Login / Register
Follow Us
0
0
Register Free!
Register Free!
Online Mock Tests & Exam Preparation | Target Exams
Select Medium / माध्यम चुनें:
Eng (English) Hindi (हिन्दी)
CBSE • Class XII • Mathematics • Ch 10
Estimated Time: 45 Mins
Study Progress: In Progress

Vector Algebra

In Class 12 Mathematics, "Vector Algebra" provides an authoritative, curriculum-verified master resource aligned with the 2026–27 NCERT syllabus.

Take Chapter CBT Test View All Chapters
Overview Core Concepts Summary Self Check Take MCQ Test

➡️ Have You Ever Wondered?

How do aerospace navigation systems plot a rocket's flight through gale-force high-altitude jet stream winds, or calculate the exact mechanical work done by a magnetic force on an electron? Vectors combine magnitude with spatial direction.

Why This Chapter Matters

In Class 12 Mathematics, "Vector Algebra" provides an authoritative, curriculum-verified master resource aligned with the 2026–27 NCERT syllabus.

Before You Begin (Prerequisites)

  • Vectors in 2D from Class 11 Physics.
  • Cartesian coordinates.
  • Dot product and cross product.

What You Will Learn (Core Objectives)

  • Distinguish between Scalar and Vector quantities; define Unit, Zero, Co-initial, and Collinear vectors.
  • Represent vectors in 3D: $\vec{r} = x\hat{i} + y\hat{j} + z\hat{k}$ and find Direction Cosines ($l, m, n$).
  • Perform Vector Addition and Multiplication by a scalar.
  • Calculate Scalar (Dot) Product: $\vec{a} \cdot \vec{b} = ab\cos\theta$; find projection of a vector on a line.
  • Calculate Vector (Cross) Product: $\vec{a} \times \vec{b} = (ab\sin\theta)\hat{n}$; find area of triangles and parallelograms.

Chapter Roadmap & Progression

1 1. Components, Direction Cosines &...
2 2. Dot Product & Projection
3 3. Cross Product & Geometric Areas

Complete Concept Guide (100% Curriculum Coverage)

1. Components, Direction Cosines & Addition

A vector with initial point $O(0, 0, 0)$ and terminal point $P(x, y, z)$ is: $$\mathbf{\vec{r} = x\hat{i} + y\hat{j} + z\hat{k}} \quad (|\vec{r}| = \sqrt{x^2 + y^2 + z^2})$$ Direction Cosines: $l = \cos\alpha = \frac{x}{|\vec{r}|}, m = \cos\beta = \frac{y}{|\vec{r}|}, n = \cos\gamma = \frac{z}{|\vec{r}|}$ with identity: $$\mathbf{l^2 + m^2 + n^2 = 1}$$

2. Dot Product & Projection

$\mathbf{\vec{a} \cdot \vec{b} = a_1 b_1 + a_2 b_2 + a_3 b_3 = |\vec{a}||\vec{b}|\cos\theta}$.
• Orthogonal vectors: $\mathbf{\vec{a} \cdot \vec{b} = 0}$.
• Projection of $\vec{a}$ on $\vec{b}$: $$\mathbf{\text{Proj}_{\vec{b}} \vec{a} = \frac{\vec{a} \cdot \vec{b}}{|\vec{b}|}}$$
• $\hat{i} \cdot \hat{i} = \hat{j} \cdot \hat{j} = \hat{k} \cdot \hat{k} = 1$.

3. Cross Product & Geometric Areas

$\mathbf{\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \end{vmatrix}}$. Non-commutative: $\vec{a} \times \vec{b} = -(\vec{b} \times \vec{a})$.
• Collinear/parallel: $\mathbf{\vec{a} \times \vec{b} = \vec{0}}$.
• Area of Parallelogram with adjacent sides $\vec{a}, \vec{b}$: $\mathbf{\text{Area} = |\vec{a} \times \vec{b}|}$.
• Area of Triangle: $\mathbf{\text{Area} = \frac{1}{2}|\vec{a} \times \vec{b}|}$.

Vector Algebra - Key Conceptual & Analytical Model

Vector Algebra - Mathematical Architecture Axiomatic & Matrix Foundations Equivalence theorems & algebraic proofs Calculus & 3D Vector Geometry Differential optimization & spatial lines High-Stakes Examination & Engineering Mastery CBSE Class 12 Board criteria, JEE Advanced problem frameworks & applications

Chapter Summary & 10 Key Takeaways

Takeaway 1
Direction Cosines: Invariant direction angles satisfying $l^2 + m^2 + n^2 = 1$.
Takeaway 2
Scalar Dot Product: Evaluates geometric projection and tests orthogonality ($a \cdot b = 0$).
Takeaway 3
Vector Cross Product: Determinant algorithm yielding normal vector perpendicular to both inputs.
Takeaway 4
Area Formulation: $|\vec{a} \times \vec{b}|$ directly measures parallelogram surface area.
Takeaway 5
Unit Vector Normal: $\hat{n} = \frac{\vec{a} \times \vec{b}}{|\vec{a} \times \vec{b}|}$ defining planar normal.

Check Your Understanding (Diagnostic Practice Questions)

Diagnostic questions testing core conceptual clarity. Answers are hidden initially — solve each problem first, then click to reveal the step-by-step verified solution.

1
Find the unit vector in the direction of the vector $\vec{a} = 2\hat{i} + 3\hat{j} + \hat{k}$.
Reveal Answer & Explanation
Answer: $|\vec{a}| = \sqrt{2^2 + 3^2 + 1^2} = \sqrt{4 + 9 + 1} = \sqrt{14}$. Unit vector $\hat{a} = \frac{\vec{a}}{|\vec{a}|} = \frac{2}{\sqrt{14}}\hat{i} + \frac{3}{\sqrt{14}}\hat{j} + \frac{1}{\sqrt{14}}\hat{k}$.
2/√14 i + 3/√14 j + 1/√14 k.
2
Find the angle between the vectors $\vec{a} = \hat{i} + \hat{j} - \hat{k}$ and $\vec{b} = \hat{i} - \hat{j} + \hat{k}$.
Reveal Answer & Explanation
Answer: $\vec{a} \cdot \vec{b} = (1)(1) + (1)(-1) + (-1)(1) = 1 - 1 - 1 = -1$. $|\vec{a}| = \sqrt{1+1+1} = \sqrt{3}, |\vec{b}| = \sqrt{1+1+1} = \sqrt{3}$. $\cos\theta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|} = \frac{-1}{\sqrt{3}\sqrt{3}} = -\frac{1}{3} \implies \theta = \cos^{-1}\left(-\frac{1}{3}\right)$.
θ = cos^-1(-1/3).
3
Find the projection of the vector $\vec{a} = 2\hat{i} + 3\hat{j} + 2\hat{k}$ on the vector $\vec{b} = \hat{i} + 2\hat{j} + \hat{k}$.
Reveal Answer & Explanation
Answer: $\text{Projection} = \frac{\vec{a} \cdot \vec{b}}{|\vec{b}|} = \frac{2(1) + 3(2) + 2(1)}{\sqrt{1^2 + 2^2 + 1^2}} = \frac{2 + 6 + 2}{\sqrt{6}} = \frac{10}{\sqrt{6}} = \frac{5\sqrt{6}}{3}$.
10/√6 (or 5√6/3).
4
Find the area of a parallelogram whose adjacent sides are given by the vectors $\vec{a} = 3\hat{i} + \hat{j} + 4\hat{k}$ and $\vec{b} = \hat{i} - \hat{j} + \hat{k}$.
Reveal Answer & Explanation
Answer: $\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 1 & 4 \\ 1 & -1 & 1 \end{vmatrix} = \hat{i}(1 - (-4)) - \hat{j}(3 - 4) + \hat{k}(-3 - 1) = 5\hat{i} + \hat{j} - 4\hat{k}$. $\text{Area} = |\vec{a} \times \vec{b}| = \sqrt{5^2 + 1^2 + (-4)^2} = \sqrt{25 + 1 + 16} = \sqrt{42}\text{ sq. units}$.
√42 sq. units.
5
Show that the vectors $\vec{a} = 2\hat{i} - \hat{j} + \hat{k}, \vec{b} = \hat{i} - 3\hat{j} - 5\hat{k},$ and $\vec{c} = 3\hat{i} - 4\hat{j} - 4\hat{k}$ form the vertices of a right-angled triangle.
Reveal Answer & Explanation
Answer: Sides: $\vec{AB} = \vec{b} - \vec{a} = -\hat{i} - 2\hat{j} - 6\hat{k} \implies |AB|^2 = 1 + 4 + 36 = 41$. $\vec{BC} = \vec{c} - \vec{b} = 2\hat{i} - \hat{j} + \hat{k} \implies |BC|^2 = 4 + 1 + 1 = 6$. $\vec{CA} = \vec{a} - \vec{c} = -\hat{i} + 3\hat{j} + 5\hat{k} \implies |CA|^2 = 1 + 9 + 25 = 35$. Notice: $|BC|^2 + |CA|^2 = 6 + 35 = 41 = |AB|^2$. By converse of Pythagoras, it is a right-angled triangle!
Verified |BC|^2 + |CA|^2 = |AB|^2 = 41.
Finished Studying This Chapter?
READY TO PRACTICE?

Timed CBT Practice Tests (Exam Simulator)

Put your concepts to the test with official curriculum-aligned Foundation and Advanced practice tests. Get instant accuracy scores, time metrics, and step-by-step verified explanations.

Foundation Test
15 MCQs • Core Concepts
Advanced Test
15 MCQs • HOTS & Traps

All Class 12 Mathematics Chapters

AI Study Friend

Instant Concept Tutor

Have a doubt in Vector Algebra? Ask our AI study tutor for rapid explanations or customized quizzes.