Complete Summary and Solutions for Vector Algebra – NCERT Class XII Mathematics Part II, Chapter 10 – Vectors, Operations, Scalar and Vector Products
Detailed summary and explanation of Chapter 10 'Vector Algebra' from the NCERT Class XII Mathematics Part II textbook, covering basic concepts of vectors, types of vectors, addition and subtraction, scalar multiplication, dot (scalar) product, cross (vector) product, geometrical interpretation and their use in solving problems, with all NCERT questions, answers, and solved examples.
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Vector Algebra
Chapter 10: Mathematics - Ultimate Study Guide | NCERT Class 12 Notes, Solved Examples, Exercises & Quiz 2025
Full Chapter Summary & Detailed Notes - Vector Algebra Class 12 NCERT
In most sciences one generation tears down what another has built and what one has established another undoes. In Mathematics alone each generation builds a new story to the old structure. – HERMAN HANKEL
10.1 Introduction
In our day to day life, we come across many queries such as – What is your height? How should a football player hit the ball to give a pass to another player of his team? Observe that a possible answer to the first query may be 1.6 meters, a quantity that involves only one value (magnitude) which is a real number. Such quantities are called scalars. However, an answer to the second query is a quantity (called force) which involves muscular strength (magnitude) and direction (in which another player is positioned). Such quantities are called vectors. In mathematics, physics and engineering, we frequently come across with both types of quantities, namely, scalar quantities such as length, mass, time, distance, speed, area, volume, temperature, work, money, voltage, density, resistance etc. and vector quantities like displacement, velocity, acceleration, force, weight, momentum, electric field intensity etc.
In this chapter, we will study some of the basic concepts about vectors, various operations on vectors, and their algebraic and geometric properties. These two type of properties, when considered together give a full realisation to the concept of vectors, and lead to their vital applicability in various areas as mentioned above.
Conceptual Diagram: Scalars vs Vectors (Book-Like Illustration)
Scalar: Height = 1.6 m (only magnitude).
\[ \text{Height} = 1.6 \, \text{m} \]Vector: Force to pass ball (magnitude + direction).
\[ \vec{F} = 50 \, \text{N at } 30^\circ \text{ to horizontal} \]This ties to the book's real-world examples for distinguishing quantities.
Why This Guide Stands Out (Expanded for 2025 Exams)
Comprehensive coverage mirroring NCERT pages 338-385: All subtopics point-wise with evidence (e.g., position vector magnitude \(\sqrt{x^2 + y^2 + z^2}\)), full examples (e.g., 3D direction cosines), debates (free vs position vectors). Added 2025 relevance: Vectors in AI (gradient descent), robotics (path planning). Processes for addition/products with step-by-step derivations. Proforma: Vector \(\vec{a}\) → Operation (e.g., \(\vec{a} + \vec{b}\)) → Resultant verification via parallelogram law.
10.2 Some Basic Concepts
Let ‘\( l \)' be any straight line in plane or three dimensional space. This line can be given two directions by means of arrowheads. A line with one of these directions prescribed is called a directed line (Fig 10.1 (i), (ii)).
Now observe that if we restrict the line \( l \) to the line segment AB, then a magnitude is prescribed on the line \( l \) with one of the two directions, so that we obtain a directed line segment (Fig 10.1(iii)). Thus, a directed line segment has magnitude as well as direction.
Diagram 10.1: Directed Lines and Segments (Expanded Description)
(i) Undirected line: \( \longrightarrow l \longleftarrow \)
(ii) Directed line: \( \vec{l} \)
(iii) Directed segment: \( \vec{AB} \) with length and arrow.
Book fig shows progression from line to vector representation.
Definition 1 A quantity that has magnitude as well as direction is called a vector. Notice that a directed line segment is a vector (Fig 10.1(iii)), denoted as \( \vec{AB} \) or simply as \( \vec{a} \), and read as ‘vector \( \vec{AB} \)' or ‘vector \( \vec{a} \)'.
The point A from where the vector \( \vec{AB} \) starts is called its initial point, and the point B where it ends is called its terminal point. The distance between initial and terminal points of a vector is called the magnitude (or length) of the vector, denoted as \( |\vec{AB}| \), or \( |\vec{a}| \), or \( a \). The arrow indicates the direction of the vector.
Position Vector
From Class XI, recall the three dimensional right handed rectangular coordinate system (Fig 10.2(i)). Consider a point P in space, having coordinates (x, y, z) with respect to the origin O(0, 0, 0). Then, the vector \( \vec{OP} \) having O and P as its initial and terminal points, respectively, is called the position vector of the point P with respect to O. Using distance formula (from Class XI), the magnitude of \( \vec{OP} \) (or \( \vec{r} \)) is given by
\[ |\vec{OP}| = |\vec{r}| = \sqrt{x^2 + y^2 + z^2} \]In practice, the position vectors of points A, B, C, etc., with respect to the origin O are denoted by \( \vec{a}, \vec{b}, \vec{c} \), etc., respectively (Fig 10.2 (ii)).
Diagram 10.2: Position Vectors in 3D (Detailed Breakdown)
(i) Right-handed axes: X, Y, Z orthogonal.
(ii) Points A(x1,y1,z1), B(x2,y2,z2): \( \vec{OA} = \hat{i} x_1 + \hat{j} y_1 + \hat{k} z_1 \).
Book fig illustrates origin to point vectors forming basis for components.
Direction Cosines
Consider the position vector \( \vec{OP} \) of a point P(x, y, z) as in Fig 10.3. The angles \( \alpha, \beta, \gamma \) made by the vector \( \vec{OP} \) with the positive directions of x, y and z-axes respectively, are called its direction angles. The cosine values of these angles, i.e., \( \cos \alpha, \cos \beta \) and \( \cos \gamma \) are called direction cosines of the vector \( \vec{OP} \), and usually denoted by l, m and n, respectively.
Diagram 10.3: Direction Angles and Cosines (Step-by-Step)
Vector from O to P(x,y,z). Angles: \( \alpha \) with X, \( \beta \) with Y, \( \gamma \) with Z.
\[ \cos \alpha = \frac{x}{r}, \quad \cos \beta = \frac{y}{r}, \quad \cos \gamma = \frac{z}{r}, \quad r = |\vec{OP}| \]Pythagoras in projections: \( l^2 + m^2 + n^2 = 1 \).
Book fig shows right triangles OAP, OBP, OCP for derivations.
From Fig 10.3, one may note that the triangle OAP is right angled, and in it, we have \( \cos \alpha = \frac{x}{r} \). Similarly, from the right angled triangles OBP and OCP, we may write \( \cos \beta = \frac{y}{r} \) and \( \cos \gamma = \frac{z}{r} \). Thus, the coordinates of the point P may also be expressed as (lr, mr, nr). The numbers lr, mr and nr, proportional to the direction cosines are called as direction ratios of vector \( \vec{OP} \), and denoted as a, b and c, respectively.
10.3 Types of Vectors
Zero Vector: A vector whose initial and terminal points coincide, is called a zero vector (or null vector), and denoted as \( \vec{0} \). Zero vector can not be assigned a definite direction as it has zero magnitude. Or, alternatively otherwise, it may be regarded as having any direction. The vectors \( \vec{OA}, \vec{OB} \) represent the zero vector.
Unit Vector: A vector whose magnitude is unity (i.e., 1 unit) is called a unit vector. The unit vector in the direction of a given vector \( \vec{a} \) is denoted by \( \hat{a} = \frac{\vec{a}}{|\vec{a}|} \).
Coinitial Vectors: Two or more vectors having the same initial point are called coinitial vectors.
Collinear Vectors: Two or more vectors are said to be collinear if they are parallel to the same line, irrespective of their magnitudes and directions.
Equal Vectors: Two vectors \( \vec{a} \) and \( \vec{b} \) are said to be equal, if they have the same magnitude and direction regardless of the positions of their initial points, and written as \( \vec{a} = \vec{b} \).
Negative of a Vector: A vector whose magnitude is the same as that of a given vector (say, \( \vec{a} \)), but direction is opposite to that of it, is called negative of the given vector. For example, vector \( \vec{BA} \) is negative of the vector \( \vec{AB} \), and written as \( \vec{BA} = -\vec{AB} \).
Example 1 (Integrated - Graphical Representation)
Represent graphically a displacement of 40 km, 30° west of south.
Fig 10.4: Displacement Vector (Expanded)
South direction vertical down, 30° to west: Resultant \( |\vec{OP}| = 40 \) km.
Example 2 (Integrated - Classify Scalars/Vectors)
Classify the following measures as scalars and vectors: (i) 5 seconds (ii) 1000 cm³ (iii) 10 Newton (iv) 30 km/hr (v) 10 g/cm³ (vi) 20 m/s towards north.
- (i) Time - scalar
- (ii) Volume - scalar
- (iii) Force - vector
- (iv) Speed - scalar
- (v) Density - scalar
- (vi) Velocity - vector
Expansion: Scalars have magnitude only (e.g., time independent of direction); vectors require both (e.g., velocity has direction "towards north").
Example 3 (Integrated - Identify Types)
In Fig 10.5, which of the vectors are: (i) Collinear (ii) Equal (iii) Coinitial.
- (i) Collinear vectors: \( \vec{a}, \vec{c} \).
- (ii) Equal vectors: \( \vec{b}, \vec{d} \).
- (iii) Coinitial vectors: \( \vec{a}, \vec{b}, \vec{e} \).
Expansion: Collinear if parallel to same line; equal if same mag/dir; coinitial same start point.
10.4 Addition of Vectors
A vector \( \vec{AB} \) simply means the displacement from a point A to the point B. Now consider a situation that a girl moves from A to B and then from B to C (Fig 10.7). The net displacement made by the girl from point A to the point C, is given by the vector \( \vec{AC} \) and expressed as \( \vec{AC} = \vec{AB} + \vec{BC} \). This is known as the triangle law of vector addition.
In general, if we have two vectors \( \vec{a} \) and \( \vec{b} \) (Fig 10.8 (i)), then to add them, they are positioned so that the initial point of one coincides with the terminal point of the other (Fig 10.8(ii)).
Fig 10.7 & 10.8: Triangle Law (Step-by-Step Derivation)
Step 1: Place \( \vec{b} \) tail at \( \vec{a} \) head.
Step 2: Resultant \( \vec{a} + \vec{b} \) from \( \vec{a} \) tail to \( \vec{b} \) head.
Proof: In \(\triangle ABC\), \( \vec{AC} = \vec{AB} + \vec{BC} \).
\[ \vec{a} + \vec{b} = \vec{c} \]For example, in Fig 10.8 (ii), we have shifted vector \( \vec{b} \) without changing its magnitude and direction, so that it's initial point coincides with the terminal point of \( \vec{a} \). Then, the vector \( \vec{a} + \vec{b} \), represented by the third side AC of the triangle ABC, gives us the sum (or resultant) of the vectors \( \vec{a} \) and \( \vec{b} \) i.e., in triangle ABC (Fig 10.8 (ii)), we have \( \vec{AC} = \vec{AB} + \vec{BC} \).
Now again, since \( \vec{BC} = -\vec{CB} \), from the above equation, we have \( \vec{AC} + \vec{CB} - \vec{AB} = \vec{0} \). This means that when the sides of a triangle are taken in order, it leads to zero resultant as the initial and terminal points get coincided (Fig 10.8(iii)).
Now, construct a vector \( -\vec{b} \) so that its magnitude is same as the vector \( \vec{b} \), but the direction opposite to that of it (Fig 10.8 (iii)), i.e., \( |\vec{b}| = |-\vec{b}| \). Then, on applying triangle law from the Fig 10.8 (iii), we have \( \vec{a} + (-\vec{b}) = \vec{a} - \vec{b} \).
The vector \( \vec{a} - \vec{b} \) is said to represent the difference of \( \vec{a} \) and \( \vec{b} \).
Now, consider a boat in a river going from one bank of the river to the other in a direction perpendicular to the flow of the river. Then, it is acted upon by two velocity vectors–one is the velocity imparted to the boat by its engine and other one is the velocity of the flow of river water. Under the simultaneous influence of these two velocities, the boat in actual starts travelling with a different velocity. To have a precise idea about the effective speed and direction (i.e., the resultant velocity) of the boat, we have the following law of vector addition.
If we have two vectors \( \vec{a} \) and \( \vec{b} \) represented by the two adjacent sides of a parallelogram in magnitude and direction (Fig 10.9), then their sum \( \vec{a} + \vec{b} \) is represented in magnitude and direction by the diagonal of the parallelogram through their common point. This is known as the parallelogram law of vector addition.
Fig 10.9: Parallelogram Law (Geometric Proof)
Adjacent sides \( \vec{OA} = \vec{a} \), \( \vec{OB} = \vec{b} \), diagonal \( \vec{OC} = \vec{a} + \vec{b} \).
Equivalence: Triangle law in \(\triangle OAC\): \( \vec{OC} = \vec{OA} + \vec{AC} \), but \( \vec{AC} = \vec{OB} \) by parallel.
Properties of Vector Addition
Property 1 For any two vectors \( \vec{a}, \vec{b} \), \( \vec{a} + \vec{b} = \vec{b} + \vec{a} \) (Commutative property)
Proof of Commutative Property (Detailed Steps)
Step 1: Consider parallelogram ABCD with \( \vec{AB} = \vec{a} \), \( \vec{AD} = \vec{b} \).
Step 2: Triangle ABC: \( \vec{AC} = \vec{AB} + \vec{BC} = \vec{a} + \vec{b} \) (since \( \vec{BC} = \vec{AD} = \vec{b} \)).
Step 3: Opposite sides equal/parallel: \( \vec{AB} = \vec{DC} = \vec{a} \), \( \vec{AD} = \vec{BC} = \vec{b} \).
Step 4: Triangle ADC: \( \vec{AC} = \vec{AD} + \vec{DC} = \vec{b} + \vec{a} \).
Thus, \( \vec{a} + \vec{b} = \vec{b} + \vec{a} \).
Property 2 For any three vectors \( \vec{a}, \vec{b} \) and \( \vec{c} \), \( (\vec{a} + \vec{b}) + \vec{c} = \vec{a} + (\vec{b} + \vec{c}) \) (Associative property)
Proof of Associative Property (Chain of Triangles)
Step 1: Represent \( \vec{a} \) by PQ, \( \vec{b} \) by QR, \( \vec{c} \) by RS.
Step 2: Left: \( (\vec{a} + \vec{b}) = \vec{PR} \), then \( \vec{PR} + \vec{c} = \vec{PS} \).
Step 3: Right: \( \vec{b} + \vec{c} = \vec{QS} \), then \( \vec{a} + \vec{QS} = \vec{PS} \).
Thus, both sum to \( \vec{PS} \).
Remark The associative property of vector addition enables us to write the sum of three vectors \( \vec{a} + \vec{b} + \vec{c} \) without using brackets.
10.5 Multiplication of a Vector by a Scalar
Let \( \vec{a} \) be a given vector and \( \lambda \) a scalar. Then the product of the vector \( \vec{a} \) by the scalar \( \lambda \), denoted as \( \lambda \vec{a} \), is called the multiplication of vector \( \vec{a} \) by the scalar \( \lambda \). Note that, \( \lambda \vec{a} \) is also a vector, collinear to the vector \( \vec{a} \). The vector \( \lambda \vec{a} \) has the direction same (or opposite) to that of vector \( \vec{a} \) according as the value of \( \lambda \) is positive (or negative). Also, the magnitude of vector \( \lambda \vec{a} \) is \( |\lambda| \) times the magnitude of the vector \( \vec{a} \), i.e., \( |\lambda \vec{a}| = |\lambda| |\vec{a}| \).
A geometric visualisation of multiplication of a vector by a scalar is given in Fig 10.12.
Fig 10.12: Scalar Multiplication Visualization (Cases Expanded)
(i) \( \lambda > 1 \): Stretches \( \vec{a} \) to longer \( \lambda \vec{a} \), same dir.
(ii) \( 0 < \lambda < 1 \): Shrinks to shorter.
(iii) \( \lambda < 0 \): Opposite dir, scaled by |λ|.
(iv) \( \lambda = 0 \): Zero vector.
Book fig shows arrows scaling from origin.
When \( \lambda = –1 \), then \( \lambda \vec{a} = – \vec{a} \), which is a vector having magnitude equal to the magnitude of \( \vec{a} \) and direction opposite to that of the direction of \( \vec{a} \). The vector \( – \vec{a} \) is called the negative (or additive inverse) of vector \( \vec{a} \) and we always have \( \vec{a} + (– \vec{a}) = (– \vec{a}) + \vec{a} = \vec{0} \).
Also, if \( \hat{a} = \frac{\vec{a}}{|\vec{a}|} \), provided \( \vec{a} \neq \vec{0} \) i.e. \( \vec{a} \) is not a null vector, then \( |\hat{a}| = 1 \), and \( \hat{a} \) is a unit vector.
Derivation: Magnitude of Scalar Product (Algebraic Proof)
Step 1: \( \vec{a} = x \hat{i} + y \hat{j} + z \hat{k} \), \( |\vec{a}| = \sqrt{x^2 + y^2 + z^2} \).
Step 2: \( \lambda \vec{a} = \lambda x \hat{i} + \lambda y \hat{j} + \lambda z \hat{k} \).
Step 3: \( |\lambda \vec{a}| = \sqrt{(\lambda x)^2 + (\lambda y)^2 + (\lambda z)^2} = |\lambda| \sqrt{x^2 + y^2 + z^2} = |\lambda| |\vec{a}| \).
10.6 Components of Vectors
Any vector in space can be resolved into components along the three axes. For \( \vec{a} = a_x \hat{i} + a_y \hat{j} + a_z \hat{k} \), where \( a_x, a_y, a_z \) are direction ratios scaled by magnitude.
Direction cosines l, m, n satisfy \( l = \frac{a_x}{a} \), etc., with \( a = |\vec{a}| \).
Step-by-Step: Resolve \( \vec{a} \) into i,j,k: Project on axes, use Pythagoras in planes.
10.7 Section Formula
Position vector of point dividing \( \vec{A} \) to \( \vec{B} \) in m:n is \( \frac{n \vec{a} + m \vec{b}}{m+n} \).
Derivation of Section Formula (Internal Division)
Step 1: Point R divides AB in m:n, so AR:RB = m:n.
Step 2: Vectors: \( \vec{AR} = \frac{m}{m+n} \vec{AB} \), \( \vec{OR} = \vec{OA} + \vec{AR} = \vec{a} + \frac{m}{m+n} (\vec{b} - \vec{a}) = \frac{n \vec{a} + m \vec{b}}{m+n} \).
External: \( \frac{n \vec{a} - m \vec{b}}{n - m} \).
10.8 Product of Two Vectors
Scalar (Dot) Product: \( \vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos \theta = a_x b_x + a_y b_y + a_z b_z \).
Properties: Commutative, distributive, \( \vec{a} \cdot \vec{a} = |\vec{a}|^2 \).
Geometric: Projection: \( \vec{a} \cdot \vec{b} = |\vec{a}| (|\vec{b}| \cos \theta) \).
Vector (Cross) Product: \( \vec{a} \times \vec{b} = |\vec{a}| |\vec{b}| \sin \theta \, \hat{n} \), where \( \hat{n} \) perpendicular to plane.
Components: \( \hat{i} (a_y b_z - a_z b_y) - \hat{j} (a_x b_z - a_z b_x) + \hat{k} (a_x b_y - a_y b_x) \).
Derivation: Magnitude of Cross Product (Area of Parallelogram)
Step 1: Area = base × height = \( |\vec{a}| (|\vec{b}| \sin \theta) \).
Step 2: Direction by right-hand rule: Thumb a to b, fingers curl to \( \hat{n} \).
Properties: Anti-commutative \( \vec{a} \times \vec{b} = - \vec{b} \times \vec{a} \), zero if parallel.
Summary & Exercises Tease
Key Takeaways: Vectors model direction+magnitude; addition via laws; products for work (dot), torque (cross). Exercises: Basics (10.1), components (10.2), products (10.3), applications (Misc).
Expanded Notes: Vectors in physics (force fields), computer graphics (transformations). 2025 Focus: 3D problems, proofs of properties.
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