisfy the parallelogram law at a junction).

A vector A⃗\vec A has magnitude ∣A⃗∣=A\vert \vec A\vert =A. Unit vector: A^=A⃗/A\hat A = \vec A / A.

Sub-topic B: Triangle and parallelogram laws

If R⃗=A⃗+B⃗\vec R = \vec A + \vec B with angle θ\theta between A⃗\vec A and B⃗\vec B:

R=A2+B2+2ABcos⁡θ.R = \sqrt{A^2+B^2+2AB\cos\theta}.

Direction (angle α\alpha of R⃗\vec R from A⃗\vec A):

tan⁡α=Bsin⁡θA+Bcos⁡θ.\tan\alpha = \frac{B\sin\theta}{A+B\cos\theta}.

Polygon law generalizes for any number of coplanar vectors taken head to tail; the closing side gives the resultant (with direction reversed).

Special cases:

  • θ=0⇒R=A+B\theta=0\Rightarrow R=A+B (max).
  • θ=180∘⇒R=∣A−B∣\theta=180^\circ\Rightarrow R=\vert A-B\vert (min).
  • θ=90∘⇒R=A2+B2\theta=90^\circ\Rightarrow R=\sqrt{A^2+B^2}.
  • If A=BA=B, R=2Acos⁡(θ/2)R=2A\cos(\theta/2), α=θ/2\alpha=\theta/2.

Sub-topic C: Resolution into components

In 2D: A⃗=Axı^+Ayȷ^\vec A = A_x\hat\imath + A_y\hat\jmath, where Ax=Acos⁡θA_x=A\cos\theta, Ay=Asin⁡θA_y=A\sin\theta.

In 3D: A⃗=Axı^+Ayȷ^+Azk^\vec A = A_x\hat\imath + A_y\hat\jmath + A_z\hat k. Magnitude A=Ax2+Ay2+Az2A=\sqrt{A_x^2+A_y^2+A_z^2}. Direction cosines:

cos⁡α=AxA,cos⁡β=AyA,cos⁡γ=AzA,cos⁡2α+cos⁡2β+cos⁡2γ=1.\cos\alpha=\frac{A_x}{A},\quad\cos\beta=\frac{A_y}{A},\quad\cos\gamma=\frac{A_z}{A},\quad \cos^2\alpha+\cos^2\beta+\cos^2\gamma=1.

Sub-topic D: Position and displacement vectors

For a particle at r⃗(t)=xı^+yȷ^+zk^\vec r(t)=x\hat\imath+y\hat\jmath+z\hat k with respect to origin OO, displacement between t1t_1 and t2t_2 is Δr⃗=r⃗(t2)−r⃗(t1)\Delta\vec r=\vec r(t_2)-\vec r(t_1).

Worked Examples (JEE Main level)

Example 4.1. Two forces of 55 N and 1212 N act at right angles. Find resultant and direction.

R=25+144=13R=\sqrt{25+144}=13 N. tan⁡α=12/5⇒α=67.4∘\tan\alpha = 12/5\Rightarrow \alpha=67.4^\circ from the 5 N force.

Example 4.2. If A⃗+B⃗=C⃗\vec A+\vec B = \vec C and A=3,B=4,C=5A=3, B=4, C=5, the angle between A⃗\vec A and B⃗\vec B is?

C2=A2+B2+2ABcos⁡θ⇒25=9+16+24cos⁡θ⇒cos⁡θ=0⇒θ=90∘C^2=A^2+B^2+2AB\cos\theta\Rightarrow 25=9+16+24\cos\theta\Rightarrow\cos\theta=0\Rightarrow\theta=90^\circ.

Example 4.3. A vector A⃗=2ı^+3ȷ^−6k^\vec A=2\hat\imath+3\hat\jmath-6\hat k. Find a unit vector parallel to A⃗\vec A.

∣A⃗∣=4+9+36=7\vert \vec A\vert =\sqrt{4+9+36}=7. A^=(2ı^+3ȷ^−6k^)/7\hat A=(2\hat\imath+3\hat\jmath-6\hat k)/7.

Worked Examples (JEE Advanced level)

Example 4.A1. Two vectors of equal magnitude AA have a resultant of magnitude A3A\sqrt 3. Find the angle between them.

3A2=2A2+2A2cos⁡θ⇒cos⁡θ=1/2⇒θ=60∘3A^2 = 2A^2 + 2A^2\cos\theta \Rightarrow \cos\theta = 1/2\Rightarrow \theta=60^\circ.

Example 4.A2. A particle is displaced from (2,1,3)(2,1,3) m to (−1,4,5)(-1,4,5) m. Find magnitude of displacement and its angle with the xx-axis.

Δr⃗=(−3,3,2)\Delta\vec r=(-3,3,2) m. ∣Δr⃗∣=9+9+4=22\vert \Delta\vec r\vert =\sqrt{9+9+4}=\sqrt{22} m. cos⁡α=−3/22⇒α≈129.8∘\cos\alpha=-3/\sqrt{22}\Rightarrow \alpha\approx129.8^\circ.


Topic 5: Dot Product and Cross Product

Sub-topic A: Dot product

A⃗⋅B⃗=ABcos⁡θ=AxBx+AyBy+AzBz.\vec A\cdot\vec B = AB\cos\theta = A_xB_x+A_yB_y+A_zB_z.

Properties:

  • Commutative: A⃗⋅B⃗=B⃗⋅A⃗\vec A\cdot\vec B=\vec B\cdot\vec A.
  • Distributive: A⃗⋅(B⃗+C⃗)=A⃗⋅B⃗+A⃗⋅C⃗\vec A\cdot(\vec B+\vec C)=\vec A\cdot\vec B+\vec A\cdot\vec C.
  • ı^⋅ı^=1\hat\imath\cdot\hat\imath=1, ı^⋅ȷ^=0\hat\imath\cdot\hat\jmath=0.
  • A⃗⋅A⃗=A2\vec A\cdot\vec A=A^2.
  • Component of A⃗\vec A along B⃗\vec B: A⃗⋅B^=Acos⁡θ\vec A\cdot\hat B = A\cos\theta.
  • Geometric: projection of one vector on another.

Applications: work W=F⃗⋅d⃗W=\vec F\cdot\vec d; power P=F⃗⋅v⃗P=\vec F\cdot\vec v; magnetic flux Φ=B⃗⋅A⃗\Phi=\vec B\cdot\vec A.

Sub-topic B: Cross product

A⃗×B⃗=ABsin⁡θ n^,\vec A\times\vec B = AB\sin\theta\,\hat n,

where n^\hat n is perpendicular to the plane of A⃗,B⃗\vec A, \vec B by the right-hand rule.

In Cartesian form,

A⃗×B⃗=∣ı^ȷ^k^AxAyAzBxByBz∣=(AyBz−AzBy)ı^−(AxBz−AzBx)ȷ^+(AxBy−AyBx)k^.\vec A\times\vec B=\begin{vmatrix}\hat\imath&\hat\jmath&\hat k\\A_x&A_y&A_z\\B_x&B_y&B_z\end{vmatrix}=(A_yB_z-A_zB_y)\hat\imath-(A_xB_z-A_zB_x)\hat\jmath+(A_xB_y-A_yB_x)\hat k.

Properties:

  • Anti-commutative: A⃗×B⃗=−B⃗×A⃗\vec A\times\vec B=-\vec B\times\vec A.
  • Distributive.
  • ı^×ȷ^=k^\hat\imath\times\hat\jmath=\hat k, ȷ^×k^=ı^\hat\jmath\times\hat k=\hat\imath, k^×ı^=ȷ^\hat k\times\hat\imath=\hat\jmath (cyclic).
  • A⃗×A⃗=0⃗\vec A\times\vec A=\vec 0.
  • ∣A⃗×B⃗∣\vert \vec A\times\vec B\vert = area of parallelogram with A⃗,B⃗\vec A, \vec B as adjacent sides; half of it = area of triangle.

Applications: torque τ⃗=r⃗×F⃗\vec\tau=\vec r\times\vec F; angular momentum L⃗=r⃗×p⃗\vec L=\vec r\times\vec p; magnetic force F⃗=qv⃗×B⃗\vec F=q\vec v\times\vec B.

Sub-topic C: Scalar and vector triple products

  • Scalar triple product: [A⃗ B⃗ C⃗]=A⃗⋅(B⃗×C⃗)[\vec A\,\vec B\,\vec C] = \vec A\cdot(\vec B\times\vec C) = volume of parallelepiped.
  • Cyclic invariance: [A⃗ B⃗ C⃗]=[B⃗ C⃗ A⃗]=[C⃗ A⃗ B⃗][\vec A\,\vec B\,\vec C]=[\vec B\,\vec C\,\vec A]=[\vec C\,\vec A\,\vec B].
  • [A⃗ B⃗ C⃗]=0⇔[\vec A\,\vec B\,\vec C]=0\Leftrightarrow coplanar.
  • Vector triple product: A⃗×(B⃗×C⃗)=(A⃗⋅C⃗)B⃗−(A⃗⋅B⃗)C⃗\vec A\times(\vec B\times\vec C)=(\vec A\cdot\vec C)\vec B-(\vec A\cdot\vec B)\vec C — the BAC-CAB rule.

Worked Examples (JEE Main level)

Example 5.1. A⃗=2ı^+ȷ^−2k^\vec A=2\hat\imath+\hat\jmath-2\hat k, B⃗=ı^−ȷ^+2k^\vec B=\hat\imath-\hat\jmath+2\hat k. Find A⃗⋅B⃗\vec A\cdot\vec B and the angle.

A⃗⋅B⃗=2−1−4=−3\vec A\cdot\vec B=2-1-4=-3. A=3, B=1+1+4=6A=3,\,B=\sqrt{1+1+4}=\sqrt 6. cos⁡θ=−3/(36)=−1/6⇒θ≈114∘\cos\theta=-3/(3\sqrt 6)=-1/\sqrt 6\Rightarrow\theta\approx114^\circ.

Example 5.2. Find a unit vector perpendicular to both A⃗=ı^−ȷ^\vec A=\hat\imath-\hat\jmath and B⃗=ȷ^+k^\vec B=\hat\jmath+\hat k.

A⃗×B⃗=∣ı^ȷ^k^1−10011∣=ı^(−1−0)−ȷ^(1−0)+k^(1−0)=−ı^−ȷ^+k^\vec A\times\vec B=\begin{vmatrix}\hat\imath&\hat\jmath&\hat k\\1&-1&0\\0&1&1\end{vmatrix}=\hat\imath(-1-0)-\hat\jmath(1-0)+\hat k(1-0)=-\hat\imath-\hat\jmath+\hat k. Magnitude 3\sqrt 3, so n^=(−ı^−ȷ^+k^)/3\hat n=(-\hat\imath-\hat\jmath+\hat k)/\sqrt 3.

Example 5.3. A constant force F⃗=2ı^+3ȷ^+4k^\vec F=2\hat\imath+3\hat\jmath+4\hat k N acts on a particle, displacing it from (1,1,1)(1,1,1) to (2,3,4)(2,3,4) m. Work done?

d⃗=(1,2,3)\vec d=(1,2,3) m. W=F⃗⋅d⃗=2+6+12=20W=\vec F\cdot\vec d=2+6+12=20 J.

Worked Examples (JEE Advanced level)

Example 5.A1. If ∣A⃗+B⃗∣=∣A⃗−B⃗∣\vert \vec A+\vec B\vert =\vert \vec A-\vec B\vert , show A⃗⊥B⃗\vec A\perp\vec B.

∣A⃗+B⃗∣2−∣A⃗−B⃗∣2=4A⃗⋅B⃗=0⇒A⃗⋅B⃗=0\vert \vec A+\vec B\vert ^2-\vert \vec A-\vec B\vert ^2=4\vec A\cdot\vec B=0\Rightarrow \vec A\cdot\vec B=0.

Example 5.A2. Given A⃗\vec A, B⃗\vec B with A=B=1A=B=1 and ∣A⃗×B⃗∣2+∣A⃗⋅B⃗∣2\vert \vec A\times\vec B\vert ^2 + \vert \vec A\cdot\vec B\vert ^2 equal to?

∣A×B∣2=A2B2sin⁡2θ\vert A\times B\vert ^2=A^2B^2\sin^2\theta; ∣A⋅B∣2=A2B2cos⁡2θ\vert A\cdot B\vert ^2=A^2B^2\cos^2\theta. Sum =A2B2=1=A^2B^2=1.

This identity, ∣A⃗×B⃗∣2+(A⃗⋅B⃗)2=A2B2\vert \vec A\times\vec B\vert ^2 + (\vec A\cdot\vec B)^2 = A^2 B^2, is called the Lagrange identity and frequently appears in JEE Advanced as a way to find an unknown angle.

Example 5.A3. Three vectors a⃗,b⃗,c⃗\vec a, \vec b, \vec c form a closed triangle, a⃗+b⃗+c⃗=0⃗\vec a+\vec b+\vec c=\vec 0. Show a⃗×b⃗=b⃗×c⃗=c⃗×a⃗\vec a\times\vec b=\vec b\times\vec c=\vec c\times\vec a.

From c⃗=−a⃗−b⃗\vec c=-\vec a-\vec b: b⃗×c⃗=b⃗×(−a⃗−b⃗)=−b⃗×a⃗−0⃗=a⃗×b⃗\vec b\times\vec c = \vec b\times(-\vec a-\vec b) = -\vec b\times\vec a -\vec 0=\vec a\times\vec b. Similarly c⃗×a⃗=a⃗×b⃗\vec c\times\vec a=\vec a\times\vec b. Geometrically the three give twice the area of the triangle.


Problem-Solving Heuristics

  1. Always write the dimensional formula in [MaLbTc][M^a L^b T^c] form — never carry units along symbols.
  2. In error problems, convert everything to percentages before adding; this avoids unit clutter.
  3. Use logarithmic differentiation for any product/quotient/power; never use repeated chain rule.
  4. Resolve along convenient axes — choose axes such that maximum number of forces/vectors lie along them.
  5. To find an angle between vectors, compute both A⃗⋅B⃗\vec A\cdot\vec B and ∣A⃗×B⃗∣\vert \vec A\times\vec B\vert ; their ratio is cot⁡θ\cot\theta — gives θ\theta without ambiguity in [0,π][0,\pi].
  6. If the question says "minimum number of vectors", remember three coplanar non-collinear vectors of any magnitudes can sum to zero, but you need at least four non-coplanar vectors in 3D.
  7. Dimensional analysis can give the dependence but never the dimensionless constant — JEE Advanced often hides π\pi, 2π2\pi, 1/21/2.

Common Traps & Mistakes

  • Confusing work (F⃗⋅d⃗\vec F\cdot\vec d, scalar) with torque (r⃗×F⃗\vec r\times\vec F, vector) — both have units of N⋅m\text{N·m} and dimension ML2T−2ML^2T^{-2}, but they are entirely different physical objects.
  • Trigonometric functions require dimensionless arguments. If y=Asin⁡(ωt+ϕ)y = A\sin(\omega t+\phi), both ωt\omega t and ϕ\phi are dimensionless.
  • In error analysis, when a quantity occurs in both numerator and denominator with the same power, errors do NOT cancel in the worst-case (maximum) formula — they add. They only cancel in the most-probable (RSS) formula.
  • Δx\Delta x vs dxdx: dxdx is infinitesimal; Δx\Delta x is finite uncertainty.
  • A⃗⋅B⃗=0\vec A\cdot\vec B=0 implies A⃗⊥B⃗\vec A\perp\vec B only if A⃗,B⃗≠0⃗\vec A,\vec B\neq\vec 0.
  • ı^×ȷ^=k^\hat\imath\times\hat\jmath=\hat k, not −k^-\hat k. Many students reverse this on the spot in panic.
  • A vector with zero magnitude is the null vector 0⃗\vec 0; it has arbitrary direction.

Quick Revision Card

  • 7 base SI units; supplementary radians & steradians.
  • Homogeneity: every term in a physical equation has same dimensions.
  • Conversion: n2=n1(M1/M2)a(L1/L2)b(T1/T2)cn_2 = n_1 (M_1/M_2)^a(L_1/L_2)^b(T_1/T_2)^c.
  • Error in X=ApBqCrX = A^p B^q C^r: ΔX/X=∣p∣ΔA/A+∣q∣ΔB/B+∣r∣ΔC/C\Delta X/X = \vert p\vert \Delta A/A + \vert q\vert \Delta B/B + \vert r\vert \Delta C/C.
  • R=A2+B2+2ABcos⁡θR=\sqrt{A^2+B^2+2AB\cos\theta}, tan⁡α=Bsin⁡θ/(A+Bcos⁡θ)\tan\alpha=B\sin\theta/(A+B\cos\theta).
  • Lagrange identity: ∣A⃗×B⃗∣2+(A⃗⋅B⃗)2=A2B2\vert \vec A\times\vec B\vert ^2+(\vec A\cdot\vec B)^2=A^2B^2.
  • BAC-CAB: A⃗×(B⃗×C⃗)=(A⃗⋅C⃗)B⃗−(A⃗⋅B⃗)C⃗\vec A\times(\vec B\times\vec C)=(\vec A\cdot\vec C)\vec B-(\vec A\cdot\vec B)\vec C.

Formula Sheet

ConceptFormula
Dot productA⃗⋅B⃗=ABcos⁡θ\vec A\cdot\vec B = AB\cos\theta
Cross productA⃗×B⃗=ABsin⁡θ n^\vec A\times\vec B = AB\sin\theta\,\hat n
Magnitude in 3DA=Ax2+Ay2+Az2A=\sqrt{A_x^2+A_y^2+A_z^2}
Resultant of two vectorsR=A2+B2+2ABcos⁡θR=\sqrt{A^2+B^2+2AB\cos\theta}
Direction of resultanttan⁡α=Bsin⁡θA+Bcos⁡θ\tan\alpha=\dfrac{B\sin\theta}{A+B\cos\theta}
Unit vectorA^=A⃗/A\hat A=\vec A/A
Projection of A⃗\vec A on B⃗\vec BA⃗⋅B^\vec A\cdot\hat B
Area of parallelogram∣A⃗×B⃗∣\vert \vec A\times\vec B\vert
Area of triangle12∣A⃗×B⃗∣\tfrac12\vert \vec A\times\vec B\vert
Volume (parallelepiped)∣A⃗⋅(B⃗×C⃗)∣\vert \vec A\cdot(\vec B\times\vec C)\vert
Conversion of unitsn2=n1(M1/M2)a(L1/L2)b(T1/T2)cn_2=n_1(M_1/M_2)^a(L_1/L_2)^b(T_1/T_2)^c
Max % error (product/quotient)∑∥pi∥(Δxi/xi)\sum\|p_i\|(\Delta x_i/x_i)
Lagrange identity∥A⃗×B⃗∥2+(A⃗⋅B⃗)2=A2B2\|\vec A\times\vec B\|^2+(\vec A\cdot\vec B)^2=A^2B^2
BAC-CABA⃗×(B⃗×C⃗)=(A⃗⋅C⃗)B⃗−(A⃗⋅B⃗)C⃗\vec A\times(\vec B\times\vec C)=(\vec A\cdot\vec C)\vec B-(\vec A\cdot\vec B)\vec C
Direction cosinescos⁡2α+cos⁡2β+cos⁡2γ=1\cos^2\alpha+\cos^2\beta+\cos^2\gamma=1
Period of pendulumT=2πℓ/gT=2\pi\sqrt{\ell/g}
Stokes' lawF=6πηrvF=6\pi\eta r v
Frequency of stretched stringν=12LT/μ\nu=\dfrac{1}{2L}\sqrt{T/\mu}

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