📘 Algebra · Chapter MN10🎯 NDA Level : High Priority
Vector Algebra is a consistently high-scoring chapter in NDA Mathematics. Vectors have both magnitude and direction, making them the natural language of physics and geometry. For NDA, the focus is squarely on the three products — dot product, cross product, and scalar triple product — along with their geometric interpretations and applications.
📌 What to expect in NDA (based on 2022–2024 papers): (1) Finding unit vector and magnitude from component form;
(2) Dot product a·b = |a||b|cosθ — finding angle between vectors;
(3) Condition for perpendicularity (a·b = 0) and parallelism (a×b = 0);
(4) Cross product magnitude |a×b| = |a||b|sinθ — area of parallelogram/triangle;
(5) Scalar triple product [abc] — volume of parallelepiped; coplanarity condition;
(6) Projection of one vector onto another;
(7) Section formula and position vectors;
(8) î, ĵ, k cross products and dot products — standard results.
Topics at a Glance
① Vector Basics
Magnitude, direction, unit vector, position vector
All vectors in NDA are expressed as a = xî + yĵ + zk — master this form
A vector has both magnitude and direction. A scalar has only magnitude. In 3D, every vector is expressed in terms of the standard unit vectors î, ĵ, k along the x, y, z axes respectively.
⚡ Vector Notation & Key Definitions
Component form: a = a₁î + a₂ĵ + a₃k or simply (a₁, a₂, a₃)
Magnitude: |a| = √(a₁² + a₂² + a₃²)
Unit vector: â = a / |a| (magnitude = 1, same direction as a)
Zero/Null vector: 0 = 0î + 0ĵ + 0k (magnitude = 0, direction undefined)
Position vector: OP = xî + yĵ + zk (from origin O to point P(x,y,z))
AB vector from A(x₁,y₁,z₁) to B(x₂,y₂,z₂):
AB = (x₂−x₁)î + (y₂−y₁)ĵ + (z₂−z₁)k
Distance |AB| = √[(x₂−x₁)² + (y₂−y₁)² + (z₂−z₁)²]
Unit vectors î, ĵ, k: |î| = |ĵ| = |k| = 1. They are mutually perpendicular: î·ĵ = ĵ·k = k·î = 0.
Types of Vectors
Unit vector: |â| = 1
Null/Zero vector:0, magnitude = 0
Equal vectors: Same magnitude and direction
Negative vector: −a, same magnitude, opposite direction
Collinear vectors: Parallel or anti-parallel (a × b = 0)
Coplanar vectors: Lie in the same plane ([abc] = 0)
Section Formula
Point P dividing AB internally in ratio m:n:
OP = (m·OB + n·OA) / (m+n)
Midpoint (m=n=1): OP = (OA + OB)/2
Externally (ratio m:n): OP = (m·OB − n·OA)/(m−n)
Centroid of triangle with vertices A, B, C: G = (A+B+C)/3
Vector Representation in 3D — Component Form
Fig 1: Vector a = a₁î + a₂ĵ + a₃k in 3D. Dashed lines show projection onto each axis. |a| is the length OP.
Triangle and parallelogram laws — drawn in diagrams, tested as MCQs
⚡ Vector Operations in Component Form
If a = a₁î+a₂ĵ+a₃k and b = b₁î+b₂ĵ+b₃k:
Addition: a + b = (a₁+b₁)î + (a₂+b₂)ĵ + (a₃+b₃)k
Subtraction: a − b = (a₁−b₁)î + (a₂−b₂)ĵ + (a₃−b₃)k
Scalar mult: λa = λa₁î + λa₂ĵ + λa₃k (stretches/reverses direction)
Triangle Law: OA + AB = OB (head-to-tail addition)
Parallelogram Law: If OA = a and OB = b, then diagonal OC = a + b
Other diagonal: BA − BB = a − b (or b − a)
Properties of addition:
Commutative: a + b = b + a
Associative: (a+b)+c = a+(b+c)
Identity: a + 0 = a
Inverse: a + (−a) = 0
Component-wise addition is the fastest calculation method. The geometric laws (triangle, parallelogram) are tested as statements or diagrams in NDA.
Fig 2: Left — Triangle Law (head to tail). Right — Parallelogram Law (both from same point; diagonal = sum; other diagonal = difference).
📋 TOPIC-WISE PYQ
Basics & Operations — NDA-Pattern Questions
Q1. Find a unit vector in the direction of a = 2î − 3ĵ + 6k.
Result is a SCALAR — the most tested product in NDA
The dot product (scalar product) of two vectors gives a number — it is NOT a vector. It is used to find angles between vectors, check perpendicularity, and calculate projections.
⚡ Dot Product — All Forms
Geometric definition:
a · b = |a| |b| cos θ where θ is angle between a and b (0 ≤ θ ≤ π)
Component form (fastest for NDA):
If a = a₁î+a₂ĵ+a₃k and b = b₁î+b₂ĵ+b₃k:
a · b = a₁b₁ + a₂b₂ + a₃b₃
Finding angle:
cos θ = (a·b) / (|a|·|b|)
Projection of b onto a:
= (a·b) / |a| (scalar projection)
= [(a·b)/|a|²] a (vector projection)
Standard results:
î·î = ĵ·ĵ = k·k = 1 (parallel unit vectors, θ=0°)
î·ĵ = ĵ·k = k·î = 0 (perpendicular unit vectors, θ=90°)
Key conditions: a · b = 0 ⟺ a ⊥ b (perpendicular). a · b = |a||b| ⟺ a ∥ b (parallel, θ=0°).
Properties of Dot Product
Commutative: a·b = b·a
Distributive: a·(b+c) = a·b + a·c
a·a = |a|²
(λa)·b = λ(a·b)
NOT associative: (a·b)·c is meaningless (scalar·vector ≠ dot product)
Physical Applications
Work done: W = F·d = |F||d|cosθ
Angle between vectors: cos θ = (a·b)/(|a||b|)
Perpendicularity check:a·b = 0
Projection of b on a: (a·b)/|a|
Component of b along a: |b|cosθ = (a·b)/|a|
Worked Example — Angle Between Vectors
Find the angle between a = î+2ĵ+2k and b = 3î+2ĵ+6k.
|a+b|² = |a−b|² (squaring both sides).
(a+b)·(a+b) = (a−b)·(a−b)
|a|²+2(a·b)+|b|² = |a|²−2(a·b)+|b|²
4(a·b) = 0 → a·b = 0 → a ⊥ b. Proved. Geometric meaning: diagonals of a parallelogram are equal iff it is a rectangle.
🧩 T2. For unit vectors â and b, if |â+b| = 1, find |â−b|.
Result is a VECTOR — perpendicular to both a and b
The cross product (vector product) of two vectors gives a vector perpendicular to both. Its magnitude equals the area of the parallelogram formed by a and b.
⚡ Cross Product — All Forms
Geometric definition:
a × b = |a||b| sin θ n where n is unit normal (right-hand rule), 0 ≤ θ ≤ π
Component form (determinant expansion):
a × b = | î ĵ k |
| a₁ a₂ a₃ |
| b₁ b₂ b₃ |
= î(a₂b₃−a₃b₂) − ĵ(a₁b₃−a₃b₁) + k(a₁b₂−a₂b₁)
Magnitude (area of parallelogram):
|a × b| = |a||b| sin θ
Area of triangle formed by a and b:
= (1/2)|a × b|
Standard unit vector results:
î×î = ĵ×ĵ = k×k = 0 (parallel → sin 0° = 0)
î×ĵ = k, ĵ×k = î, k×î = ĵ (cyclic order → positive)
ĵ×î = −k, k×ĵ = −î, î×k = −ĵ (anti-cyclic → negative)
Key conditions: a × b = 0 ⟺ a ∥ b (parallel or anti-parallel). |a × b| is maximum when θ = 90° (perpendicular vectors).
Standard Cross Product Results — Cyclic Order
î × ĵ
= k
→
ĵ × k
= î
→
k × î
= ĵ
|
ĵ × î
= −k
→
k × ĵ
= −î
→
î × k
= −ĵ
î × î
= 0
ĵ × ĵ
= 0
k × k
= 0
Memory: î → ĵ → k → î is the positive cyclic order. Going backwards gives the negative.
Q8. The area of the triangle with vertices O(0,0,0), A(1,2,3), B(3,2,1) is:
(a) 2√6 (b) √6 (c) 3√2 (d) 2√3
Answer: (a) 2√6 OA = î+2ĵ+3k, OB = 3î+2ĵ+k. OA×OB = î(2−6)−ĵ(1−9)+k(2−6) = −4î+8ĵ−4k.
|OA×OB| = √(16+64+16) = √96 = 4√6.
Area of triangle = (1/2)×4√6 = 2√6.
Q9. Which of the following is TRUE about the cross product?
(a) a×b = b×a (b) a×b = −b×a (c) a×b is always zero (d) a×b is parallel to both
Answer: (b) a×b = −b×a
Cross product is anti-commutative: reversing the order reverses the sign (direction). (a) is the property of dot product (which IS commutative).
🔥 TRICKY QUESTIONS
Cross Product — Area & Parallelism Traps
🧩 T3. For what value of λ are a = 2î+λĵ+3k and b = 4î+6ĵ+9k parallel?
Parallel ⟺ a×b = 0 ⟺ a₁/b₁ = a₂/b₂ = a₃/b₃.
2/4 = λ/6 = 3/9 → 1/2 = λ/6 = 1/3.
Wait: 2/4 = 1/2 and 3/9 = 1/3. These are unequal, so there is no λ that makes them parallel by ratios alone.
Using cross product condition: î(9λ−18) − ĵ(18−12) + k(12−4λ) = 0.
9λ−18=0 → λ=2. Check ĵ: 18−12=6 ≠ 0. Vectors are NOT parallel for any λ since ratios 2/4 ≠ 3/9. Trap: Students use only one ratio. For parallelism ALL three ratios must be equal. Here a₁/b₁=1/2 but a₃/b₃=1/3, so these vectors can NEVER be parallel regardless of λ.
🧩 T4. If a×b = c×d and a×c = b×d, show that (a−d) is parallel to (b−c).
Know the differences cold — both are tested in the same NDA question set
Dot Product (a · b)
Cross Product (a × b)
Result is a scalar (a number)
Result is a vector (has direction)
= |a||b| cos θ
magnitude = |a||b| sin θ
Commutative: a·b = b·a
Anti-commutative: a×b = −b×a
= 0 ⟺ perpendicular (θ=90°)
= 0 ⟺ parallel (θ=0° or 180°)
Max when θ = 0° (parallel)
Max when θ = 90° (perpendicular)
î·î=1, î·ĵ=0
î×î=0, î×ĵ=k
Application: Work (F·d), angle
Application: Torque (r×F), area
5. Scalar Triple Product — [abc]
5.1
Definition, Formula & Applications
Volume of parallelepiped & coplanarity — tested directly in NDA
The scalar triple product [abc] = a·(b×c) gives a scalar. Geometrically it represents the signed volume of the parallelepiped with edges a, b, c.
⚡ Scalar Triple Product — Definition & Properties
Definition: [abc] = a·(b×c) = (a×b)·c
Component form (3×3 determinant):
If a=(a₁,a₂,a₃), b=(b₁,b₂,b₃), c=(c₁,c₂,c₃):
[abc] = |a₁ a₂ a₃|
|b₁ b₂ b₃|
|c₁ c₂ c₃|
Volume of parallelepiped with edges a, b, c:
V = |[abc]| (absolute value)
Volume of tetrahedron with edges a, b, c:
V = (1/6)|[abc]|
Properties:
[abc] = [bca] = [cab] (cyclic permutation → same value)
[abc] = −[acb] (interchange of any two → sign change)
[abc] = 0 ⟺ a, b, c are COPLANAR (most important application)
Coplanarity is the single most tested application of scalar triple product in NDA. If [abc] = 0, the three vectors lie in the same plane.
Worked Example — Volume of Parallelepiped
Find the volume of the parallelepiped with edges a=î+2ĵ, b=2ĵ+k, c=î+3k.
[abc] = |1 2 0|
|0 2 1|
|1 0 3|
= 1(2×3−1×0)−2(0×3−1×1)+0 = 1(6)−2(−1) = 6+2 = 8.
Volume = |8| = 8 cubic units.
Worked Example — Coplanarity Test
Are vectors a=2î−ĵ+k, b=î+2ĵ−3k, c=3î−4ĵ+5k coplanar?
[abc] = |2 −1 1| = 2(10−12)+1(5+9)+1(−4−6)
|1 2 −3|
|3 −4 5|
= 2(−2)+1(14)+1(−10) = −4+14−10 = 0.
Since [abc] = 0, the vectors are coplanar.
📋 TOPIC-WISE PYQ
Scalar Triple Product — NDA-Pattern Questions
Q10. The volume of a parallelepiped with edges a=î, b=ĵ, c=k is:
🧩 T5. If a, b, c are mutually perpendicular unit vectors, find [abc]².
[abc] = a·(b×c).
Since a, b, c are mutually perpendicular unit vectors: b×c = ±a (by the right-hand rule, since they form an orthonormal basis like î,ĵ,k).
So [abc] = a·(±a) = ±|a|² = ±1.
[abc]² = 1. This is the volume of a unit cube (= 1). Any orthonormal triple has |STP| = 1.
🧩 T6. The four points A(1,2,0), B(2,3,1), C(3,4,0), D(4,5,1) — are they coplanar?
Find vectors from A: AB=(1,1,1), AC=(2,2,0), AD=(3,3,1).
[ABACAD] = |1 1 1| = 1(2−0)−1(2−0)+1(6−6) = 2−2+0 = 0.
|2 2 0|
|3 3 1|
Since [ABACAD] = 0, points A, B, C, D are coplanar. Method: To test coplanarity of 4 points, form 3 vectors from one point, then compute their scalar triple product.
📋 Master Formula Sheet — MN10 Vector Algebra
All critical formulae for rapid pre-exam revision.
📐 Basics
|a| = √(a₁²+a₂²+a₃²)
Unit vector â = a/|a|
AB = B − A (position vectors)
Midpoint: (OA+OB)/2
Direction cosines: l²+m²+n² = 1
· Dot Product
a·b = a₁b₁+a₂b₂+a₃b₃
= |a||b|cosθ → cosθ = (a·b)/(|a||b|)
a⊥b ⟺ a·b = 0
Projection b on a = (a·b)/|a|
a·a = |a|²; î·î=1, î·ĵ=0
× Cross Product
a×b = determinant (î ĵ k / a₁ a₂ a₃ / b₁ b₂ b₃)
|a×b| = |a||b|sinθ
a∥b ⟺ a×b = 0
Area of ∥gram = |a×b|
Area of △ = (1/2)|a×b|
× Unit Vector Results
î×ĵ=k, ĵ×k=î, k×î=ĵ (positive cyclic)
ĵ×î=−k, k×ĵ=−î, î×k=−ĵ (negative)
î×î=ĵ×ĵ=k×k=0
Anti-commutative: a×b=−b×a
[ ] Scalar Triple Product
[abc] = a·(b×c) = 3×3 determinant
Volume of ∥piped = |[abc]|
Volume of tetrahedron = (1/6)|[abc]|
Coplanar ⟺ [abc] = 0
Cyclic: [abc]=[bca]=[cab]
⚖ Key Conditions
Perpendicular: a·b = 0
Parallel: a×b = 0 (or ratios equal)
Coplanar: [abc] = 0
Work = F·d; Torque = r×F
|a+b|² = |a|²+2(a·b)+|b|²
⚡ Quick Revision Booster — MN10 Vector Algebra
📐 Basics Quick-Ref
|2î−3ĵ+6k| = √(4+9+36) = 7
Unit vector = vector ÷ magnitude
AB = OB − OA (always)
|â| = 1 by definition
Direction cosines: l²+m²+n²=1
· Dot Product Rules
a·b: multiply components, add
= 0 → perpendicular (90°)
>0 → acute angle
<0 → obtuse angle
Commutative: a·b = b·a
× Cross Product Rules
Result is a vector (NOT scalar)
= 0 → parallel
Anti-commutative: a×b = −b×a
î→ĵ→k→î (cyclic = positive)
Area of △ = (1/2)|a×b|
[ ] Triple Product
Use 3×3 determinant directly
= 0 → coplanar
Cyclic permutation: same value
Swap two: change sign
4 points coplanar: form 3 vectors from one point, check STP