Vector Addition by Analytical Method with Formula Made Easy for Class 11 Students

The analytical method of vector addition resolves vectors into components along coordinate axes, sums the components, and shows the resultant vector's magnitude and direction. This addition, which you may recall from early in the Motion in a Plane 11 notes, either follows the parallelogram or triangle law. 

Importance of Vector Addition using the Analytical Method

Here are some essentials to help you understand why vector addition is important when using the analytical method. 

• It applies to any number of vectors.
• Complex geometric problems can be simplified into algebraic ones.
• Essential for equilibrium problems (net force =0).
• Used in navigation and kinematics problems.

Learning about vector addition using the analytical method is also essential for the JEE Main. 

In case you missed it, check out our article on scalar and vector quantities. 

For vectors $\vec{A}$  and $\vec{B}$ , the resultant $\vec{R}=\vec{A}+\vec{B}$  is found by resolving each into components along $x$  and $y$ axes.

If $\vec{A}=A_x\hat{i}+A_y\hat{j},\vec{B}=B_x\hat{i}+B_y\hat{j}$ , then: $\vec{R}=\left(A_x+B_x\right)\hat{i}+\left(A_y+B_y\right)\hat{j}$

Magnitude: $\left|\vec{R}\right|=\sqrt[]{{\left(A_x+B_x\right)}^2+{\left(A_y+B_y\right)}^2}$ .

Direction: $\mathrm{t}\mathrm{a}\mathrm{n}\theta =\frac{A_y+B_y}{A_x+B_x}$ .

 

The analytical method involves resolving vectors into components, summing them, and reconstructing the resultant vector. 

1. You can resolve each vector into $x$ - and $y$ -components using $A_x=A\mathrm{c}\mathrm{o}\mathrm{s}\theta ,A_y=A\mathrm{s}\mathrm{i}\mathrm{n}\theta$ .
2. After vector resolution, the next step is to sum the components $R_x=\sum A_x,R_y=\sum A_y$ .
3. Then we compute the resultant magnitude $\left|\vec{R}\right|=\sqrt[]{R_x^2+R_y^2}$ .
4. After that, we can find the direction using $\theta ={\mathrm{t}\mathrm{a}\mathrm{n}}^{-1}\left(\frac{R_y}{R_x}\right)$ . This will help adjust the correct quadrant.

 

Let's say, you take a walk of 3 metres in the east direction. Then you go up 4 metres north. 

The displacement components in the situation are (3,4). 

The total displacement magnitude is √(3² + 4²) = 5 metres at an angle of tan⁻¹(4/3) ≈ 53° north of east.

You know this component method works only because vectors follow the parallelogram rule.  

To illustrate how we apply the vector addition formula for two displacement vectors, consider the following snapshot. 

Vector A (walking east): A = 3î + 0ĵ

Ax = 3, Ay = 0

Vector B (walking north): B = 0î + 4ĵ

Bx = 0, By = 4

Using the vector addition formula:

Resultant R = A + B:

Rx = Ax + Bx = 3 + 0 = 3

Ry = Ay + By = 0 + 4 = 4

So R = 3î + 4ĵ

Magnitude:

|R| = √[(Ax + Bx)² + (Ay + By)²] = √(3² + 4²) = √(9 + 16) = √25 = 5 meters

Direction:

θ = tan⁻¹[(Ay + By)/(Ax + Bx)] = tan⁻¹(4/3) ≈ 53° north of east

The formula gives us the net displacement: instead of walking 7 meters total (3 + 4), you end up 5 meters from your starting point in a straight line. 

Here are some sample exercises to practice when you are preparing for competitive tests. 

Example 1: Two vectors $\vec{A}=3\hat{i}+4\hat{j}\mathrm{m}$  and $\vec{B}=5\hat{i}-2\hat{j}\mathrm{m}$ represent displacements. Find the resultant displacement's magnitude and direction.

Solution:

$\vec{R}=\vec{A}+\vec{B}=(3+5)\hat{i}+(4-2)\hat{j}=8\hat{i}+2\hat{j}\mathrm{m}$

Magnitude: $\left|\vec{R}\right|=\sqrt[]{8^2+2^2}=\sqrt[]{68}\approx 8.25\mathrm{m}$ . Direction: $\mathrm{t}\mathrm{a}\mathrm{n}\theta =\frac{2}{8}=0.25$ , so $\theta \approx {\mathrm{t}\mathrm{a}\mathrm{n}}^{-1}\left(0.25\right)\approx$   ${14.0}^{\circ }$  (first quadrant).

Diagram:

Diagram Description: The diagram shows $\vec{A}$  (blue), $\vec{B}$  (green) added head-to-tail, and resultant $\vec{R}$  (red) from origin to the tip of $\vec{B}$ .

Example 2: A boat moves at $6\mathrm{m}/\mathrm{s}$  north, and the river flows at $8\mathrm{m}/\mathrm{s}$ east. Find the boat's resultant velocity relative to the ground.

Solution: Boat: ${\vec{v}}_B=6\hat{j}\mathrm{m}/\mathrm{s}$ . River: ${\vec{v}}_R=8\hat{i}\mathrm{m}/\mathrm{s}$ .

$\vec{R}={\vec{v}}_R+{\vec{v}}_B=8\hat{i}+6\hat{j}\mathrm{m}/\mathrm{s}$

Magnitude: $\left|\vec{R}\right|=\sqrt[]{8^2+6^2}=\sqrt[]{100}=10\mathrm{m}/\mathrm{s}$ . Direction: $\mathrm{t}\mathrm{a}\mathrm{n}\theta =\frac{6}{8}=0.75$ , so $\theta \approx$ ${\mathrm{t}\mathrm{a}\mathrm{n}}^{-1}\left(0.75\right)\approx {36.9}^{\circ }$    north of east.

Diagram:

Diagram Description: The diagram shows ${\vec{v}}_R$  (blue) along east, ${\vec{v}}_B$  (green) along north, and resultant $\vec{R}$  (red) at angle $\theta$ .

Example 3: Three forces act on a particle: ${\vec{F}}_1=10\hat{i}\mathrm{N},{\vec{F}}_2=-5\hat{i}+5\hat{j}\mathrm{N},{\vec{F}}_3=3\hat{j}\mathrm{N}$ . Find the net force.

Solution:

$\vec{R}={\vec{F}}_1+{\vec{F}}_2+{\vec{F}}_3=(10-5+0)\hat{i}+(0+5+3)\hat{j}=5\hat{i}+8\hat{j}\mathrm{N}$

Magnitude: $\left|\vec{R}\right|=\sqrt[]{5^2+8^2}=\sqrt[]{89}\approx 9.43$  N. Direction: $\mathrm{t}\mathrm{a}\mathrm{n}\theta =\frac{8}{5}=1.6$ , so $\theta \approx {\mathrm{t}\mathrm{a}\mathrm{n}}^{-1}\left(1.6\right)\approx$ ${58.0}^{\circ }$ .

Diagram:

Diagram Description: The diagram shows ${\vec{F}}_1$  (blue), ${\vec{F}}_2$  (green), ${\vec{F}}_3$  (purple), and resultant $\vec{R}$ (red) summed head-to-tail.

Check out these links to revise all your Physics chapters easily before exams. 

Units and Measurements Class 11 Notes	Mechanical Properties of Solids Class 11 Notes
Motion in a Straight Line Class 11 Notes	Mechanical Properties of Fluids Class 11 Notes
NCERT Class 11 Notes for Motion in a Plane	Thermal Properties of Matter Class 11 Notes
Laws of Motion Class 11 Notes	Thermodynamics Class 11 Notes
Work, Energy, and Power Class 11 Notes	Kinetic Theory of Gas Class 11 Notes
System of Particles and Rotational Motion Class 11 Notes	Oscillations Class 11 Notes
Gravitation Class 11 Notes	Waves Class 11 Notes

Some more Science stream notes for Class 11 CBSE to check.

NCERT Class 11 Notes for PCM

NCERT Class 11 Physics Notes

 

Find the NCERT Solutions for all the chapters in Class 11 right here.   

Units and Measurements Class 11 NCERT Solutions	Mechanical Properties of Solids Class 11 NCERT Solutions
Motion in a Straight Line Class 11 NCERT Solutions	Mechanical Properties of Fluids Class 11 NCERT Solutions
Motion in a Plane Class 11 NCERT Solutions	Thermal Properties of Matter Class 11 NCERT Solutions
Laws of Motion Class 11 NCERT Solutions	Thermodynamics Class 11 NCERT Solutions
Work, Energy, and Power Class 11 NCERT Solutions	Kinetic Theory Class 11 NCERT Solutions
System of Particles and Rotational Motion Class 11 NCERT Solutions	Oscillations Class 11 NCERT Solutions
Gravitation Class 11 NCERT Solutions	Waves Class 11 NCERT Solutions

Keep exploring the regularly updated NCERT Solutions for Physics Class 11!