VECTORVERSE
Ashu • Class 12 Science Non-Medical
Mathematics • Class XII • Senior Secondary

VECTORVERSE

AN INTERACTIVE 3D EXPLORATION OF VECTORS
Continuous Spatial Mechanics & Solved Class 12 Vector Algebra

Prepared By
ASHU
Class 12th Science — Non-Medical
Swami Dayanand Saraswati School
Submitted To
MR. CHIRAG SHARMA
Department of Mathematics
CBSE Class XII Academic Project
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Academic Submission • 2026

MATHEMATICS PROJECT

CHAPTER: VECTORS
Student Name
ASHU
Class 12th Science — Non-Medical
Submitted To
MR. CHIRAG SHARMA
Lecturer in Mathematics
Institution
Swami Dayanand Saraswati School
Senior Secondary Science Stream
Subject & Focus
Mathematics (Class XII)
Vector Algebra, Projections & 3D Geometry

A rigorous interactive exploration synthesizing mathematical theory, 3D geometric visualization, and step-by-step solved textbook questions.

Spatial Foundation

The Architecture of Space

Mathematics models 3D physical reality through mutually perpendicular Cartesian coordinate axes intersecting at Origin O(0, 0, 0). Every spatial direction is calibrated through unit basis vectors:

Right-Handed Basis Frame
î = (1,0,0) • ĵ = (0,1,0) • k̂ = (0,0,1)
Mutually orthogonal unit vectors satisfying î × ĵ = k̂ via the Right-Hand Screw Rule.
X-Axis Electric Cyan: Transverse horizontal axis (î)
Y-Axis Mint Emerald: Longitudinal ground axis (ĵ)
Z-Axis Radiant Violet: Normal vertical altitude (k̂)
Core Principle

What Does a Vector Represent?

A vector is not simply a number. It is an ordered physical entity requiring both magnitude (scalar size) and direction (orientation in reference space).

Distance vs Displacement Inequality
|Δr| ≤ s (Direct Vector ≤ Curved Path)
Path length s ≈ 9.8 m (scalar, red curve) whereas displacement |Δr| ≈ 6.65 m (cyan straight vector).
Scalars Mass, time, temperature, distance — obey standard arithmetic.
Vectors Displacement, velocity, force — obey geometric vector laws.
Component Resolution

Decomposing 3D Vectors

Any directed vector in space can be resolved along orthogonal axes. In 3D space, position vector OP from origin O(0,0,0) to P(x, y, z) forms the interior space diagonal of a rectangular prism.

Position Vector Form
r = x î + y ĵ + z k̂
Point P(3, 4, 5): x=3 along î, y=4 along ĵ, z=5 along k̂. Ground projection is P'(3, 4, 0).
x = 3 Orthogonal projection on X-axis
y = 4 Orthogonal projection on Y-axis
z = 5 Vertical altitude along Z-axis
Solved CBSE Question • Topic 01

Magnitude of a Vector

The magnitude of vector a = xî + yĵ + zk̂ represents its true Euclidean length in 3D space, derived from successive applications of the Pythagorean Theorem.

3D Euclidean Magnitude Formula
|a| = √(x² + y² + z²)
QUESTION 1
Find the magnitude of the vector: a = 3î + 4ĵ + 12k̂
Step 1 Identify Components
x = 3, y = 4, z = 12
Step 2 Substitute into Magnitude Formula
|a| = √(3² + 4² + 12²)
Step 3 Calculate Squares & Sum
|a| = √(9 + 16 + 144) = √169
Final Answer |a| = 13 units
Follow-up Example • b = 2î - 3ĵ + 6k̂
|b| = √(2² + (-3)² + 6²) = √(4 + 9 + 36) = √49 = 7 units
Solved CBSE Question • Topic 02

Unit Vector Normalization

A unit vector retains the exact orientation of a non-zero vector while scaling its magnitude strictly to unity (1). It isolates pure directional heading.

Unit Vector Formula
â = a / |a| where |â| = 1
QUESTION 1
Find a unit vector in the direction of: a = 2î - 3ĵ + 6k̂
Step 1 Calculate Magnitude of a
|a| = √(2² + (-3)² + 6²) = √(4 + 9 + 36) = √49 = 7
Step 2 Divide Vector by its Magnitude
â = (2î - 3ĵ + 6k̂) / 7
Final Unit Vector â = (2/7)î - (3/7)ĵ + (6/7)k̂
✓ Magnitude Check: √((2/7)² + (-3/7)² + (6/7)²) = √(49/49) = 1.0
Follow-up 2D Example • b = 3î + 4ĵ
|b| = √(3² + 4²) = 5 ⟹ b̂ = (3/5)î + (4/5)ĵ
Solved CBSE Question • Topic 03

Direction Ratios & Cosines

If a vector makes angles α, β, γ with the positive X, Y, and Z axes, its direction cosines are l = cos α, m = cos β, n = cos γ. Direction ratios (a, b, c) are any numbers proportional to direction cosines.

Direction Cosine Identity & Formulas
l² + m² + n² = 1 • l = x/|r|, m = y/|r|, n = z/|r|
QUESTION 1
Find the direction cosines of the vector: a = 2î - 3ĵ + 6k̂
Step 1 Find Magnitude |a|
|a| = √(2² + (-3)² + 6²) = √49 = 7
Step 2 Direction Cosines (l, m, n)
l = 2/7, m = -3/7, n = 6/7
Direction Cosines ( 2/7, -3/7, 6/7 )
✓ Invariant: (2/7)² + (-3/7)² + (6/7)² = 4/49 + 9/49 + 36/49 = 49/49 = 1
Direction Ratios Example • b = 3î + 4ĵ + 12k̂
Direction ratios = 3 : 4 : 12 (|b| = 13 ⟹ Cosines = 3/13, 4/13, 12/13)
Geometric Law

Parallelogram Law of Addition

Vectors do not add like simple scalars. Two co-initial vectors A and B form the adjacent sides of a parallelogram. The interior space diagonal represents their resultant sum R.

Vector Addition Formula
R = A + B = (Ax + Bx)î + (Ay + By)ĵ + (Az + Bz)k̂
Let A = (3, 2, 1) and B = (1, 2, 3)
R = (3+1)î + (2+2)ĵ + (1+3)k̂ = 4î + 4ĵ + 4k̂
Resultant Magnitude |R| = √(4² + 4² + 4²) = √48 = 4√3 ≈ 6.928
Relative Motion

Subtraction: Adding the Negative

Vector subtraction is defined as adding the opposite vector: A - B = A + (-B). Geometrically, negating B inverts its directional arrow by 180° through the origin.

Subtraction Formulation
D = A - B = (Ax - Bx)î + (Ay - By)ĵ + (Az - Bz)k̂
Let A = (4, 3, 1) and B = (1, 2, 3) ⟹ -B = (-1, -2, -3)
D = (4-1)î + (3-2)ĵ + (1-3)k̂ = 3î + 1ĵ - 2k̂
Difference Magnitude |D| = √(3² + 1² + (-2)²) = √14 ≈ 3.742
Solved CBSE Question • Topic 04

Scalar (Dot) Product & Angle

The dot product of two vectors yields a pure real scalar quantity. It quantifies how much two vectors reinforce each other along a shared heading, and tests orthogonality.

Dot Product & Angle Formulas
a · b = ax bx + ay by + az bz = |a||b| cos θ
cos θ = (a · b) / (|a| |b|)
QUESTION 1
Given a = 2î + 3ĵ + 6k̂ and b = î + 2ĵ + 2k̂. Find the dot product a · b and the angle θ between them.
Step 1 Calculate Algebraic Dot Product
a · b = (2)(1) + (3)(2) + (6)(2) = 2 + 6 + 12 = 20
Step 2 Compute Magnitudes |a| and |b|
|a| = √(4 + 9 + 36) = 7 • |b| = √(1 + 4 + 4) = 3
Step 3 Solve for Angle θ
cos θ = (a · b) / (|a||b|) = 20 / (7 × 3) = 20/21
Final Answers a · b = 20 • θ = cos⁻¹(20/21) ≈ 17.75°
Solved CBSE Question • Topic 05

Scalar & Vector Projection

Crucial distinction: Scalar Projection is a signed numerical length representing the shadow of vector a along b. Vector Projection is a directed vector along the line of b.

Projection Formulas
Scalar: comp_b(a) = (a · b) / |b|
Vector: proj_b(a) = ((a · b) / |b|²) b
QUESTION 1
Given a = 2î + 3ĵ + 6k̂ and b = î + 2ĵ + 2k̂. Find:
1. The scalar projection of a on b.
2. The vector projection of a on b.
Step 1 Compute Dot Product & |b|
a · b = (2)(1) + (3)(2) + (6)(2) = 20 • |b| = √(1+4+4) = 3
Step 2 Scalar Projection of a on b
comp_b(a) = (a · b) / |b| = 20 / 3 units
Step 3 Vector Projection of a on b
proj_b(a) = ((a · b) / |b|²) b = (20 / 3²) (î + 2ĵ + 2k̂) = (20/9)(î + 2ĵ + 2k̂)
Vector Projection proj = (20/9)î + (40/9)ĵ + (40/9)k̂
✓ Scalar Result = 20/3 ≈ 6.67 (Scalar) • Vector Result has direction b̂ (Vector)
Solved CBSE Question • Topic 06

The Vector (Cross) Product

Unlike the scalar dot product, the cross product produces a brand-new vector that is strictly perpendicular (orthogonal) to both input vectors via the Right-Hand Rule.

Determinant Expansion Formula
a × b = (ay bz - az by)î + (az bx - ax bz)ĵ + (ax by - ay bx)k̂
QUESTION 1
Given a = î + 2ĵ + 3k̂ and b = 2î - ĵ + k̂. Find a × b.
Components a = (1, 2, 3) • b = (2, -1, 1)
î-Component (ay bz - az by) = (2)(1) - (3)(-1) = 2 + 3 = 5
ĵ-Component (az bx - ax bz) = (3)(2) - (1)(1) = 6 - 1 = 5
k̂-Component (ax by - ay bx) = (1)(-1) - (2)(2) = -1 - 4 = -5
Final Cross Product a × b = 5î + 5ĵ - 5k̂
✓ Orthogonality Check: (a × b) · a = 5(1)+5(2)-5(3) = 0 • (a × b) · b = 5(2)+5(-1)-5(1) = 0
Geometric Theorem

Vectors Measure Surface Area

A fundamental discovery in vector geometry: the scalar magnitude of the cross product |a × b| exactly equals the geometric area of the parallelogram spanned by vectors a and b.

Parallelogram & Triangle Area Theorems
Area of Parallelogram = |a × b|
Area of Triangle = (1/2)|a × b|
For a = 4î and b = 1.5î + 3ĵ in XY-plane:
|a × b| = |(4)(3)k̂| = 12 sq units. Triangle area = 6 sq units.
Enclosed Parallelogram Area Area = |a × b| = 12 sq units
Interactive Evaluation

Compact Mixed Practice

Three diagnostic evaluation checks to test your understanding of magnitude, dot product, and cross product:

1. Find the magnitude of vector: a = 5î + 12ĵ
|a| = √(5² + 12²) = √(25 + 144) = √169 = 13 units
2. Find the dot product of a = î + 2ĵ + 3k̂ and b = 2î + ĵ - k̂
a · b = (1)(2) + (2)(1) + (3)(-1) = 2 + 2 - 3 = 1
3. Find the cross product of standard basis vectors: a = î, b = ĵ
î × ĵ = k̂ (perpendicular along positive Z-axis via Right-Hand Rule)
Academic Synthesis • Conclusion

Direction Into Mathematics

Vectors bridge the gap between physical intuition and coordinate algebra, allowing nature's continuous directional laws to be computed with exact precision.

Presented By
ASHU
Class 12th Science — Non-Medical
Swami Dayanand Saraswati School
Submitted To
MR. CHIRAG SHARMA
Lecturer in Mathematics
Senior Secondary Mathematics Project
THANK YOU • QUESTIONS & DISCUSSIONS WELCOME