As you know, a vector is a quantity that has both magnitude (size) and direction. Examples include displacement, velocity, acceleration, and force. When we talk about adding vectors, we're essentially trying to find the net effect or the resultant of two or more vectors acting together.
Imagine you're trying to push a heavy box. If your friend pushes it with 50 Newtons of force to the east, and you push it with 30 Newtons of force to the east, the box will experience a total of 80 Newtons of force to the east. Here, both forces are in the same direction, so we just add their magnitudes. This is a special, simple case.
Now, what if your friend pushes with 50 Newtons east, and you push with 30 Newtons north? The box won't go purely east, nor purely north. It will move somewhere in between! To figure out exactly where it goes and with what effective force, we need vector addition.
So, the core idea is: when directions are involved, simple scalar addition (just adding numbers) simply doesn't cut it. We need special rules, and those rules are what we call vector addition laws.
Graphical methods are fantastic for building intuition. They allow us to visualize how vectors combine. There are two primary graphical methods:
This is probably the most intuitive way to add two vectors graphically. It's often called the "head-to-tail" method.
Let's illustrate with an example:
Example 1: Displacement
Imagine a person walks 4 km East, then turns and walks 3 km North. What is their total displacement from the starting point?
Solution:
If you measure this graphically, you'll find $vec{R}$ is 5 km long, pointing North-East (at an angle of approximately 36.87° North of East).
So, Resultant displacement = 5 km, 36.87° North of East.
This method clearly shows that a 4 km walk and a 3 km walk don't always add up to 7 km in terms of net displacement. Direction matters!
The parallelogram law is another great graphical method, especially useful when two vectors are originating from the same point (acting on the same object). It's essentially a variation of the triangle law.
Example 2: Forces on an Object
Two forces act on a small object at the origin. Force $vec{F_1}$ is 5 N acting purely along the positive X-axis. Force $vec{F_2}$ is 5 N acting purely along the positive Y-axis. Find the resultant force.
Solution:
Graphically, you'd measure the length of the diagonal. Using the Pythagorean theorem (which is essentially what we're doing here), the magnitude of the resultant force will be $sqrt{5^2 + 5^2} = sqrt{25 + 25} = sqrt{50} approx 7.07$ N. The direction will be at 45° to both the X and Y axes (North-East).
So, Resultant force = 7.07 N, 45° from the positive X-axis.
What if you have three, four, or even more vectors to add? The polygon law is a straightforward extension of the triangle law. You simply connect all the vectors head-to-tail in sequence. The resultant vector is then drawn from the tail of the very first vector to the head of the very last vector, forming an open polygon.
For example, to add $vec{A} + vec{B} + vec{C}$:
Just like numbers, vector addition follows certain rules that make calculations predictable and consistent. These are crucial for both graphical and analytical methods.
This means the order in which you add vectors doesn't matter. If you add $vec{A}$ to $vec{B}$, you get the same resultant as adding $vec{B}$ to $vec{A}$. You can visualize this with the parallelogram law: whether you go $vec{A}$ then $vec{B}$ (forming one side of the parallelogram) or $vec{B}$ then $vec{A}$ (forming the other side), the diagonal (resultant) is the same.
This means if you have three or more vectors, you can group them in any way for addition. You can first add $vec{A}$ and $vec{B}$, and then add $vec{C}$ to their resultant, OR you can first add $vec{B}$ and $vec{C}$, and then add $vec{A}$ to their resultant. The final answer will be the same. This is very useful when dealing with multiple vectors using the polygon law.
Adding a zero vector (a vector with zero magnitude and arbitrary direction, often thought of as no displacement) to any vector leaves the original vector unchanged. The zero vector acts like the number zero in scalar addition.
For every vector $vec{A}$, there exists an additive inverse vector, $-vec{A}$. This vector has the same magnitude as $vec{A}$ but points in the exactly opposite direction. When you add a vector to its inverse, the result is the zero vector.
While graphical methods are great for understanding, they have limitations, especially when precision is required. Drawing to perfect scale and measuring angles accurately can be tricky. This is where analytical methods come into play, allowing us to calculate the resultant vector precisely using trigonometry and coordinate geometry.
The most common and powerful analytical method involves using the component form of vectors. We'll delve much deeper into this in subsequent sections, but let's get a taste of the basic idea.
Every vector in a 2D or 3D space can be broken down into its components along the coordinate axes (X, Y, and Z). For example, a vector $vec{A}$ can be written as $vec{A} = A_xhat{i} + A_yhat{j}$ in 2D, where $A_x$ is its component along the x-axis, $A_y$ is its component along the y-axis, and $hat{i}$, $hat{j}$ are unit vectors along those axes.
The beauty of component form is that adding vectors becomes as simple as adding their corresponding components!
If you have two vectors:
Then their resultant vector $vec{R} = vec{A} + vec{B}$ is simply:
$vec{R} = (A_x + B_x)hat{i} + (A_y + B_y)hat{j}$
It's like adding apples to apples and oranges to oranges!
Example 3: Adding Vectors in Component Form
Given two vectors:
Solution:
Using the component addition rule:
$vec{R} = (3 + 2)hat{i} + (4 + (-1))hat{j}$
$vec{R} = 5hat{i} + (4 - 1)hat{j}$
$vec{R} = 5hat{i} + 3hat{j}$
This method is incredibly powerful because it turns complex directional sums into simple algebraic sums for each component. We can then find the magnitude and direction of the resultant vector from its components (using Pythagorean theorem and trigonometry), which you'll learn in detail soon!
Today, we've laid the groundwork for understanding how to combine vectors. We started by realizing that vector addition isn't like scalar addition because direction plays a crucial role. We explored the graphical methods (Triangle, Parallelogram, and Polygon laws) which help us visualize the resultant vector, and we touched upon the analytical method of adding vectors using their components, which is precise and mathematical.
In the upcoming sections, we'll dive deeper into the analytical methods, learn how to find the magnitude and direction of a resultant vector using mathematical formulas (like the Law of Cosines for two vectors), and explore subtraction of vectors, which is just a special case of addition!
Keep practicing these fundamental concepts, as they are the building blocks for almost everything else in vector algebra and its applications in physics and engineering!
Magnitude of Resultant:
$|vec{R}| = sqrt{A^2 + B^2 + 2AB cos heta}$
Magnitude of Resultant:
$|vec{R}| = sqrt{A^2 + B^2 + 2AB cos heta}$

$vec{R} = vec{A} + vec{B} = (A_x + B_x) hat{i} + (A_y + B_y) hat{j} + (A_z + B_z) hat{k}$
| Aspect | CBSE Focus | JEE Focus |
|---|---|---|
| Basic Laws | Understanding and direct application of Triangle and Parallelogram Laws. Simple problems with 2 vectors. | Assumes strong understanding of laws. Problems involving multiple vectors, 3D scenarios, and often requires analytical component method. |
| Calculations | Mostly 2D problems. Direct use of magnitude and direction formulas. | Complex 2D and 3D problems. Requires efficient use of component method. Often involves finding unknown magnitudes/directions when resultant is known. |
| Conceptual Depth | Basic understanding of properties (commutative, associative). | Deeper application of properties. Problems might involve relative velocity, forces in equilibrium, and vector geometry where these properties are implicitly used. |
| Problem Complexity | Straightforward application of formulas. Single-step solutions. | Multi-step problems often integrating with other physics concepts (e.g., equilibrium, motion, work). Requires deriving component vectors from given information. |
This is the most straightforward and frequently used method in JEE problems. No complex mnemonic is needed, just a simple rule:
For two vectors $vec{A}$ and $vec{B}$ with angle $ heta$ between them, the magnitude of their resultant $vec{R}$ is given by:
$|vec{R}| = sqrt{|vec{A}|^2 + |vec{B}|^2 + 2|vec{A}||vec{B}|cos heta}$.
If $alpha$ is the angle the resultant $vec{R}$ makes with vector $vec{A}$, then:
$ analpha = frac{|vec{B}|sin heta}{|vec{A}| + |vec{B}|cos heta}$
Keep practicing these concepts; the mnemonics will reinforce your understanding and speed up recall during exams.
Tip: Think "head-to-tail". Place the tail of the second vector at the head of the first. The resultant connects the tail of the first to the head of the second.
Tip: Think "tail-to-tail". This law is equivalent to the triangle law; you essentially complete a parallelogram. It's often used when vectors originate from the same point.
Tip: Connect vectors head-to-tail sequentially. The resultant vector goes from the tail of the first vector to the head of the last vector.
For JEE Main, vector addition is most efficiently performed using their rectangular components. If $vec{A} = A_xhat{i} + A_yhat{j} + A_zhat{k}$ and $vec{B} = B_xhat{i} + B_yhat{j} + B_zhat{k}$, then their sum is:
$vec{R} = vec{A} + vec{B} = (A_x + B_x)hat{i} + (A_y + B_y)hat{j} + (A_z + B_z)hat{k}$
Tip: Simply add the corresponding components (i.e., add i-components, j-components, and k-components separately). This method is universally applicable and less prone to geometric errors when dealing with 3D vectors.
If $ heta$ is the angle between two vectors $vec{A}$ and $vec{B}$, the magnitude of their resultant $vec{R} = vec{A} + vec{B}$ is given by:
$|vec{R}| = |vec{A} + vec{B}| = sqrt{|vec{A}|^2 + |vec{B}|^2 + 2|vec{A}||vec{B}|cos heta}$
JEE Specific: This formula is critically important for many problems, especially those involving forces or velocities. Remember to use the angle $ heta$ between the vectors when they are placed tail-to-tail.
Tip: For resultant magnitude, if vectors are perpendicular ($ heta = 90^circ$), $cos heta = 0$, so $|vec{R}| = sqrt{|vec{A}|^2 + |vec{B}|^2}$. This is a common special case.
Vector subtraction $vec{A} - vec{B}$ can be viewed as the addition of $vec{A}$ and $(-vec{B})$.
$vec{A} - vec{B} = vec{A} + (-vec{B})$
Where $-vec{B}$ is a vector with the same magnitude as $vec{B}$ but opposite in direction.
Important: The magnitude formula for subtraction is different: $|vec{A} - vec{B}| = sqrt{|vec{A}|^2 + |vec{B}|^2 - 2|vec{A}||vec{B}|cos heta}$. Note the minus sign before $2|vec{A}||vec{B}|cos heta$.
Master these tips to tackle vector addition problems effectively and confidently in your exams!
Understanding vector addition intuitively is crucial for grasping many physics and mathematics concepts. Unlike scalar quantities (like mass or time) where simple numerical addition suffices, vectors have both magnitude and direction. Therefore, adding vectors is about combining their "effects" or "displacements" to find a single resultant vector that represents the net outcome.
Imagine you perform a series of actions, each represented by a vector. Vector addition tells you the overall, single action that would achieve the same final result. Think of it as finding the "net displacement," "net force," or "net velocity."
The simplest way to understand vector addition is through displacement:
This simple example highlights that vector addition depends entirely on the directions of the individual vectors, not just their magnitudes.
This law directly visualizes the "successive movement" analogy:
It's like tracing your path: you start at the tail of the first, follow it to its head, then follow the second vector from its tail (which is at the first's head) to its head. The resultant is the shortcut from your absolute start to absolute finish.
This law is useful when two vectors originate from the same point, like two forces acting on an object simultaneously:
Intuitively, the parallelogram law achieves the same result as the triangle law. If you shift one of the vectors in the parallelogram so its tail meets the other's head, you'll see it forms the same triangle, yielding the identical resultant.
For both JEE and CBSE exams, a strong intuitive understanding of vector addition is fundamental. It underpins concepts like relative velocity, resultant forces, and electric/magnetic fields. Remember:
Mastering this intuitive grasp will make subsequent topics involving vector manipulation much easier to understand and apply.
Vector addition isn't just a mathematical operation; it's a fundamental concept that underpins our understanding and interaction with the physical world. Many quantities in physics and engineering are vector quantities, meaning they have both magnitude and direction. To combine the effects of multiple such quantities, we employ vector addition. Understanding these applications enhances your conceptual grasp and problem-solving skills, which is crucial for both CBSE board exams and JEE Main.
Here are some practical scenarios where vector addition is indispensable:
Understanding these real-world examples solidifies the conceptual foundation of vector addition. While direct "real-world application" questions are rare in JEE Main, the underlying principles of vector addition are tested extensively in physics problems related to mechanics, electromagnetism, and more. A strong conceptual understanding helps in visualizing the problem and forming the correct vector equations.
Imagine you start at point A, walk 3 km East to reach point B, and then walk 4 km North to reach point C. Your total travel distance is 3 km + 4 km = 7 km (scalar addition).
However, your final displacement from your starting point A to your end point C is not 7 km. It's the straight-line distance and direction from A to C. This final displacement is the vector sum of your two individual displacement vectors (3 km East and 4 km North). Geometrically, this forms the hypotenuse of a right-angled triangle, resulting in a displacement of 5 km in a North-East direction (approximately 53.13° North of East).
This analogy perfectly illustrates the Triangle Law of Vector Addition, where the "resultant vector" (final displacement) closes the triangle formed by the individual displacement vectors.
Consider an object being pushed or pulled by multiple forces simultaneously. If one person pushes a box with a force of 10 N East and another person pushes it with a force of 10 N North, the box doesn't move 20 N in total. Instead, the box experiences a single net force that is the vector sum of these two forces.
This net force will be approximately 14.14 N (10√2 N) in the North-East direction. This scenario perfectly models the Parallelogram Law of Vector Addition, where the diagonal represents the resultant force.
JEE Relevance: This analogy is extremely important in Physics, where concepts like net force, equilibrium, and motion under multiple forces are fundamental. Understanding how forces combine vectorially is crucial for solving problems in mechanics.
Imagine a boat trying to cross a river. The boat has its own velocity relative to the water, and the river water itself has a velocity relative to the bank. The actual velocity of the boat relative to the river bank (the ground frame) is the vector sum of these two velocities.
If the boat aims straight across the river (e.g., 5 m/s North) and the river flows East (e.g., 3 m/s East), the boat's resultant velocity relative to the bank will be a diagonal path. It will effectively travel North-East, and its speed relative to the bank will be √(5² + 3²) = √34 m/s, not simply 5+3 = 8 m/s.
This analogy helps in understanding relative velocity problems, which are common in kinematics.
Navigating vector addition in exams requires precision, as common pitfalls can significantly impact your score. This section highlights typical traps students fall into, particularly relevant for both JEE Main and CBSE Board exams.
By being aware of these common traps and practicing meticulously, you can avoid unnecessary errors and secure full marks on vector addition problems.
Understanding vector addition is fundamental to vector algebra and its applications in physics. This section summarizes the most crucial points you need to remember for both board exams and competitive exams like JEE Main.
Remember, vector addition is distinctly different from scalar addition because vectors have both magnitude and direction. Always be mindful of the vector nature of quantities in your calculations.
Mastering vector addition is fundamental for solving a wide range of problems in physics and mathematics, from mechanics to geometry. A systematic approach helps in tackling complex scenarios efficiently.
The choice between the geometric (graphical) and analytical (component) method depends on the number of vectors, their representation, and the complexity of the problem.
By consistently applying these approaches, you can systematically solve vector addition problems in your exams.
For the CBSE board examinations, understanding vector addition is foundational. The focus is primarily on conceptual clarity, direct application of laws, and solving problems involving the geometric and algebraic interpretation of vectors. Mastering these aspects will ensure strong performance in this section.
The CBSE syllabus emphasizes two primary geometric laws for vector addition and their algebraic counterparts:
CBSE questions frequently test the understanding and verification of these fundamental properties:
CBSE Exam Tip: Be prepared to prove these properties using either geometric arguments or component form, depending on the question's phrasing.
Expect questions that require you to:
For CBSE, a strong grasp of the visual representation of vector addition and the ability to perform operations accurately in component form are crucial. While JEE might delve into more complex applications and integration with other topics, CBSE focuses on solidifying these fundamental building blocks.
Vector addition is a foundational concept in vector algebra and frequently appears in JEE Main, often integrated into problems across various topics like mechanics, electrodynamics, and geometry. A thorough understanding of its laws and applications, particularly the component method, is crucial for success.
These properties are fundamental and useful for simplifying vector expressions:
| Aspect | CBSE Board Exam | JEE Main |
|---|---|---|
| Emphasis | Understanding laws (Triangle, Parallelogram), basic component addition. | Application of laws, especially component method in 3D, integration with other topics (geometry, physics), finding magnitudes and directions of resultants, multi-vector addition. |
| Problem Type | Direct calculation, simple proofs of properties. | Complex problems requiring strategic use of component method, finding specific conditions (e.g., when resultant is minimum/maximum or has a certain direction), geometric vector problems. |
For JEE, practice problems that involve finding unknown vectors given their resultant, determining the angle between vectors, and applying vector addition in 3D space. Master the component method as it's the most versatile tool.
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Incorrect Approach: Adding magnitudes directly: 10 N + 5000 dynes = 5010 (incorrect unit and magnitude).
Correct Approach: First convert dynes to Newtons (1 N = 10⁵ dynes).
5000 dynes = 5000 / 10⁵ N = 0.05 N.
Now, add the forces: Resultant Force = 10 N + 0.05 N = 10.05 N (along the x-axis).
Acosθ, Asinθ) but fail to adjust the sign based on the quadrant. A making an angle θ with the positive X-axis (measured counter-clockwise from positive X-axis), its components are Ax = A cosθ and Ay = A sinθ. Ensure θ is the standard angle, or explicitly determine signs for cosθ and sinθ based on the quadrant. For JEE, understanding the component method thoroughly is crucial. Ax = 10 cos(30°) = 8.66Ay = 10 sin(30°) = 5 (Mistakenly taking Y-component as positive)Ax = 10 cos(30°) = 8.66Ay = -10 sin(30°) = -5θ = 360° - 30° = 330°:Ax = 10 cos(330°) = 10 (√3/2) = 8.66Ay = 10 sin(330°) = 10 (-1/2) = -5cosθ and sinθ.Given two vectors A = 3i and B = 4j.
Incorrect calculation: The student finds individual magnitudes |A| = 3 and |B| = 4. Then, mistakenly concludes that |A + B| = |A| + |B| = 3 + 4 = 7.
Given two vectors A = 3i and B = 4j.
Correct calculation:
A common minor error in vector addition, especially in JEE Advanced problems, is performing arithmetic operations on vector components or magnitudes that are expressed in inconsistent units. Students might add a component given in meters (m) to another given in centimeters (cm) directly, or combine vector magnitudes with different unit systems. Vector addition is only physically meaningful and mathematically correct when all quantities involved share the same physical dimension and are expressed in a uniform system of units.
Always convert all vector magnitudes and their respective components to a single, consistent system of units (preferably SI units like meters, kilograms, seconds, Newtons) before performing any vector addition. This ensures that the resultant vector accurately represents the physical quantity.
Consider two displacement vectors:
Vector A: &vec;A = 50 cm î + 2 m ĵ
Vector B: &vec;B = 1 m î + 150 cm ĵ
Incorrect Addition: &vec;R = &vec;A + &vec;B = (50+1) î + (2+150) ĵ = 51 î + 152 ĵ (This result is meaningless due to mixed units).
Using the same vectors as above, first convert all components to meters:
&vec;A = 0.5 m î + 2 m ĵ (since 50 cm = 0.5 m)&vec;B = 1 m î + 1.5 m ĵ (since 150 cm = 1.5 m)Correct Addition:&vec;R = &vec;A + &vec;B = (0.5 + 1) î + (2 + 1.5) ĵ = 1.5 î + 3.5 ĵ (in meters).
This approach yields a physically correct and interpretable result. This applies equally to CBSE and JEE Advanced problems, although JEE Advanced problems might involve more complex unit conversions.
Two forces, F1 = 3N and F2 = 4N, act on a particle. A student incorrectly calculates the resultant force magnitude as |F1 + F2| = 3N + 4N = 7N, regardless of the angle between them.
If F1 = 3N and F2 = 4N act perpendicularly (θ = 90°), the correct resultant magnitude is:
|R| = √(32 + 42 + 2×3×4×cos90°) = √(9 + 16 + 0) = √25 = 5N.
If they act in the same direction (θ = 0°), |R| = 7N. If they act in opposite directions (θ = 180°), |R| = |3 - 4| = 1N.
A force F₁ = 5 N acts on an object. Another force F₂ = 200 dynes acts in the same direction. What is the total force?
Wrong Calculation: Total Force = F₁ + F₂ = 5 N + 200 dynes = 205 N (or 205 dynes). This is incorrect as units are inconsistent.
A force F₁ = 5 N acts on an object. Another force F₂ = 200 dynes acts in the same direction. What is the total force?
Correct Calculation: First, convert F₂ to Newtons.
Now, add the forces:
| Quadrant | X-component | Y-component |
|---|---|---|
| I | + | + |
| II | - | + |
| III | - | - |
| IV | + | - |
The magnitude of the resultant R of two vectors A and B with angle θ is:
R = &sqrt;(A² + B² + 2ABcosθ)
Question: Two forces, F&sub1; of magnitude 3N acting East and F&sub2; of magnitude 4N acting North, act on a body. Find the magnitude of the resultant force.
Wrong Calculation:
Resultant magnitude = 3N + 4N = 7N.
Correct Calculation (using Component Method / Parallelogram Law for perpendicular vectors):
Since F&sub1; acts East and F&sub2; acts North, the angle between them is θ = 90°.
The correct magnitude of the resultant force is 5N.
To correctly add vectors and find the resultant's magnitude:
Problem: Two forces, F1 of 3 N and F2 of 4 N, act on an object at an angle of 90° to each other. Find the magnitude of the resultant force.
Incorrect Approach: Student adds magnitudes directly: |R| = |F1| + |F2| = 3 N + 4 N = 7 N.
Correct Approach: Using the formula for resultant magnitude:
|R| = √(|F1|2 + |F2|2} + 2|F1||F2|cosθ)
|R| = √(32 + 42 + 2 × 3 × 4 × cos90°)
|R| = √(9 + 16 + 24 × 0)
|R| = √(25)
|R| = 5 N
A common mistake is to attempt vector addition directly without converting units, for example, by thinking of it as (10, 2) in components, or just adding the numbers 10 + 2 if it were a scalar sum (which it isn't).
Incorrect Component approach: R = (10 m)i + (2 km)j. This directly mixes units and is fundamentally flawed.
Step 1: Convert to a consistent unit. Let's use meters (SI unit).
Vector A = 10 m (East)
Vector B = 2 km = 2 × 1000 m = 2000 m (North)
Step 2: Perform vector addition using consistent units.
Using component method:
R = (10 m)i + (2000 m)jMagnitude of resultant R = √( (10)² + (2000)² ) = √(100 + 4000000) = √4000100 ≈ 2000.025 m
Direction θ = tan⁻¹(2000/10) = tan⁻¹(200) ≈ 89.71° North of East.
Acosθ/Asinθ without considering quadrant or component direction.|R| = sqrt(|A|^2 + |B|^2 + 2|A||B| cosθ)
|R| = sqrt(10^2 + 10^2 + 2 * 10 * 10 * cos(2°))
|R| = sqrt(200 + 200 * cos(2°))
|R| ≈ sqrt(200 + 200 * 0.9994) = sqrt(200 + 199.88) = sqrt(399.88) ≈ 19.997 N
The magnitude of the resultant vector R = A + B must be calculated using the parallelogram law of vector addition:
|R| = |A + B| = √(|A|2 + |B|2 + 2|A||B|cosθ), where θ is the angle between vectors A and B.
Alternatively, if vectors are given in component form (e.g., A = Axî + Ayï + Azí), sum the corresponding components first to get R = (Ax+Bx)î + (Ay+By)ï + (Az+Bz)í, and then find its magnitude: |R| = √((Ax+Bx)2 + (Ay+By)2 + (Az+Bz)2).
Given: Two vectors, A with |A| = 3 units and B with |B| = 4 units. The angle between them is θ = 90°.
Incorrect Calculation: |A + B| = |A| + |B| = 3 + 4 = 7 units.
Given: Vectors A (|A| = 3), B (|B| = 4), and θ = 90°.
Correct Calculation (using parallelogram law):
|A + B| = √(|A|2 + |B|2 + 2|A||B|cosθ)
= √(32 + 42 + 2(3)(4)cos90°)
= √(9 + 16 + 0)
= √(25) = 5 units.
(Note: 5 ≠ 7, clearly showing the error in direct scalar addition.)
If Vector A has magnitude 3 units and Vector B has magnitude 4 units, and they are perpendicular (θ = 90°).
Wrong Calculation: |R| = |A| + |B| = 3 + 4 = 7 units.
If Vector A has magnitude 3 units and Vector B has magnitude 4 units, and they are perpendicular (θ = 90°).
Correct Calculation:
|R| = √(|A|2 + |B|2 + 2|A||B|cosθ)
|R| = √(32 + 42 + 2(3)(4)cos90°)
|R| = √(9 + 16 + 0)
|R| = √(25) = 5 units.
Problem: Two forces, F1 = 10 N and F2 = 10 N, act on an object at an angle of θ = 6 degrees with each other.
Incorrect Approximation: Resultant magnitude R ≈ F1 + F2 = 10 + 10 = 20 N.
(Here, the student assumes θ is practically 0, treating vectors as perfectly parallel, and uses cos θ ≈ 1 without checking for precision.)
For the same problem (F1 = 10 N, F2 = 10 N, θ = 6 degrees):
Correct Calculation using Law of Cosines:
R = √(F1² + F2² + 2F1F2 cosθ)
R = √(10² + 10² + 2 * 10 * 10 * cos(6°))
R = √(100 + 100 + 200 * 0.9945)
R = √(200 + 198.9) = √(398.9)
R ≈ 19.97 N
The approximated result of 20 N from the 'wrong example' is numerically close but could be considered significantly different in JEE, where a difference of 0.03 N might lead to choosing the wrong option. More importantly, this highlights the principle of avoiding unjustified approximations.
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