Fundamentals
The core facts every aspirant should own — each a titled nugget with a real-world story, the concept in plain words, and a memory trick. Works even when the internet doesn't.
1,200 fundamentals
“The volume of a skewed box built from three edge vectors equals their scalar triple product.”
The scalar triple product a·(b×c) equals the signed volume of the parallelepiped formed by a, b, c. Cyclic permutations give the same value; swapping two vectors changes the sign. Zero indicates the three vectors lie in one plane.
[a b c]
Memory trick: a·(b×c)=volume; zero→coplanar.
“A flashlight beam hitting a wall shows the beam’s shadow length – that shadow is the projection of the beam onto the wall.”
The scalar projection of a on b equals (a·b)/|b|, giving the length of a along b’s direction. From a·b=|a||b|cosθ, dividing by |b| leaves |a|cosθ. In JEE, remember the denominator is |b|, not |b|² (that belongs to the vector projection).
(a·b)/|b|
Memory trick: Scalar proj = dot ÷ magnitude → ‘dot over mag’
“A ladder leaning against a wall forms angles with floor and wall; its direction cosines describe its 3-D tilt.”
Direction cosines l, m, n are cos α, cos β, cos γ where α,β,γ are the angles a line makes with the x, y, z axes. They always obey l²+m²+n²=1. Mistake: treating them as simple ratios without the cosine sign or forgetting the unit-vector condition.
Memory trick: L²+M²+N²=1
“When two book covers intersect, the angle you feel between them is the same as the angle between their normal arrows.”
The angle between two planes equals the angle between their normal vectors. Since normals are perpendicular to the planes, rotating one plane to the other is equivalent to rotating its normal to the other’s. Remember to take the acute angle; using the supplement gives a wrong answer.
Memory trick: Use normals; take acute θ
“Flipping a sheet of music notes over its diagonal swaps the staff lines, just like transposing swaps rows and columns.”
The transpose Aᵀ of a matrix A is obtained by interchanging its rows with columns; element aᵢⱼ becomes aⱼᵢ. This operation is useful for dot products and symmetry checks. Remember that (Aᵀ)ᵀ = A.
Memory trick: Rows ↔ columns, so (i,j)→(j,i)
“A perfectly still lake reflects the sky unchanged (symmetric), whereas a rotating fan blade gives a reversed image (skew-symmetric).”
A matrix is symmetric if Aᵀ = A, meaning it equals its mirror image across the main diagonal. It is skew-symmetric if Aᵀ = –A, which forces all diagonal entries to be zero. In JEE, verify symmetry by checking aᵢⱼ=aⱼᵢ and skewness by aᵢⱼ=–aⱼᵢ.
Memory trick: Sym: aij = aji; Skew: aij = –aji, diag 0
“A GPS map corrects coordinates only if its transformation matrix has a non-zero determinant.”
For a square matrix A, A⁻¹ = adj(A)/det(A). The inverse exists only when det(A) ≠ 0. Compute adjugate by cofactors and transpose. A common trap is to forget that a negative determinant still yields an inverse, only the sign of each entry changes.
A⁻¹=adjA/detA
Memory trick: A⁻¹ = adj/det, det≠0.
“In a 3-wire circuit, Cramer’s rule gives each current by swapping the voltage column into the conductance matrix.”
Cramer’s rule solves Ax = b by x_i = det(A_i)/det(A), where A_i replaces column i of A with vector b. It works only if det(A) ≠ 0; otherwise the system is either dependent or inconsistent. The exam trap is using the rule when the main determinant vanishes.
Memory trick: x_i = det(replace i)/det(main).
“The area of a triangular garden plot equals half the absolute value of the 3×3 determinant of its corner coordinates.”
For vertices (x1,y1), (x2,y2), (x3,y3), area = ½ | x1(y2−y3)+x2(y3−y1)+x3(y1−y2) | = ½ |det| of the 3×3 matrix with a column of 1’s. Forgetting the absolute value can give a negative area; ordering of points only changes sign.
Memory trick: Area = |det|/2 for 3 points.
“Rolling a die, the chance of getting a 2 or any even number uses the addition rule.”
For any two events, P(A∪B)=P(A)+P(B)−P(A∩B). Adding the separate probabilities counts the overlap twice, so subtract it once. Forgetting the subtraction can produce a probability larger than 1, which is impossible.
P(A∪B)=P(A)+P(B)−P(A∩B)
Memory trick: Add A and B, then subtract their overlap
“Two independent coin tosses both landing heads have probability (1/2)*(1/2)=1/4.”
When events are independent, the joint probability equals the product of the individual probabilities: P(A∩B)=P(A)P(B). If dependence exists, the product rule does not hold and you must use P(A|B)P(B). Applying multiplication blindly is a frequent error.
Memory trick: Indep events → product of probabilities
“Counting heads when flipping a fair coin 10 times is a binomial experiment.”
The binomial distribution gives the probability of exactly k successes in n independent trials with constant success probability p. It follows from multiplying p^k (1-p)^(n-k) and the combination nCk. Remember the trial count must be fixed; using it for dependent events is a common error.
Memory trick: B-n-p: Binomial needs fixed n and constant p.
“The average winnings in a game of rolling a die and receiving the face value is the expected value.”
Expected value E(X)= Σ x_i·P(x_i) is the long-run average of a random variable. It works because each outcome contributes proportionally to its chance. A trap: mixing up sample mean with expected value; use probabilities, not observed frequencies.
Memory trick: E = Σ(value × chance).
“When you slide a set of parallel contour lines over a map, the first touch point on the feasible region is where the optimum lies.”
A linear objective reaches its extreme on the boundary of the feasible region. Usually the extreme occurs at a vertex, but if the objective line is parallel to a constraint edge, every point on that edge is optimal; a constant objective makes the whole region optimal.
Memory trick: Check vertices; if objective line aligns with an edge, the whole edge is optimal.
“Like climbing a ladder, proving the first rung and then showing each step leads to the next secures the whole climb.”
Induction shows P(n) holds for all natural n by (1) proving P(1) (base) and (2) assuming P(k) true to prove P(k+1) (inductive step). The hypothesis applies only to a single k, not all n. Skipping the base case or using an unjustified jump in the inductive step is a common exam trap.
Memory trick: Base = first rung; if you can move from any rung to the next, you can reach any height.
“|x|<a is like a person staying within a fence a meters from the house; |x|>a is stepping outside the fence.”
|x|<a (a>0) means –a<x<a, a single open interval; |x|>a means x<–a or x>a, two disjoint rays. The direction of inequality flips when dividing by a negative inside absolute value. In JEE, watch the strict vs. non-strict signs; ≤ gives closed intervals.
Memory trick: '<' gives interval, '>' gives two rays.
“A metronome’s tick repeats every beat, just like a periodic function repeats its values each period.”
A periodic function satisfies f(x+T)=f(x) for all x, where T>0 is called the period. The smallest such T is the fundamental period; any integer multiple also works. In JEE problems you must pick the least T, otherwise you may over-estimate the answer.
Memory trick: Fundamental period = smallest T with f(x+T)=f(x).
“A traffic light shows red, yellow, green – the signum function tells if a number is negative, zero, or positive.”
sgn(x) = -1 for x<0, 0 for x=0, and +1 for x>0. It is a piecewise constant function used to capture the sign of a number. A common trap is forgetting that sgn(0)=0, so sgn²(x) equals 1 only for x≠0.
Memory trick: sgn: negative → -1, zero → 0, positive → +1.
“When two speakers emit sound waves at angles A and B, the combined pressure can be rewritten as two sine waves with angles (A+B) and (A-B).”
2 sin A cos B = sin(A+B)+sin(A-B). It follows from the sine addition formulas by multiplying and rearranging. A common trap is dropping the factor 2 or ignoring that sin(A-B)=-sin(B-A) when B>A, which changes the sign.
Memory trick: 2 sin A cos B → two sines (A+B, A-B).
“A swinging pendulum with horizontal drive a and vertical drive b reaches its peak displacement equal to √(a²+b²).”
The expression a sinθ + b cosθ can be written as R sin(θ+φ) where R=√(a²+b²); thus its maximum magnitude is ±R. The trap is to think the max is a+b or |a|+|b| – only the resultant vector length matters.
√(a²+b²)
Memory trick: R is hypotenuse of a-b right triangle.
“A swing’s rope at the midpoint of its swing makes angle θ/2, and the height rise R(1-cosθ) follows from the half-angle identity.”
The half-angle identities are sin(θ/2)=±√[(1-cosθ)/2] and cos(θ/2)=±√[(1+cosθ)/2]; the sign is chosen from the quadrant of θ/2. They are derived from the double-angle formulas. A frequent error is to drop the ± and always take the positive root, giving wrong values outside the first quadrant.
Memory trick: ½θ: sin → √[(1-cosθ)/2]; cos → √[(1+cosθ)/2]; keep the ± sign.
“Like points of a pizza cut into equal slices, the n roots lie on a circle equally spaced.”
A non-zero complex number z = re^(iθ) has exactly n distinct nth roots: r^(1/n) e^(i(θ+2kπ)/n), k=0…n−1. They sit on a circle of radius r^(1/n) and are separated by 2π/n radians. Missing the 2kπ term loses some roots.
Memory trick: Root angles = (θ+2kπ)/n, step = 360°/n
“When two rotating vectors combine, the resulting vector’s direction is the sum of their angles.”
If z₁ = r₁e^(iα) and z₂ = r₂e^(iβ), then arg(z₁z₂)=α+β (mod 2π). Add the angles and, if the sum exceeds 2π, subtract 2π. A common trap is forgetting the modulo-2π reduction.
Memory trick: Multiply → add angles, wrap at 360°
“The brightness of a fading LED that dims linearly while its power supply decays exponentially follows an arithmetico-geometric pattern.”
An arithmetico-geometric progression has term Tₖ=(a+(k-1)d)·r^(k-1), product of an AP term and a GP term. Sum is found by multiplying the series by r and subtracting to cancel. Trap: using pure AP or GP sum formulas gives wrong result.
Memory trick: AGP = (AP term)×(GP term)