NCERT Plus Two (Class 12) Mathematics • Chapter 10

Vector Algebra

Scalars vs vectors, position vectors, unit vectors & direction cosines, vector addition & section formula, dot product & projection, and cross product & geometric areas.

📄 Official NCERT Textbook PDF

Topic 10.2 – 10.3: Scalars, Position Vectors, Unit Vectors & Direction Cosines

Tell a Kochi boat captain "Sail 10 km!" and they'll stare blankly — 10 km WHERE?! North, West, or straight into the Arabian Sea? Quantities with ONLY size (temperature, mass, speed, time) are Scalars. Quantities carrying BOTH magnitude + direction (displacement, velocity, force) are Vectors — drawn as directed arrows starting from the origin O(0,0,0) to any point P(x, y, z)!
POSITION VECTOR & LENGTHOP = r = x î + y ĵ + z k̂
|r| = √(x² + y² + z²)
UNIT VECTOR (LENGTH = 1)â = a|a|
Vector of length k along a = k â!
DIRECTION COSINES (l, m, n)l = cos α = x/r, m = y/r, n = z/r
l² + m² + n² = 1!
🧠 Memory Trick: Unit vector = a / |a|, and its coefficients ARE the Direction Cosines (l, m, n)! Check DCs in 2 seconds: l² + m² + n² MUST equal 1!
💡 Simple Kid-Check: Think of Direction Ratios (a, b, c) = (x, y, z) as raw GPS coordinates like (1, 2, 2). Divide each coordinate by total distance r = √(1² + 2² + 2²) = 3 to shrink the arrow down to length 1 — those normalized numbers (1/3, 2/3, 2/3) are the exact Direction Cosines!
  • 1
    Types of Vectors: Zero vector 0 (length 0, arbitrary direction), Unit vector (length 1), Coinitial (same tail), Collinear (parallel to same line — even opposite arrows!), Equal (identical length AND direction).
  • 2
    Unit Vector Formula: Divide vector by its own length: â = a|a|.
  • 3
    DCs from Angles: If r makes angles α, β, γ with +X, +Y, +Z axes, l = cos α, m = cos β, n = cos γ and cos²α + cos²β + cos²γ = 1.
🧭 NCERT Ex 7 & 9 Classic: Given a = î + 2ĵ + 3k̂, find its Direction Cosines AND a vector of magnitude 14 along a.
• Magnitude: |a| = √(1² + 2² + 3²) = √14. Direction Ratios: 1, 2, 3.
• Direction Cosines: (1√14, 2√14, 3√14). (Check: 1+4+914 = 1 ✓)
• Vector of length 14 along a: 14 · â = √14(î + 2ĵ + 3k̂)!
Watch THREE classic 1-mark traps! (1) Mixing up Direction Ratios (1, 2, 3) with Direction Cosines (1/√14, 2/√14, 3/√14) — DCs MUST satisfy l² + m² + n² = 1! (2) Thinking two vectors pointing in OPPOSITE directions (like a and −2a) are NOT collinear — in NCERT, parallel vectors in EITHER direction ARE collinear! (3) Remembering that two vectors with equal magnitudes (|a| = |b|) need NOT be equal vectors if they point differently!

Topic 10.4 – 10.5: Vector Addition Laws, Components (î, ĵ, k̂) & Section Formula

Walk from Hostel A to Canteen B (AB), then from Canteen B to Class C (BC). Your net displacement from Hostel to Class is the shortcut arrow AC = AB + BC! That head-to-tail relay is the Triangle Law — and walking all the way back to Hostel A closes the triangle loop to 0!
TRIANGLE & PARALLELOGRAMAB + BC = AC
Closed loop: AB + BC + CA = 0!
VECTOR JOINING P → QPQ = OQ − OP (Head minus Tail!)
(x₂−x₁)î + (y₂−y₁)ĵ + (z₂−z₁)k̂
SECTION FORMULA (m : n)Internal: mb + nam + n  |  Mid: a + b2
External: mb − nam − n
🧠 Memory Trick: For PQ, remember "End minus Start" (Q − P)! And in Section Formula, cross-multiply weights: m hugs the FAR vector b, n hugs a!
💡 Simple Kid-Check: Adding vectors in î, ĵ, k̂ component form is as easy as counting apples, bananas, and mangoes separately — add î with î, ĵ with ĵ, and k̂ with k̂!
  • 1
    Triangle vs Parallelogram: Put vectors head-to-tail for Triangle Law; put both tails at one corner for Parallelogram Law (diagonal = a + b).
  • 2
    Vector Between Two Points: From P(x₁,y₁,z₁) to Q(x₂,y₂,z₂), subtract Start from End: PQ = (x₂−x₁)î + (y₂−y₁)ĵ + (z₂−z₁)k̂.
  • 3
    Collinearity Test: Three points A, B, C (or two vectors a, b) are collinear if b = λa (same component ratios b₁/a₁ = b₂/a₂ = b₃/a₃)!
📍 NCERT Ex 10 & 11 Combo:
1. Vector from P(2, 3, 0) to Q(−1, −2, −4): PQ = (−1−2)î + (−2−3)ĵ + (−4−0)k̂ = −3î − 5ĵ − 4k̂.
2. Point R dividing P(a) and Q(b) in ratio 2 : 1:
   • Internally: 2b + 1a2 + 1 = a + 2b3  |  Externally: 2b − 1a2 − 1 = 2b − a!
Watch TWO classic traps! (1) Subtracting P − Q instead of Q − P when finding PQ (reverses every sign!). (2) Famous Triangle Loop MCQ (Ex 10.2 Q18): In ΔABC, AB + BC + CA = 0 and AB + BC − AC = 0 are TRUE, whereas AB + BC + AC = 0 is FALSE (it equals 2AC)!

Topic 10.6.1: Scalar (Dot) Product (a·b = |a||b|cos θ), Angle & Orthogonal Test

Imagine pulling a heavy school bag across the floor using a tilted strap at angle θ! Only the forward part of your pull (|F| cos θ) helps move the bag forward; pulling straight up at 90° gives ZERO forward push (cos 90° = 0)! That teamwork score between two vectors is the Scalar (Dot) Product — outputting one plain number!
DOT PRODUCT (OUTPUTS SCALAR!)a · b = |a||b| cos θ
= a₁b₁ + a₂b₂ + a₃b₃
PERPENDICULAR (⊥) TESTa ⊥ b ⇔ a · b = 0
Angle: cos θ = a · b|a||b|
BASIS & SELF DOTî·î = ĵ·ĵ = k̂·k̂ = 1
î·ĵ = ĵ·k̂ = k̂·î = 0  |  a·a = |a|²
🧠 Memory Trick: Dot loves twins (î·î = 1), kills strangers (î·ĵ = 0)! Whenever you see |a + b| or a + b + c = 0, SQUARE IT using |v|² = v · v!
💡 Simple Kid-Check: Dot product is a "Teamwork Meter"! Positive dot (θ < 90°) = pulling together; Zero dot (θ = 90°) = perpendicular (zero overlap!); Negative dot (θ > 90°) = pulling against each other! And notice: a · b = b · a (order doesn't matter!).
Super-Useful Squaring Identities:
• |a ± b|² = |a|² + |b|² ± 2(a · b)
• (a + b) · (a − b) = |a|² − |b|²  |  If |a + b| = |a − b| ⇒ 4(a·b) = 0 ⇒ a ⊥ b!
📐 Two High-Yield Board Examples:
1. Angle between a = î + ĵ − k̂ and b = î − ĵ + k̂:
   a·b = 1(1) + 1(−1) + (−1)(1) = −1; |a| = |b| = √3 ⇒ cos θ = −13 ⇒ θ = cos⁻¹(−13)!
2. If a, b, c are unit vectors with a + b + c = 0, find a·b + b·c + c·a:
   Square |a+b+c|² = 0 ⇒ 1 + 1 + 1 + 2(a·b + b·c + c·a) = 0 ⇒ −32!
Watch THREE deadly traps! (1) Leaving î, ĵ, k̂ inside a dot product answer — a·b is a pure number, NEVER a vector! (2) Thinking a·b = 0 forces a = 0 or b = 0 — NO, non-zero perpendicular vectors ALSO give 0! (3) NCERT Ex 10.3 Q16: a·b ≥ 0 only when 0 ≤ θ ≤ π/2 (acute/right, NOT (0, π)!).

Topic 10.6.2: Projection of One Vector on Another

Stand under midday sun holding a tilted bamboo pole a above a flat road pointing along b. Look straight down at the shadow a casts onto the road! That signed shadow length along b is the Scalar Projection of a on b!
SCALAR PROJECTION OF a ON ba · b|b| = a · b̂
(= |a| cos θ, signed number!)
PROJECTION VECTOR OF a ON bMultiply scalar projection by b̂:
(a · b|b|²) b
ORDER MATTERS!Projection of b on a divides by |a|, NOT |b|!
🧠 Memory Trick: Divide by the BOTTOM ("ON") vector's length! Think of b as the bottom floor catching the shadow: Projection of a ON b = (a · b) / |b|!
💡 Simple Kid-Check: Drop a perpendicular straight down from the tip of arrow a onto the line of arrow b. The right triangle gives shadow length = |a| cos θ = |a||b| cos θ|b| = a · b|b|!
  • 1
    Compute the numerator dot product a · b (same for both directions!).
  • 2
    Find magnitude of ONLY the target vector after the word "on" (so |b| for "on b").
  • 3
    Divide a · b|b|. If the question says "projection vector", multiply once more by unit vector b̂ = b/|b|!
☀️ NCERT Example 16 & Ex 10.3 Q4: Find the projection of a = î + 3ĵ + 7k̂ on b = 7î − ĵ + 8k̂.
• Dot product: a · b = 1(7) + 3(−1) + 7(8) = 7 − 3 + 56 = 60.
• Magnitude of floor vector b: |b| = √(7² + (−1)² + 8²) = √(49 + 1 + 64) = √114.
• Projection = 60√114!
Watch TWO sneaky 1–2 mark traps! (1) Dividing by |a| instead of |b|! Always look at which vector comes AFTER "on": Projection of a ON b puts |b| in the denominator! (2) If a · b = 0 (e.g., î − ĵ on î + ĵ), the projection is 0 (perpendicular stick casts a zero shadow at noon!).

Topic 10.6.3: Vector (Cross) Product (a × b), Right-Hand Rule & sin θ

Try tightening a bicycle wheel nut with a spanner a by applying force b! Pulling straight out along the handle gives ZERO spin (sin 0° = 0), whereas pushing sideways at 90° drives the bolt straight IN or OUT perpendicular to BOTH the handle and your push! That twisting 3D perpendicular arrow is the Vector (Cross) Product a × b!
CROSS PRODUCT (VECTOR!)a × b = |a||b| sin θ n̂
a × b = −(b × a) (Flips sign!)
PARALLEL (∥) & SELF TESTa ∥ b ⇔ a × b = 0
a × a = 0  |  î×î = ĵ×ĵ = k̂×k̂ = 0
CYCLIC CLOCK & UNIT ⊥î×ĵ = k̂, ĵ×k̂ = î, k̂×î = ĵ (reverse = −)
Unit ⊥ to both: ± a × b|a × b|
🧠 Memory Trick: Cross Clock î → ĵ → k̂ → î: Forward around circle = +, Backward (î × k̂) = −ĵ! Twins vanish (î × î = 0)! And to compute in components, build a 3×3 determinant with Row 1 = î, ĵ, k̂!
💡 Simple Kid-Check: Point your right-hand fingers along a and curl them toward b — your thumb shoots straight out perpendicular to the page along n̂! Flipping order to b × a forces your thumb to flip 180° backward (−)!
Component Determinant & Lagrange's Identity:
• a × b = îĵk̂a₁a₂a₃b₁b₂b₃ = (a₂b₃−a₃b₂)î − (a₁b₃−a₃b₁)ĵ + (a₁b₂−a₂b₁)k̂
• Lagrange's Identity: |a × b|² + (a · b)² = |a|² |b|² (relates sin²θ + cos²θ = 1!)
🔧 NCERT Ex 23 & Ex 10.4 Q19:
1. Unit vector ⊥ to BOTH a = 2î + ĵ + 3k̂ and b = 3î + 5ĵ − 2k̂:
   Determinant gives a × b = (−2−15)î − (−4−9)ĵ + (10−3)k̂ = −17î + 13ĵ + 7k̂.
   Length = √(289 + 169 + 49) = √507 ⇒ Unit vector = ± −17î + 13ĵ + 7k̂√507!
2. When is |a · b| = |a × b|? |a||b|cos θ = |a||b|sin θ ⇒ tan θ = 1 ⇒ θ = π4!
Watch THREE classic traps! (1) Swapping rows 2 and 3 in the determinant — remember a × b ≠ b × a! (2) Classic NCERT Misc Q18: Find î·(ĵ×k̂) + ĵ·(î×k̂) + k̂·(î×ĵ). Watch the middle term! î × k̂ = −ĵ (reverse cycle!), so 1 + (−1) + 1 = 1 (NOT 3)! (3) Remember to divide a × b by |a × b| when asked for a unit perpendicular vector!

Topic 10.6.3: Geometric Areas of Triangle (½|a × b|) & Parallelogram

Tie two wooden sticks a and b at a hinge to frame a parallelogram kite! Base length is |a| and perpendicular height is |b| sin θ — multiplying base × height gives |a||b| sin θ, which is LITERALLY |a × b|! Slice the kite diagonally in half, and you get the area of a triangle: 12|a × b|!
PARALLELOGRAM (SIDES a, b)Area = |a × b|
(NO ½ when adjacent sides are given!)
PARALLELOGRAM (DIAGONALS d₁, d₂)Area = 12 |d₁ × d₂|
(WITH ½ for diagonals!)
TRIANGLE ΔABCAdjacent sides: 12|a × b|
Vertices A,B,C: 12|AB × AC|
🧠 Memory Trick: Adjacent sides Parallelogram = Full Kite (|a × b|, NO ½); Triangle OR Diagonals Parallelogram = Half Kite (½ |... × ...|, WITH ½)!
💡 Simple Kid-Check: Every triangle is half of a parallelogram, so its area has ½! Why does a parallelogram given by diagonals d₁ = a + b and d₂ = b − a ALSO wear a ½? Because expanding (a + b) × (b − a) = 2(a × b) — doubling the cross product, so ½ brings it right back to |a × b|!
Shape & What Is GivenExact Area FormulaWatch For
Parallelogram (adjacent sides a, b)|a × b|No ½ factor!
Parallelogram (diagonals d₁, d₂)12 |d₁ × d₂|Include ½ factor!
Triangle (adjacent sides a, b)12 |a × b|Include ½ factor!
Triangle (3 vertices A, B, C)12 |AB × AC|Build edge vectors AB, AC FIRST!
📐 NCERT Example 24 & Ex 25:
1. Area of ΔABC with A(1,1,1), B(1,2,3), C(2,3,1):
   Edge vectors: AB = 0î + ĵ + 2k̂ and AC = î + 2ĵ + 0k̂.
   AB × AC = −4î + 2ĵ − k̂ ⇒ |AB × AC| = √(16 + 4 + 1) = √21 ⇒ Area = √212 sq. units!
2. Parallelogram with adjacent sides a = 3î + ĵ + 4k̂, b = î − ĵ + k̂:
   a × b = 5î + ĵ − 4k̂ ⇒ Area = √(25 + 1 + 16) = √42 sq. units!
Watch TWO HUGE BOARD TRAPS! (1) Given three vertex points A, B, C, NEVER take ½ |OA × OB|! You MUST first compute two edge vectors starting at the same vertex, like AB = B − A and AC = C − A! (2) In parallelogram questions, read carefully: "adjacent sides" means |a × b|, whereas "diagonals" means ½ |d₁ × d₂|!

Solved Previous Year Questions (Kerala DHSE & NCERT)

Q1 (Kerala HSE March • 2 Marks): Find a vector in the direction of a = 5î − ĵ + 2k̂ which has magnitude 8 units.

|a| = √(5² + (−1)² + 2²) = √30.
Unit vector: â = 5î − ĵ + 2k̂√30.
Required vector = 8 â = 8√30(5î − ĵ + 2k̂).

Q2 (Kerala HSE Say • 2 Marks): Find the direction ratios and direction cosines of the vector joining points A(1, 2, −3) and B(−1, −2, 1) directed from A to B.

AB = (−1−1)î + (−2−2)ĵ + (1−(−3))k̂ = −2î − 4ĵ + 4k̂.
• Direction Ratios: −2, −4, 4. Magnitude |AB| = √(4 + 16 + 16) = 6.
• Direction Cosines: (−2/6, −4/6, 4/6) = (−13, −23, 23).

Q3 (Kerala HSE March • 2 Marks): If a = 2î + 2ĵ + 3k̂, b = −î + 2ĵ + k̂, and c = 3î + ĵ are such that a + λb is perpendicular to c, find λ.

a + λb = (2 − λ)î + (2 + 2λ)ĵ + (3 + λ)k̂.
Perpendicular to c ⇒ (a + λb) · c = 0:
3(2 − λ) + 1(2 + 2λ) + 0 = 0 ⇒ 8 − λ = 0 ⇒ λ = 8.

Q4 (Kerala HSE Model • 2 Marks): Find the projection of the vector î + 3ĵ + 7k̂ on the vector 7î − ĵ + 8k̂.

Let a = î + 3ĵ + 7k̂ and b = 7î − ĵ + 8k̂.
a · b = 7 − 3 + 56 = 60; |b| = √(49 + 1 + 64) = √114.
Projection of a on b = a·b|b| = 60√114.

Q5 (Kerala HSE March • 3 Marks): Three vectors a, b, c satisfy a + b + c = 0. Evaluate μ = a·b + b·c + c·a if |a| = 3, |b| = 4, and |c| = 2.

Square |a + b + c|² = 0:
|a|² + |b|² + |c|² + 2(a·b + b·c + c·a) = 0
9 + 16 + 4 + 2μ = 0 ⇒ 29 + 2μ = 0 ⇒ μ = −292.

Q6 (Kerala HSE March • 3 Marks): Find a unit vector perpendicular to each of the vectors a + b and a − b, where a = 3î + 2ĵ + 2k̂ and b = î + 2ĵ − 2k̂.

u = a + b = 4î + 4ĵ, v = a − b = 2î + 4k̂.
u × v = 16î − 16ĵ − 8k̂ ⇒ |u × v| = √(256 + 256 + 64) = √576 = 24.
Unit ⊥ vector = ± 16î − 16ĵ − 8k̂24 = ± (23î − 23ĵ − 13k̂).

Q7 (NCERT / DHSE Say • 3 Marks): Using vectors, find the area of triangle ABC with vertices A(1, 1, 2), B(2, 3, 5), and C(1, 5, 5).

Build edge vectors from A: AB = î + 2ĵ + 3k̂, AC = 0î + 4ĵ + 3k̂.
AB × AC = (6 − 12)î − (3 − 0)ĵ + (4 − 0)k̂ = −6î − 3ĵ + 4k̂.
Area = 12|AB × AC| = 12√(36 + 9 + 16) = √612 sq. units.

Q8 (Kerala HSE Model • 2 Marks): Let |a| = 3, |b| = √23 and a × b is a unit vector. Find the angle between a and b.

Given |a × b| = 1:
|a||b| sin θ = 1 ⇒ 3 · √23 sin θ = 1 ⇒ sin θ = 1√2 ⇒ θ = π4 (or 45°).