Vector Algebra

2.1 Core concepts

• A scalar has only magnitude (e.g. length, mass, time, speed, volume), while a vector has both magnitude and direction (e.g. displacement, velocity, force, momentum) (NCERT §10.1, p. 338).
• A directed line segment from initial point A to terminal point B is a vector denoted AB; its magnitude |AB| is the distance between A and B, and is never negative (NCERT §10.2, p. 339).
• For a point P(x, y, z), the position vector OP has magnitude |OP| = √(x² + y² + z²) with respect to origin O (NCERT §10.2 Position Vector, p. 339).
• The direction angles α, β, γ are the angles OP makes with the positive x-, y-, z-axes; l = cos α, m = cos β, n = cos γ are the direction cosines, and lr, mr, nr (proportional to l, m, n) are the direction ratios a, b, c (NCERT §10.2 Direction Cosines, p. 340).
• The identity l² + m² + n² = 1 always holds, but in general a² + b² + c² ≠ 1 (NCERT §10.2 Note, p. 341).
• Types of vectors — Zero vector (coincident initial/terminal points, no definite direction); Unit vector (magnitude 1, denoted â); Coinitial vectors (same initial point); Collinear vectors (parallel to the same line); Equal vectors (same magnitude and direction); Negative of a vector (same magnitude, opposite direction) (NCERT §10.3, p. 341).
• Free vectors — any vector may be displaced parallel to itself without changing its magnitude or direction; throughout this topic we deal with free vectors only (NCERT §10.3 Remark, p. 341).
• Triangle law of vector addition — if a girl moves A → B → C, then AB + BC = AC (NCERT §10.4, p. 343).
• Parallelogram law — if two vectors are represented by two adjacent sides of a parallelogram, their sum is the diagonal through their common point; the two laws are equivalent (NCERT §10.4, p. 344).
• Properties of vector addition — commutative (a + b = b + a) and associative ((a + b) + c = a + (b + c)); zero vector is the additive identity (NCERT §10.4 Properties, pp. 344–346).
• Multiplication by scalar λ — λa is collinear to a, has magnitude |λ||a|, and the same direction as a if λ > 0 or opposite direction if λ < 0; when λ = 1/|a|, λa gives the unit vector â = a/|a| (NCERT §10.5, pp. 346–347).
• Component form — any vector r with terminal point P(x, y, z) is r = xî + yĵ + zk̂, where x, y, z are scalar components and |r| = √(x² + y² + z²) (NCERT §10.5.1, pp. 347–348).
• Two vectors a = a₁î + a₂ĵ + a₃k̂ and b = b₁î + b₂ĵ + b₃k̂ are equal iff a₁ = b₁, a₂ = b₂, a₃ = b₃; they are collinear iff b₁/a₁ = b₂/a₂ = b₃/a₃ = λ (NCERT §10.5.1, pp. 348–349).
• Vector joining two points — P₁P₂ = (x₂ − x₁)î + (y₂ − y₁)ĵ + (z₂ − z₁)k̂ with magnitude √((x₂ − x₁)² + (y₂ − y₁)² + (z₂ − z₁)²) (NCERT §10.5.2, p. 351).
• Section formula (internal) — point R dividing PQ internally in ratio m:n has position vector (mb + na)/(m + n); (external) — (mb − na)/(m − n); midpoint = (a + b)/2 (NCERT §10.5.3, pp. 352–353).
• Scalar (dot) product — a · b = |a||b| cos θ, where θ is the angle between a and b; result is a scalar (NCERT §10.6.1, p. 355).
• Dot-product observations — a · b = 0 iff a ⊥ b (for nonzero vectors); a · a = |a|²; commutative (a · b = b · a); distributive over addition; î · î = ĵ · ĵ = k̂ · k̂ = 1 and î · ĵ = ĵ · k̂ = k̂ · î = 0 (NCERT §10.6.1, pp. 356–357).
• Component form of dot product — a · b = a₁b₁ + a₂b₂ + a₃b₃; angle cos θ = (a · b)/(|a||b|) (NCERT §10.6.1, p. 357).
• Projection of a vector — projection of a on a directed line with unit vector p̂ is a · p̂; projection of a on vector b is (a · b)/|b| (NCERT §10.6.2, pp. 357–358).
• Vector (cross) product — a × b = |a||b| sin θ n̂, where n̂ is a unit vector perpendicular to both a and b such that (a, b, n̂) forms a right-handed system; result is a vector (NCERT §10.6.3, p. 363).
• Cross-product observations — a × b = 0 iff a ∥ b (for nonzero vectors); not commutative — a × b = − b × a (anti-commutative); distributive over addition; î × î = ĵ × ĵ = k̂ × k̂ = 0; î × ĵ = k̂, ĵ × k̂ = î, k̂ × î = ĵ (NCERT §10.6.3, pp. 363–365).
• Area applications — area of triangle with adjacent sides a, b is ½|a × b|; area of parallelogram with adjacent sides a, b is |a × b| (NCERT §10.6.3 Observations 8–9, p. 365).
• Determinant form — a × b = | î ĵ k̂ ; a₁ a₂ a₃ ; b₁ b₂ b₃ |; angle via sin θ = |a × b|/(|a||b|) (NCERT §10.6.3, pp. 363, 366).

2.2 Definitions to memorise

2.3 Diagrams / processes to remember

• Fig 10.1 (p. 339): Directed line, directed line segment AB showing magnitude with arrow direction.
• Fig 10.2 (p. 340): Three-dimensional right-handed rectangular coordinate system showing position vectors of points A, B, C with respect to origin O.
• Fig 10.3 (p. 340): Direction angles α, β, γ of position vector OP with x-, y-, z-axes; right-angled triangles OAP, OBP, OCP used to derive l = x/r, m = y/r, n = z/r.
• Fig 10.7–10.8 (pp. 343–344): Triangle law and polygon-extension for vector addition; subtraction as addition of negative.
• Fig 10.9–10.10 (pp. 344–345): Parallelogram law of vector addition; proof of commutativity via parallelogram ABCD.
• Fig 10.13–10.14 (pp. 347–348): Unit vectors î, ĵ, k̂ along OX, OY, OZ; component-form derivation by dropping perpendicular foot P₁.
• Fig 10.16–10.17 (pp. 352–353): Internal and external section formula construction.
• Fig 10.19 (p. 356): Angle θ between two vectors for scalar product.
• Fig 10.20 (p. 357): Projection vector AC for the four cases 0 < θ < 90°, 90° < θ < 180°, 180° < θ < 270°, 270° < θ < 360°.
• Fig 10.22–10.24 (pp. 363–364): Right-hand rule for cross product; orientation of î, ĵ, k̂ in a right-handed system.
• Fig 10.26–10.27 (p. 365): Area of triangle and parallelogram in terms of |a × b|.

2.5 Key formulas & theorems

2.6 Solved examples (NCERT-grounded)

Example A (NCERT Example 14, p. 359). Angle between a = î + ĵ − k̂ and b = î − ĵ + k̂.

Step 1 — dot product: 1·1 + 1·(−1) + (−1)·1 = −1. Step 2 — magnitudes: |a| = |b| = √3. Step 3 — cosine: cos θ = −1/3 ⇒ θ = cos⁻¹(−1/3).

Example B (NCERT Example 24, p. 367). Area of triangle with vertices A(1,1,1), B(1,2,3), C(2,3,1).

Step 1 — vectors: AB = (0,1,2); AC = (1,2,0). Step 2 — cross product: AB × AC = (1·0 − 2·2, 2·1 − 0·0, 0·2 − 1·1) = (−4, 2, −1); magnitude = √21. Step 3 — area: ½ · √21 = (1/2)√21 sq units.

Example C (NCERT Exercise 10.3 Q4, p. 361). Projection of a = î + 3ĵ + 7k̂ on b = 7î − ĵ + 8k̂.

Step 1 — dot product: 7 − 3 + 56 = 60. Step 2 — |b|: √(49 + 1 + 64) = √114. Step 3 — projection: 60/√114.

Example D (NCERT Exercise 10.2 Q14, p. 354). Show î + ĵ + k̂ is equally inclined to axes.

Step 1 — magnitude: √3. Step 2 — DCs: (1/√3, 1/√3, 1/√3). Step 3 — conclude: DCs equal ⇒ direction angles equal ⇒ equally inclined.

Example E (Internal section formula). Find R dividing P (a = î + 2ĵ + 3k̂) and Q (b = 4î + 5ĵ + 6k̂) in ratio 2:1.

Step 1 — apply formula: R = (2·b + 1·a)/(2 + 1). Step 2 — numerator: 2(4î + 5ĵ + 6k̂) + (î + 2ĵ + 3k̂) = 9î + 12ĵ + 15k̂. Step 3 — divide by 3: R = 3î + 4ĵ + 5k̂.

2.4 Common confusions / NTA trap points

• Students confuse direction cosines (l, m, n) with direction ratios (a, b, c); remember l² + m² + n² = 1 always, but a² + b² + c² ≠ 1 in general (NCERT §10.2 Note, p. 341).
• "Two collinear vectors are always equal in magnitude" — FALSE. Collinearity needs only parallelism, not equal magnitude (NCERT Exercise 10.1 Q5, p. 342).
• Internal vs external section formula sign — internal uses mb + na over m + n, while external uses mb − na over m − n. NTA likes to swap these (NCERT §10.5.3, pp. 352–353).
• Dot product is commutative (a · b = b · a), but cross product is anti-commutative (a × b = − b × a). A frequent distractor reverses this (NCERT §10.6.1 Observation 7, p. 356; §10.6.3 Observation 6, p. 364).
• The converse of a × b = 0 ⇒ a or b is zero is NOT true — a × b can vanish when a is parallel to b without either being zero (NCERT Exercise 10.4 Q8, p. 368).
• Projection of a on b is (a · b)/|b|, NOT (a · b)/|a|. Students often pick the wrong denominator (NCERT §10.6.2, p. 358).
• For triangle area, the factor is ½|a × b|, not |a × b| (which is for parallelogram) — easy to mix up (NCERT §10.6.3, p. 365).