This MCQ module is based on: Introduction to Vectors
TOPIC 9 OF 33
Introduction to Vectors
🎓 Class 11
Physics
CBSE
Theory
Ch 3 – Motion in a Plane
⏱ ~14 min
🌐 Language: [gtranslate]
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Introduction to Vectors
3.1 Introduction
In the previous chapter, we developed the concepts of position, displacement, velocity and acceleration that are needed to describe the motion of an object along a straight line. We found that the directional aspect of these quantities can be taken care of by + and − signs, as in one dimension only two directions are possible. But in order to describe motion of an object in two dimensions (a plane) or three dimensions (space), we need to use vectors to describe the direction-mentioned physical quantities. Therefore, it is first necessary to learn the language of vectors.
What is a vector? How to add, subtract and multiply vectors? What is the result of multiplying a vector by a real number? We shall learn this to enable us to use vectors for defining velocity and acceleration in a plane. We then discuss motion of an object in a plane. As a simple case of motion in a plane, we shall discuss motion with constant acceleration and treat in detail the projectile motion. Circular motion is a familiar class of motion that has a special significance in daily-life situations. We shall discuss uniform circular motion in some detail.
The equations developed in this chapter for motion in a plane can be easily extended to the case of three dimensions.
3.2 Scalars and Vectors
In physics, we can classify quantities as scalars or vectors. Basically, the difference is that a direction is associated with a vector but not with a scalar.
Scalar: A scalar quantity is a quantity with magnitude only. It is specified completely by a single number, along with the proper unit. Examples are: distance, mass, temperature, time, speed, work and energy.
Vector: A vector quantity has both magnitude and direction and obeys the triangle law of addition (or, equivalently, the parallelogram law of addition). So, a vector is specified by giving its magnitude by a number and its direction. Examples are: displacement, velocity, acceleration and force.
Vector: A vector quantity has both magnitude and direction and obeys the triangle law of addition (or, equivalently, the parallelogram law of addition). So, a vector is specified by giving its magnitude by a number and its direction. Examples are: displacement, velocity, acceleration and force.
Scalars can be added, subtracted, multiplied and divided just as the ordinary numbers. For example, if the length and breadth of a rectangle are 1.0 m and 0.5 m respectively, then its perimeter is the sum of the lengths of the four sides, 1.0 m + 0.5 m + 1.0 m + 0.5 m = 3.0 m. The length of each side is a scalar and the perimeter is also a scalar.
A vector is represented by a bold-face letter or by an arrow placed over a letter. Thus, we can write the velocity vector by \(\vec{v}\) or by v. Since printing of bold-face letters is inconvenient, we will use the notation with an arrow on top throughout this chapter.
The magnitude of a vector is often called its absolute value, indicated by \(|\vec{v}|\) or simply \(v\). Thus, a vector is represented by a bold-face letter or by an arrow on top of a letter, and the magnitude (which is always positive) is shown by the same letter without bold or arrow.
3.2.1 Position and Displacement Vectors
To describe the position of an object moving in a plane, we need to choose a convenient point, say O, as origin. Let P and P' be the positions of the object at time \(t\) and \(t'\) respectively (Fig. 3.1a). We join O and P by a straight line. Then, OP is the position vector of the object at time \(t\). An arrow is marked at the head of this line. It is represented by a symbol \(\vec{r}\), i.e. \(\vec{r} = \vec{OP}\). Point P' is represented by another position vector, \(\vec{OP'}\), denoted by \(\vec{r}'\).
The length of the vector \(\vec{r}\) represents the magnitude of the vector and its direction is the direction in which P lies as seen from O. If the object moves from P to P', the vector \(\vec{PP'}\) (with tail at P and head at P') is called the displacement vector corresponding to motion from point P (at time \(t\)) to point P' (at time \(t'\)).
Crucial Point: The displacement vector depends only on the initial and final positions, NOT on the actual path taken. So even if the object follows a curved path between P and P', the displacement is still the straight arrow from P to P'.
3.2.2 Equality of Vectors
Two vectors \(\vec{A}\) and \(\vec{B}\) are said to be equal if, and only if, they have the same magnitude and the same direction.
The position of the tail of a vector is irrelevant. A vector can be shifted to a parallel position without changing the vector itself. This means a vector retains its identity under parallel transport.
Why direction matters: Two displacement vectors of the same magnitude (say, 5 m) but pointing in different directions correspond to physically different motions. Equality requires both magnitude AND direction to match.
🎯 Interactive Simulation: Position Vector Explorer
Drag the sliders below to change the position of point P in the plane. Observe how the position vector \(\vec{r}\) and its magnitude \(|\vec{r}|\) update in real-time.
Observe: The vector's magnitude is \(|\vec{r}| = \sqrt{x^2 + y^2}\), and its direction is \(\theta = \tan^{-1}(y/x)\) measured from the positive x-axis.
3.3 Multiplication of Vectors by Real Numbers
Multiplying a vector \(\vec{A}\) with a positive number \(\lambda\) gives a vector whose magnitude is changed by the factor \(\lambda\) but the direction is the same as that of \(\vec{A}\):
\[|\lambda \vec{A}| = \lambda |\vec{A}|, \quad \text{if } \lambda > 0\]
For example, if \(\vec{A}\) is multiplied by 2, the resulting vector \(2\vec{A}\) is in the same direction as \(\vec{A}\) and has a magnitude twice that of \(|\vec{A}|\).
Multiplying a vector \(\vec{A}\) by a negative number \(\lambda\) gives a vector \(\lambda \vec{A}\) whose direction is opposite to the direction of \(\vec{A}\) and whose magnitude is \(|\lambda||\vec{A}|\). For example, multiplying a given vector \(\vec{A}\) by negative numbers, say −1 and −1.5, gives vectors as shown:
The factor \(\lambda\) by which a vector \(\vec{A}\) is multiplied could be a scalar having its own physical dimension. Then, the dimension of \(\lambda\vec{A}\) is the product of the dimensions of \(\lambda\) and \(\vec{A}\). For example, when you multiply a constant velocity vector by time, you get a displacement vector:
\[\vec{r} = \vec{v} \cdot t \quad \text{[m] = [m/s]} \cdot \text{[s]}\]
Worked Example 1: Position Vector Magnitude
A particle is located at coordinates (3 m, 4 m) relative to the origin O. (a) Write down its position vector \(\vec{r}\). (b) Find the magnitude \(|\vec{r}|\). (c) Find the angle that \(\vec{r}\) makes with the positive x-axis.
Solution:
Given: P = (3, 4) m, origin O = (0, 0).
(a) Position vector: \[\vec{r} = 3\hat{i} + 4\hat{j} \text{ m}\] where \(\hat{i}\) and \(\hat{j}\) are unit vectors along x and y axes.
(b) Magnitude: \[|\vec{r}| = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = \boxed{5 \text{ m}}\]
(c) Angle with x-axis: \[\tan\theta = \frac{y}{x} = \frac{4}{3}\] \[\theta = \tan^{-1}\left(\frac{4}{3}\right) \approx \boxed{53.13°}\]
Given: P = (3, 4) m, origin O = (0, 0).
(a) Position vector: \[\vec{r} = 3\hat{i} + 4\hat{j} \text{ m}\] where \(\hat{i}\) and \(\hat{j}\) are unit vectors along x and y axes.
(b) Magnitude: \[|\vec{r}| = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = \boxed{5 \text{ m}}\]
(c) Angle with x-axis: \[\tan\theta = \frac{y}{x} = \frac{4}{3}\] \[\theta = \tan^{-1}\left(\frac{4}{3}\right) \approx \boxed{53.13°}\]
Worked Example 2: Multiplication by Scalar
A car has velocity \(\vec{v} = 20\) m/s pointing east. After the driver applies the brakes, the velocity becomes \(\vec{v'} = 0.4 \vec{v}\). (a) Find the new velocity. (b) Find the velocity vector \(-\vec{v}\) and explain its physical meaning.
Solution:
(a) Multiplying by positive scalar 0.4 keeps direction same, scales magnitude: \[\vec{v'} = 0.4 \times 20 = \boxed{8 \text{ m/s, east}}\]
(b) Multiplying by −1 reverses direction without changing magnitude: \[-\vec{v} = \boxed{20 \text{ m/s, west}}\] Physical meaning: a car moving with the same speed in the opposite direction.
(a) Multiplying by positive scalar 0.4 keeps direction same, scales magnitude: \[\vec{v'} = 0.4 \times 20 = \boxed{8 \text{ m/s, east}}\]
(b) Multiplying by −1 reverses direction without changing magnitude: \[-\vec{v} = \boxed{20 \text{ m/s, west}}\] Physical meaning: a car moving with the same speed in the opposite direction.
📐 Activity 3.1 — Visualizing Vector Equality
L3 Apply
Materials: Graph paper, ruler, protractor, coloured pencils.
Procedure:
- Draw an arrow of length 5 cm at angle 30° from the horizontal at the origin (0,0). Label this vector \(\vec{A}\).
- Now, starting at point (3 cm, 2 cm), draw another arrow of length 5 cm also at angle 30° from horizontal. Label this \(\vec{B}\).
- Starting again at origin, draw a third arrow of length 5 cm but at angle 60° from horizontal. Label this \(\vec{C}\).
- Measure the magnitude and angle of each vector with your ruler and protractor.
Predict before measuring: Which two vectors are equal? Which two have the same magnitude but different directions?
Observation:
- \(\vec{A}\) and \(\vec{B}\): Both have magnitude 5 cm and direction 30°. Different starting points, but same magnitude and direction → equal vectors.
- \(\vec{A}\) and \(\vec{C}\): Same magnitude (5 cm), but different directions (30° vs 60°) → not equal.
Conclusion: A vector is unchanged by parallel translation. Equality of vectors requires both magnitude AND direction to match. Position of the tail is irrelevant.
🎯 Competency-Based Questions
A surveyor at point O records the position of two trees A and B. Tree A is at coordinates (4 m, 3 m) relative to O. Tree B is at (-3 m, 4 m) from O. The surveyor wants to express these positions as vectors and analyze the relationships between them.
Q1. Identify the position vector of Tree A in component form.L3 Apply
Answer: (b). The x-coordinate is 4 m and y-coordinate is 3 m, so \(\vec{r}_A = 4\hat{i} + 3\hat{j}\) m. Magnitude = \(\sqrt{16+9} = 5\) m.
Q2. The surveyor compares the magnitudes of \(\vec{r}_A\) and \(\vec{r}_B\). Determine whether they are equal vectors. L4 Analyse
Answer: \(|\vec{r}_A| = \sqrt{4^2 + 3^2} = 5\) m. \(|\vec{r}_B| = \sqrt{(-3)^2 + 4^2} = 5\) m. Both have equal magnitudes (5 m), but their directions are different — \(\vec{r}_A\) points into Q1 (NE), \(\vec{r}_B\) points into Q2 (NW). So they are NOT equal vectors. Equality requires same magnitude AND same direction.
Q3. State whether the following statement is true or false: "If two displacement vectors have the same magnitude, they must be equal." Justify your answer with a counter-example. L5 Evaluate
Answer: FALSE. Counter-example: Walking 5 m east and walking 5 m north both have magnitude 5 m, but they are different vectors because directions differ. Vector equality requires BOTH same magnitude AND same direction. This is precisely why we need vectors instead of just scalars in 2D and 3D physics — direction is an essential part of the description.
Q4. Fill in the blank: A vector multiplied by −1 has the same _______ but opposite _______. L2 Understand
Answer: magnitude; direction. Multiplication by −1 is called the negative of a vector. \(-\vec{A}\) has \(|\vec{A}|\) units of magnitude but points in the exactly opposite direction.
Q5. HOT (Higher Order Thinking): Design a method using only a graph paper and a protractor to verify whether two given displacement arrows represent equal vectors. List the minimum measurements required. L6 Create
Answer (Sample Solution):
- Measure magnitude: Use ruler to find the length of each arrow. If lengths differ, vectors are unequal.
- Measure direction: Use protractor to measure the angle each arrow makes with a fixed reference (e.g., positive x-axis).
- Compare: Vectors are equal ONLY if BOTH measurements match.
🧠 Assertion–Reason Questions
Choose the correct option for each pair:
(A) Both Assertion and Reason are true; Reason is the correct explanation of Assertion.
(B) Both Assertion and Reason are true; Reason is NOT the correct explanation of Assertion.
(C) Assertion is true; Reason is false. (D) Assertion is false; Reason is true.
Assertion (A): The displacement vector between two points depends only on the initial and final positions, not on the path travelled.
Reason (R): Displacement is defined as the straight-line vector from the initial to the final position.
Answer: (A). Both statements are true and R correctly explains A. Displacement is purely a function of endpoints, regardless of curved or straight motion in between.
Assertion (A): Two vectors of equal magnitude are always equal vectors.
Reason (R): Vectors have direction in addition to magnitude.
Answer: (D). The Assertion is FALSE — equal magnitudes alone do not make equal vectors; direction must also match. The Reason is TRUE and actually shows why the assertion is wrong (R contradicts A).
Assertion (A): Multiplying a vector by −2 produces a vector that is twice as long and points in the opposite direction.
Reason (R): Negative scalars reverse the direction of a vector while their magnitude scales the length.
Answer: (A). Both A and R are true, and R correctly explains A. \(|-2\vec{A}| = 2|\vec{A}|\) and direction reverses.
Frequently Asked Questions - Introduction to Vectors
What is the main concept covered in Introduction to Vectors?
In NCERT Class 11 Physics Chapter 3 (Motion in a Plane), "Introduction to Vectors" covers core principles and equations needed for board exam success. The MyAiSchool lesson explains the topic with definitions, derivations, worked examples, and interactive simulations. Key formulas and dimensional analysis are included to build conceptual depth and problem-solving skills aligned with the CBSE 2025-26 syllabus.
How is Introduction to Vectors useful in real-life applications?
Real-life applications of Introduction to Vectors from NCERT Class 11 Physics Chapter 3 include engineering design, satellite mechanics, sports biomechanics, transportation safety, and electrical/electronic devices. The MyAiSchool lesson links every concept to a tangible example so students see physics as a problem-solving framework for the physical world, not as abstract formulas.
What are the key formulas in Introduction to Vectors?
Key formulas in Introduction to Vectors (NCERT Class 11 Physics Chapter 3 Motion in a Plane) are derived step-by-step in the MyAiSchool lesson. Students should memorize the final formula AND understand its derivation for full board marks. Each formula is listed with its dimensional formula, SI unit, applicability range, and common pitfalls. The Summary section at the end of each part includes a quick-reference formula card.
How does this part connect to other parts of Chapter 3?
NCERT Class 11 Physics Chapter 3 (Motion in a Plane) is structured so each part builds on the previous one. Introduction to Vectors connects directly to neighbouring parts via shared definitions, units, and methodology. The MyAiSchool lesson cross-references related concepts with internal links so students can navigate the whole chapter as one connected story rather than disconnected fragments.
What types of CBSE board questions come from Introduction to Vectors?
CBSE board questions from Introduction to Vectors typically include: (1) 1-mark MCQs on definitions and formulas, (2) 2-mark short-answer derivations or applications, (3) 3-mark numerical problems with units, (4) 5-mark long-answer derivations followed by application. The MyAiSchool lesson tags each Competency-Based Question (CBQ) with Bloom level (L1-L6) so students know how to study for each weight.
How can students use the interactive simulation effectively?
The interactive simulation in the Introduction to Vectors lesson allows students to adjust input parameters (sliders or selectors) and see physical quantities update in real time. To use it effectively: (1) try extreme values to understand limiting cases, (2) compare with the analytical formula, (3) check unit consistency, (4) test special configurations from worked examples. The simulation reinforces conceptual intuition that pure formula manipulation cannot.
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