This MCQ module is based on: Vector Operations
TOPIC 10 OF 33
Vector Operations
🎓 Class 11
Physics
CBSE
Theory
Ch 3 – Motion in a Plane
⏱ ~14 min
🌐 Language: [gtranslate]
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Vector Operations
3.4 Addition and Subtraction of Vectors — Graphical Method
As mentioned earlier, vectors, by definition, obey the triangle law of addition. We shall now describe this law of addition using the graphical method. Let us consider two vectors \(\vec{A}\) and \(\vec{B}\) that lie in a plane as shown in Fig. 3.4(a). The lengths of the line segments representing these vectors are proportional to the magnitude of the vectors.
To find the sum \(\vec{A} + \vec{B}\), we place vector \(\vec{B}\) so that its tail is at the head of the vector \(\vec{A}\), as in Fig. 3.4(b). Then, we join the tail of \(\vec{A}\) to the head of \(\vec{B}\). This line OQ represents a vector \(\vec{R}\), that is, the sum of the vectors \(\vec{A}\) and \(\vec{B}\). Since, in this procedure of vector addition, vectors are arranged head-to-tail, this method is called the head-to-tail method.
Properties of Vector Addition
Vector addition is commutative:
\[\vec{A} + \vec{B} = \vec{B} + \vec{A}\]
Vector addition obeys the associative law:
\[(\vec{A} + \vec{B}) + \vec{C} = \vec{A} + (\vec{B} + \vec{C})\]
The result of adding vectors \(\vec{A}\), \(\vec{B}\), and \(\vec{C}\) does not depend on the order in which they are added.
Null Vector (Zero Vector)
What is the result of adding two vectors \(\vec{A}\) and \(-\vec{A}\)? The vectors \(\vec{A}\) and \(-\vec{A}\) have equal magnitudes but opposite directions. Their head-to-tail addition gives a closed path. The resulting "vector" has zero length and is called a null vector or zero vector:
\[\vec{A} + (-\vec{A}) = \vec{0}\]
The null vector \(\vec{0}\) has the following properties:
- \(\vec{A} + \vec{0} = \vec{A}\)
- \(\lambda \vec{0} = \vec{0}\) for any scalar \(\lambda\)
- \(0 \vec{A} = \vec{0}\)
Subtraction of Vectors
The subtraction of vector \(\vec{B}\) from \(\vec{A}\) is defined as the addition of vector \(-\vec{B}\) to \(\vec{A}\):
\[\vec{A} - \vec{B} = \vec{A} + (-\vec{B})\]
Parallelogram Law of Vector Addition
An equivalent method for adding vectors is the parallelogram law of addition. The two vectors \(\vec{A}\) and \(\vec{B}\) are drawn with a common origin O. The parallelogram is then constructed by drawing lines parallel to each vector through the head of the other. The diagonal of the parallelogram from O represents the resultant \(\vec{R}\).
Magnitude of Resultant (Parallelogram Law):
\[|\vec{R}| = \sqrt{A^2 + B^2 + 2AB\cos\theta}\]
where \(A = |\vec{A}|\), \(B = |\vec{B}|\), and \(\theta\) is the angle between \(\vec{A}\) and \(\vec{B}\).
Direction of Resultant (angle α with \(\vec{A}\)): \[\tan\alpha = \frac{B\sin\theta}{A + B\cos\theta}\]
Direction of Resultant (angle α with \(\vec{A}\)): \[\tan\alpha = \frac{B\sin\theta}{A + B\cos\theta}\]
🎯 Interactive Simulation: Parallelogram Law Explorer
Adjust the magnitudes of vectors \(\vec{A}\) and \(\vec{B}\) and the angle θ between them. Observe how the resultant \(\vec{R}\) changes.
Try this: Set θ = 0° (vectors aligned) → |R| is maximum (A + B). Set θ = 180° (opposite) → |R| is minimum (|A − B|). Set θ = 90° → |R| = √(A² + B²).
3.5 Resolution of Vectors
Let \(\vec{a}\) and \(\vec{b}\) be any two non-zero vectors in a plane with different directions and let \(\vec{A}\) be another vector in the same plane. \(\vec{A}\) can be expressed as a sum of two vectors — one obtained by multiplying \(\vec{a}\) by a real number and the other obtained by multiplying \(\vec{b}\) by another real number. To see this, let O and P be the tail and head of the vector \(\vec{A}\). Then, through O, draw a straight line parallel to \(\vec{a}\), and through P, a straight line parallel to \(\vec{b}\). Let them intersect at Q. Then, we have:
\[\vec{A} = \vec{OP} = \vec{OQ} + \vec{QP}\]
But since \(\vec{OQ}\) is parallel to \(\vec{a}\) and \(\vec{QP}\) is parallel to \(\vec{b}\), we can write \(\vec{OQ} = \lambda \vec{a}\) and \(\vec{QP} = \mu \vec{b}\), where \(\lambda\) and \(\mu\) are real numbers. Therefore:
\[\vec{A} = \lambda \vec{a} + \mu \vec{b}\]
We say that \(\vec{A}\) has been resolved into two component vectors \(\lambda \vec{a}\) and \(\mu \vec{b}\) along \(\vec{a}\) and \(\vec{b}\) respectively. Using this method one can resolve a given vector into two component vectors along a set of two vectors — all the three lie in the same plane.
Unit Vectors
It is convenient to resolve a general vector along the axes of a rectangular coordinate system using vectors of unit magnitude called unit vectors. A unit vector is a vector of unit magnitude and points in a particular direction. It has no dimension and unit. It is used to specify a direction only.
The unit vectors along the x-, y-, and z-axes of a rectangular coordinate system are denoted by \(\hat{i}, \hat{j}, \hat{k}\) respectively. These three unit vectors are perpendicular to each other:
\[|\hat{i}| = |\hat{j}| = |\hat{k}| = 1\]
Resolution Equations (2D): If \(\vec{A}\) makes angle θ with the x-axis:
\[A_x = A\cos\theta, \quad A_y = A\sin\theta\]
\[\vec{A} = A_x \hat{i} + A_y \hat{j}\]
\[A = \sqrt{A_x^2 + A_y^2}, \quad \tan\theta = \frac{A_y}{A_x}\]
3.6 Vector Addition — Analytical Method
Although the graphical method of adding vectors helps us in visualising the vectors and the resultant vector, it is sometimes tedious and has limited accuracy. It is much easier to add vectors by combining their respective components. Consider two vectors \(\vec{A}\) and \(\vec{B}\) in the x-y plane with components \(A_x, A_y\) and \(B_x, B_y\):
\[\vec{A} = A_x\hat{i} + A_y\hat{j}\]
\[\vec{B} = B_x\hat{i} + B_y\hat{j}\]
Then their sum \(\vec{R} = \vec{A} + \vec{B}\) has components:
\[\vec{R} = (A_x + B_x)\hat{i} + (A_y + B_y)\hat{j}\]
\[R_x = A_x + B_x, \quad R_y = A_y + B_y\]
The magnitude and direction of the resultant are then:
\[|\vec{R}| = \sqrt{R_x^2 + R_y^2}, \quad \tan\theta = \frac{R_y}{R_x}\]
Worked Examples
Worked Example 1 (NCERT Example 3.1): Resultant of Two Vectors
Find the resultant of two vectors \(\vec{A} = 3\hat{i} + 4\hat{j}\) and \(\vec{B} = 5\hat{i} - 2\hat{j}\). Also find its magnitude and direction.
Solution:
\[\vec{R} = \vec{A} + \vec{B} = (3+5)\hat{i} + (4-2)\hat{j}\] \[\vec{R} = 8\hat{i} + 2\hat{j}\]
Magnitude: \[|\vec{R}| = \sqrt{8^2 + 2^2} = \sqrt{64+4} = \sqrt{68} \approx \boxed{8.25}\]
Direction: \[\tan\theta = \frac{2}{8} = 0.25\] \[\theta = \tan^{-1}(0.25) \approx \boxed{14.04°} \text{ from x-axis}\]
\[\vec{R} = \vec{A} + \vec{B} = (3+5)\hat{i} + (4-2)\hat{j}\] \[\vec{R} = 8\hat{i} + 2\hat{j}\]
Magnitude: \[|\vec{R}| = \sqrt{8^2 + 2^2} = \sqrt{64+4} = \sqrt{68} \approx \boxed{8.25}\]
Direction: \[\tan\theta = \frac{2}{8} = 0.25\] \[\theta = \tan^{-1}(0.25) \approx \boxed{14.04°} \text{ from x-axis}\]
Worked Example 2 (NCERT Example 3.2): Resultant Velocity of Boat in River
A motorboat is racing towards north at 25 km/h and the water current in that region is 10 km/h in the direction 60° east of south. Find the resultant velocity of the boat.
Given:
\(v_b\) = 25 km/h (north) → \(\vec{v_b} = 25\hat{j}\)
\(v_c\) = 10 km/h, 60° east of south
Step 1: Resolve \(\vec{v_c}\) into components.
The current direction makes 60° east of south, so: \[v_{cx} = 10\sin60° = 10 \times 0.866 = 8.66 \text{ km/h (east)}\] \[v_{cy} = -10\cos60° = -10 \times 0.5 = -5 \text{ km/h (south)}\] \[\vec{v_c} = 8.66\hat{i} - 5\hat{j}\]
Step 2: Add vectors. \[\vec{v_R} = \vec{v_b} + \vec{v_c} = 8.66\hat{i} + (25-5)\hat{j} = 8.66\hat{i} + 20\hat{j}\]
Step 3: Magnitude and direction. \[|\vec{v_R}| = \sqrt{8.66^2 + 20^2} = \sqrt{75 + 400} = \sqrt{475} \approx \boxed{21.8 \text{ km/h}}\] \[\tan\theta = \frac{8.66}{20} = 0.433 \Rightarrow \theta \approx 23.4°\] The resultant velocity is approximately \(\boxed{21.8}\) km/h directed at \(\approx 23.4°\) east of north.
\(v_b\) = 25 km/h (north) → \(\vec{v_b} = 25\hat{j}\)
\(v_c\) = 10 km/h, 60° east of south
Step 1: Resolve \(\vec{v_c}\) into components.
The current direction makes 60° east of south, so: \[v_{cx} = 10\sin60° = 10 \times 0.866 = 8.66 \text{ km/h (east)}\] \[v_{cy} = -10\cos60° = -10 \times 0.5 = -5 \text{ km/h (south)}\] \[\vec{v_c} = 8.66\hat{i} - 5\hat{j}\]
Step 2: Add vectors. \[\vec{v_R} = \vec{v_b} + \vec{v_c} = 8.66\hat{i} + (25-5)\hat{j} = 8.66\hat{i} + 20\hat{j}\]
Step 3: Magnitude and direction. \[|\vec{v_R}| = \sqrt{8.66^2 + 20^2} = \sqrt{75 + 400} = \sqrt{475} \approx \boxed{21.8 \text{ km/h}}\] \[\tan\theta = \frac{8.66}{20} = 0.433 \Rightarrow \theta \approx 23.4°\] The resultant velocity is approximately \(\boxed{21.8}\) km/h directed at \(\approx 23.4°\) east of north.
Worked Example 3 (NCERT Example 3.3): Magnitude and Angle Calculations
Rain is falling vertically downward with a speed of 35 m/s. Wind starts blowing after sometime with a speed of 12 m/s in the east-west direction. In which direction should a boy waiting at a bus stop hold his umbrella?
Given:
\(v_r\) = 35 m/s (vertical down) → \(\vec{v_r} = -35\hat{j}\)
\(v_w\) = 12 m/s (east, say) → \(\vec{v_w} = 12\hat{i}\)
Resultant velocity of rain (relative to boy): \[\vec{v_R} = 12\hat{i} - 35\hat{j}\] \[|\vec{v_R}| = \sqrt{12^2 + 35^2} = \sqrt{144+1225} = \sqrt{1369} = 37 \text{ m/s}\]
Direction: The rain appears to come from the direction the resultant points TO. Angle with vertical: \[\tan\theta = \frac{v_w}{v_r} = \frac{12}{35} = 0.343\] \[\theta = \tan^{-1}(0.343) \approx \boxed{18.9°}\] The boy should hold the umbrella tilted at about \(\boxed{19°}\) east of vertical (towards the wind direction).
\(v_r\) = 35 m/s (vertical down) → \(\vec{v_r} = -35\hat{j}\)
\(v_w\) = 12 m/s (east, say) → \(\vec{v_w} = 12\hat{i}\)
Resultant velocity of rain (relative to boy): \[\vec{v_R} = 12\hat{i} - 35\hat{j}\] \[|\vec{v_R}| = \sqrt{12^2 + 35^2} = \sqrt{144+1225} = \sqrt{1369} = 37 \text{ m/s}\]
Direction: The rain appears to come from the direction the resultant points TO. Angle with vertical: \[\tan\theta = \frac{v_w}{v_r} = \frac{12}{35} = 0.343\] \[\theta = \tan^{-1}(0.343) \approx \boxed{18.9°}\] The boy should hold the umbrella tilted at about \(\boxed{19°}\) east of vertical (towards the wind direction).
📐 Activity 3.2 — Vector Addition with Paper Cutouts
L3 Apply
Materials: Stiff paper, scissors, ruler, protractor.
Procedure:
- On stiff paper, draw two arrows: \(\vec{A}\) of length 6 cm and \(\vec{B}\) of length 8 cm, making 90° between them. Cut them out.
- Place \(\vec{A}\) on graph paper. Then place the tail of \(\vec{B}\) at the head of \(\vec{A}\) (head-to-tail method).
- Draw an arrow from the tail of \(\vec{A}\) to the head of \(\vec{B}\). Measure its length and angle.
- Now reverse the order: place \(\vec{B}\) first, then \(\vec{A}\) at its head. Compare the two resultants.
Predict: What will be the magnitude of the resultant? Will the order matter?
Observation: The resultant has magnitude \(\sqrt{6^2 + 8^2} = 10\) cm regardless of order. The angle from \(\vec{A}\) is \(\tan^{-1}(8/6) = 53.13°\).
Conclusion: Vector addition is commutative: \(\vec{A} + \vec{B} = \vec{B} + \vec{A}\). The path traced by head-to-tail differs (first east-then-north vs. first-north-then-east), but the resultant arrow is the same in both cases.
🎯 Competency-Based Questions
An aeroplane is flying due north with velocity 200 km/h relative to still air. A wind starts blowing from west to east at 60 km/h. The pilot wants to determine the actual velocity (ground velocity) of the plane.
Q1. Calculate the magnitude of the ground velocity of the aeroplane. L3 Apply
Answer: Plane velocity vector = \(200\hat{j}\) km/h. Wind = \(60\hat{i}\) km/h. Ground velocity = \(60\hat{i} + 200\hat{j}\). Magnitude = \(\sqrt{60^2 + 200^2} = \sqrt{3600 + 40000} = \sqrt{43600} \approx \boxed{208.8\text{ km/h}}\).
Q2. Find the angle between the ground velocity and the original heading (north). L4 Analyse
Answer: \(\tan\theta = \frac{60}{200} = 0.3\); \(\theta = \tan^{-1}(0.3) \approx \boxed{16.7°}\) east of north. The pilot must turn slightly west of intended bearing to compensate.
Q3. The pilot wants to actually move due north relative to the ground despite the wind. In which direction (relative to north) must they steer the plane? L5 Evaluate
Answer: To cancel the eastward wind, the plane's velocity must have an x-component of −60 km/h. If plane heading makes angle α west of north: \(200\sin\alpha = 60 \Rightarrow \sin\alpha = 0.3 \Rightarrow \alpha = \boxed{17.46° \text{ west of north}}\). The northward speed becomes \(200\cos17.46° \approx 190.7\) km/h.
Q4. Two vectors of magnitudes 5 N and 12 N act at right angles. Find the magnitude of their resultant. L3 Apply
Answer: (b) 13 N. \(R = \sqrt{5^2 + 12^2} = \sqrt{169} = 13\) N. (This is the famous 5-12-13 Pythagorean triple.)
Q5. HOT: A student claims: "If I add three vectors of magnitudes 3, 4, and 5 units, the resultant must always have magnitude 12." Critique this claim. Construct a configuration where the resultant is zero. L6 Create
Answer: The claim is FALSE. The resultant magnitude depends on the directions, not just the magnitudes. Maximum resultant = 3+4+5 = 12 (only when all are parallel). Minimum can be as low as 0.
Zero-resultant configuration: A 3-4-5 right triangle! Take \(\vec{A} = 3\hat{i}\), \(\vec{B} = 4\hat{j}\), \(\vec{C} = -3\hat{i} - 4\hat{j}\). Then \(|\vec{C}| = 5\) and \(\vec{A} + \vec{B} + \vec{C} = 0\). This is the principle behind force equilibrium in static structures.
Zero-resultant configuration: A 3-4-5 right triangle! Take \(\vec{A} = 3\hat{i}\), \(\vec{B} = 4\hat{j}\), \(\vec{C} = -3\hat{i} - 4\hat{j}\). Then \(|\vec{C}| = 5\) and \(\vec{A} + \vec{B} + \vec{C} = 0\). This is the principle behind force equilibrium in static structures.
🧠 Assertion–Reason Questions
Choose: (A) Both A and R true, R explains A. (B) Both true, R doesn't explain A. (C) A true, R false. (D) A false, R true.
A: The resultant of two vectors of equal magnitude depends on the angle between them.
R: By parallelogram law, \(|R| = \sqrt{2A^2(1 + \cos\theta)}\) when both magnitudes equal A.
Answer: (A). Both true; R correctly derives the magnitude formula showing dependence on θ.
A: A vector can have a component greater than its magnitude.
R: Components are projections of the vector onto the axes, calculated using sine and cosine.
Answer: (D). A is FALSE: components are bounded — \(|A_x| = |A\cos\theta| \le A\) and \(|A_y| = |A\sin\theta| \le A\). R is TRUE: this is exactly why components can never exceed the vector's own magnitude.
A: Vector addition is commutative and associative.
R: The position of the tail does not affect the identity of a vector.
Answer: (B). Both A and R are true. R is independently true (parallel-translation invariance) but does not directly explain commutativity. The deeper reason is that head-to-tail addition forms a closed parallelogram irrespective of order.
Frequently Asked Questions - Vector Operations
What is the main concept covered in Vector Operations?
In NCERT Class 11 Physics Chapter 3 (Motion in a Plane), "Vector Operations" 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 Vector Operations useful in real-life applications?
Real-life applications of Vector Operations 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 Vector Operations?
Key formulas in Vector Operations (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. Vector Operations 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 Vector Operations?
CBSE board questions from Vector Operations 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 Vector Operations 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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