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Mathematics Grade XII

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Chapter 10: Vector Algebra

Class 12 • Mathematics • Chapter 10

Vector Algebra

Quantities with both size and direction, and the two products that unlock geometry in space.

CBSE / NCERT Aligned

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Chapter Roadmap

Scalars and Vectors • Types of Vectors • Addition and Components • Magnitude and Unit Vectors • Dot Product • Cross Product

Size, and Also Direction

Some quantities need only a number: mass, time, temperature. These are scalars. Others make no sense without a direction as well: displacement, velocity, force. These are vectors. A vector is drawn as an arrow whose length shows the size, or magnitude, and whose arrowhead shows the direction.

Vector algebra gives precise rules for combining such arrows. You will add and scale vectors, break them into components along the i, j and k directions, and measure them. Two special products then appear: the dot product, which extracts angles and projections, and the cross product, which produces a perpendicular vector whose length is an area. These tools are the language of physics and three-dimensional geometry.

Foundation

A vector a = a₁i + a₂j + a₃k has magnitude |a| = √(a₁² + a₂² + a₃²). The dot product a·b = |a||b| cosθ gives a scalar; the cross product a×b is a vector with |a×b| = |a||b| sinθ.

Key Terms You Must Know

Term

Meaning

Example

Scalar

A quantity with size only.

mass, time

Vector

A quantity with size and direction.

velocity, force

Unit vector

A vector of magnitude 1 in a given direction.

â = a/|a|

Position vector

The vector from the origin to a point.

point (x, y, z)

Dot product

a·b = |a||b| cosθ, a scalar.

measures angle and projection

Cross product

a×b, a vector perpendicular to both.

length is the parallelogram area

Core Concepts, Step by Step

1. Scalars and Vectors

A scalar is fully described by a single number with its unit. A vector needs a magnitude and a direction, so it is drawn as a directed arrow. In components we write a vector along the three axes as a₁i + a₂j + a₃k, where i, j and k are the unit vectors pointing along the x, y and z axes.

2. Types of Vectors

Several special vectors recur: the zero vector has no length; a unit vector has length 1; equal vectors have the same magnitude and direction; collinear vectors are parallel; and the position vector of a point runs from the origin to that point. Recognising these types speeds up almost every problem.

3. Addition, Scaling and Components

Vectors add tip to tail by the triangle law, or as the diagonal of a parallelogram. In components we simply add matching parts: (a₁ + b₁)i + (a₂ + b₂)j + (a₃ + b₃)k. Multiplying by a scalar stretches or shrinks the arrow and reverses it if the scalar is negative.

The triangle law: placing b at the tip of a gives the resultant a + b

Triangle law of vector addition with vector a then vector b giving the resultant a plus b

4. Magnitude and Unit Vectors

The magnitude of a = a₁i + a₂j + a₃k is √(a₁² + a₂² + a₃²), found by Pythagoras in three dimensions. To get a unit vector in the direction of a, divide a by its own magnitude. Unit vectors are the standard way to record a pure direction.

5. The Dot Product

The dot product a·b equals |a||b| cosθ and is computed in components as a₁b₁ + a₂b₂ + a₃b₃. It returns a scalar and is the tool for finding the angle between two vectors and the projection of one onto another. Crucially, two non-zero vectors are perpendicular exactly when their dot product is zero.

6. The Cross Product

The cross product a×b is a vector perpendicular to both a and b, found from the determinant of i, j, k over the two component rows. Its magnitude |a||b| sinθ equals the area of the parallelogram with a and b as sides, so it vanishes when the vectors are parallel. The cross product is essential for areas, torques and normals to planes.

Key Results with Proofs

Result 1: Angle from the Dot Product

Statement. The angle θ between non-zero vectors satisfies cosθ = (a·b)/(|a||b|).

Proof

The angle is recovered simply by rearranging the dot-product definition.

By definition a·b = |a||b| cosθ. – start from the definition

Divide both sides by |a||b|, which is non-zero. – isolate cosθ

cosθ = (a·b)/(|a||b|). – the formula

Computing a·b in components then gives θ directly. – practical use

This is the standard way to find the angle between two vectors or two lines in space.

Result 2: Test for Perpendicular Vectors

Statement. Two non-zero vectors are perpendicular if and only if a·b = 0.

Proof

A zero dot product can only come from a right angle when the lengths are non-zero.

a·b = |a||b| cosθ with |a|, |b| non-zero. – definition

So a·b = 0 forces cosθ = 0. – since the magnitudes are non-zero

cosθ = 0 means θ = 90°. – the angle is a right angle

Hence a zero dot product is exactly perpendicularity. – conclusion

This single test is used constantly to check perpendicularity.

Result 3: Cross Product Gives Area

Statement. The magnitude |a×b| equals the area of the parallelogram with sides a and b.

Proof

The sine in the cross product is precisely the height factor of the parallelogram.

A parallelogram with sides a and b has area base × height. – area of a parallelogram

Take |a| as the base; the height is |b| sinθ. – drop a perpendicular

So the area is |a||b| sinθ. – multiply

This is exactly |a×b| by the definition of the cross product. – match the definition

Half of this, ½|a×b|, is the area of the triangle with sides a and b.

Worked Examples

Example 1

Question: Find the magnitude of a = 2i + 3j + 6k.

▶ Show full working

Use the three-dimensional Pythagoras.

|a| = √(2² + 3² + 6²) = √(4 + 9 + 36). – square and add

= √49 = 7. – root

Answer: |a| = 7.

Example 2

Question: Find the unit vector in the direction of 3i + 4j.

▶ Show full working

Divide by the magnitude.

|3i + 4j| = √(9 + 16) = 5. – magnitude

Unit vector = (3i + 4j)/5 = (3/5)i + (4/5)j. – divide

Answer: (3/5)i + (4/5)j.

Example 3

Question: If a = i + 2j and b = 3i − j, find a + b.

▶ Show full working

Add matching components.

i parts: 1 + 3 = 4; j parts: 2 + (−1) = 1. – add components

Answer: 4i + j.

Example 4

Question: Find (i + 2j + 3k)·(2i − j + k).

▶ Show full working

Multiply matching parts and add.

1(2) + 2(−1) + 3(1) = 2 − 2 + 3. – component products

= 3. – sum

Answer: 3.

Example 5

Question: Find the angle between i and j.

▶ Show full working

Use the dot product.

i·j = 0 and both have magnitude 1. – compute

cosθ = 0/(1×1) = 0, so θ = 90°. – angle

Answer: 90°.

Example 6

Question: Show that 2i + 3j and 3i − 2j are perpendicular.

▶ Show full working

Check the dot product.

(2)(3) + (3)(−2) = 6 − 6 = 0. – dot product is zero

A zero dot product means a right angle. – interpret

Answer: Yes, they are perpendicular.

Example 7

Question: Find (i + j + k)×(i − j + k).

▶ Show full working

Expand the determinant of i, j, k.

i: (1)(1) − (1)(−1) = 2; j: −[(1)(1) − (1)(1)] = 0; k: (1)(−1) − (1)(1) = −2. – cofactor expansion

So the cross product is 2i + 0j − 2k. – assemble

Answer: 2i − 2k.

Example 8

Question: Find the area of the parallelogram with adjacent sides a = i + j and b = j + k.

▶ Show full working

Area is |a×b|.

a×b = i(1·1 − 0·1) − j(1·1 − 0·0) + k(1·1 − 1·0) = i − j + k. – cross product

|a×b| = √(1 + 1 + 1) = √3. – magnitude

Answer: √3 square units.

Example 9

Question: Find the projection of a = 2i + j on b = i + 2j + 2k.

▶ Show full working

Projection = (a·b)/|b|.

a·b = 2(1) + 1(2) + 0(2) = 4. – dot product

|b| = √(1 + 4 + 4) = 3. – magnitude of b

Projection = 4/3. – divide

Answer: 4/3.

Example 10

Question: Find λ so that 2i + λj + k and i − 2j + 3k are perpendicular.

▶ Show full working

Set the dot product to zero.

Dot product = 2(1) + λ(−2) + 1(3) = 5 − 2λ. – compute

Set 5 − 2λ = 0, so λ = 5/2. – solve

Answer: λ = 5/2.

Example 11

Question: Find the position vector of the midpoint of A(1, 2, 3) and B(3, 4, 5).

▶ Show full working

Average the position vectors.

Midpoint = ((1 + 3)/2, (2 + 4)/2, (3 + 5)/2). – average each coordinate

= (2, 3, 4), so 2i + 3j + 4k. – write as a vector

Answer: 2i + 3j + 4k.

Example 12

Question: Find a unit vector perpendicular to both i + j and j + k.

▶ Show full working

Take the cross product, then normalise.

(i + j)×(j + k) = i(1·1 − 0·1) − j(1·1 − 0·0) + k(1·1 − 1·0) = i − j + k. – cross product

Its magnitude is √3, so the unit vector is (i − j + k)/√3. – normalise

Answer: (i − j + k)/√3.

Where You Meet This in Real Life

Physics of forces

Forces add as vectors, so engineers use the triangle and parallelogram laws to find the net push on a structure.

Navigation

An aircraft’s true path is the vector sum of its heading and the wind, found by vector addition.

Computer graphics

Surface normals, computed with cross products, decide how light reflects off three-dimensional models.

Robotics

The torque on a robot joint is a cross product of the position and force vectors.

Work in mechanics

The work done by a force is the dot product of the force and the displacement vectors.

Practice Sets A–D

Practice Set A – Basics

A1. Find the magnitude of 3i + 4j.

▶ Reveal full working

Pythagoras.

√(9 + 16) = 5.

Answer: 5.

A2. Find (i + 2j + k) + (2i − j + 3k).

▶ Reveal full working

Add components.

3i + j + 4k.

Answer: 3i + j + 4k.

A3. Find i·i.

▶ Reveal full working

Dot product of a unit vector with itself.

Answer: 1.

A4. Find i×i.

▶ Reveal full working

Cross product of parallel vectors.

The zero vector.

Answer: 0 (the zero vector).

Practice Set B – Conceptual

B1. Why does a zero dot product mean perpendicular vectors?

▶ Reveal full working

Use the cosine.

a·b = |a||b| cosθ; with non-zero lengths, a·b = 0 forces cosθ = 0. – reason

cosθ = 0 gives θ = 90°. – right angle

Answer: Because it forces cosθ = 0, a right angle.

B2. What does |a×b| represent geometrically?

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Area.

It is the area of the parallelogram with sides a and b.

Answer: The parallelogram area.

B3. How do you make a unit vector from a?

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Normalise.

Divide a by its magnitude |a|.

Answer: Divide a by |a|.

B4. When is a cross product the zero vector?

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Parallel case.

When the two vectors are parallel, since sinθ = 0.

Answer: When the vectors are parallel.

Practice Set C – Application / Numerical

C1. Find (2i + 3j + k)·(i − j + 4k).

▶ Reveal full working

Component products.

2(1) + 3(−1) + 1(4) = 2 − 3 + 4 = 3.

Answer: 3.

C2. Find the angle between i + j and i − j.

▶ Reveal full working

Dot then cosine.

Dot product = 1 − 1 = 0, and both have magnitude √2. – compute

cosθ = 0, so θ = 90°. – angle

Answer: 90°.

C3. Find the area of the triangle with sides a = i + j and b = j + k.

▶ Reveal full working

Half the cross-product length.

a×b = i − j + k, magnitude √3. – cross product

Triangle area = ½√3. – half

Answer: ½√3 square units.

C4. Find λ so that i + λj + 2k and 2i + j − k are perpendicular.

▶ Reveal full working

Dot product zero.

2 + λ − 2 = λ; set λ = 0. – solve

Answer: λ = 0.

Practice Set D – HOTS / Multi-step

D1. Show that the vectors i + 2j + 3k, 2i + 3j + 4k and 3i + 4j + 5k are coplanar.

▶ Reveal full working

Coplanar when the scalar triple product is 0.

Form the determinant with these as rows. – scalar triple product

|1 2 3; 2 3 4; 3 4 5| expands to 1(15 − 16) − 2(10 − 12) + 3(8 − 9) = −1 + 4 − 3 = 0. – evaluate

A zero triple product means the three vectors are coplanar. – interpret

Answer: Yes, the triple product is 0, so they are coplanar.

D2. Find the angle between a = i + j + k and b = i + j − 2k.

▶ Reveal full working

Dot product and magnitudes.

a·b = 1 + 1 − 2 = 0. – dot product

Since the dot product is 0, the angle is 90°. – angle

Answer: 90°.

D3. Find a vector of magnitude 5 in the direction of 3i + 4j.

▶ Reveal full working

Unit vector times 5.

Unit vector = (3i + 4j)/5. – normalise

Multiply by 5: 3i + 4j. – scale

Answer: 3i + 4j.

D4. Find the area of the parallelogram whose diagonals are 2i and 2j.

▶ Reveal full working

Area is half the magnitude of the cross product of the diagonals.

For diagonals d₁ and d₂, area = ½|d₁×d₂|. – diagonal formula

d₁×d₂ = (2i)×(2j) = 4(i×j) = 4k, magnitude 4. – cross product

Area = ½(4) = 2. – half

Answer: 2 square units.

Chapter Summary

Everything in One Glance

Scalars vs Vectors

Scalars have size only; vectors have size and direction, written a₁i + a₂j + a₃k.

Operations

Add component by component; scaling stretches or reverses the arrow.

Magnitude & Unit

|a| = √(a₁² + a₂² + a₃²); divide by |a| for a unit vector.

Dot Product

a·b = |a||b| cosθ; zero means perpendicular; gives angles and projections.

Cross Product

a×b is perpendicular to both; |a×b| is the parallelogram area; zero means parallel.

Triangle Area

Half of |a×b| is the area of the triangle with sides a and b.

Are You Exam-Ready?

8-Point Exam Quick-Check

Find the magnitude of i + 2j + 2k.

Find a unit vector along 6i + 8j.

Find (i + j)·(i − j).

Find the angle between i and k.

Are 3i + 4j and 4i − 3j perpendicular?

Find (i + j + k)×(i + j + k).

Find the area of the parallelogram with sides i and j.

Find the projection of i + j on i.

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Class 12 Mathematics Chapter 10: Vector Algebra, Complete Notes and Practice

These free Class 12 Maths Vector Algebra notes follow the NCERT 2026 to 27 syllabus and cover scalars and vectors, types of vectors, addition and components, magnitude and unit vectors, the dot product and the cross product, with twelve worked examples and sixteen graded practice questions. Includes worked examples, step-by-step proofs and graded practice, free on SchoolRevise.com.