1^{st} ROUND | |||||

Sl. No | Registration/ Payment & Choice Filling | Choice Filling/ Locking | Processing of Seat Allotment | Result | Reporting |
---|---|---|---|---|---|

1 | 25^{th} June to 01^{st} July, 2019 up to 05:00 PM of 01^{st} July, 2019 only as per Server Time* Payment facility will be available till 2 ^{nd} July, 2019 up to 12:00 Noon as per Server Time | 02^{nd} July, 2019 (10:00am to till 05:00 PM) as per Server Time | 03^{rd} July, 2019 | 04^{th} July, 2019 | 05^{th} July, 2019 to 15^{th} July, 2019. |

(7 Days) | (1 Day) | (1 Day) | (1 Day) | (11 Days) | |

2^{nd} ROUND | |||||

2 | 24^{th} July, 2019 to 27^{th} July, 2019 till 05:00 PM as per Server Time* Payment facility will be available till 28 ^{th} July, 2019 till 12:00 Noon as per Server Time | 28^{th} July, 2019 (10:00 to till 05:00 PM) as per Server Time | 29^{th} July, 2019 to 30^{th} July, 2019 | 31^{st} July, 2019 | 1^{st} August, 2019 to 10^{th} August, 2019 |

(4 Days) | (1 Day) | (2 Days) | (1 Day) | (10 Days) | |

3 | Transfer of Non-Reporting & Non-Joining Vacant seats to State Quota (Before 12:00 Noon) 11^{th} August (Only 15% All India Quota Govt./ Govt Aided/ Private) | ||||

MOP-UP ROUND for Central/ Deemed University/ National Institutes Only | |||||

4 | Display of Seats Matrix for Mop-up 21^{st} to 22^{nd} August, 2019 (2 Days) | ||||

5 | 23^{rd} August, 2019 to 26^{th} August, 2019 05:00 PM as per Server Time* Payment facility will be available till 27^{th} August, 2019 upto 12:00 Noon as per Server Time | 27^{th} August, 2019 (10:00 to till 05:00 PM) as per Server Time | 28^{th} August, 2019 | 29^{th} August, 2019 | 30^{th} August, 2019 to 07^{th} September, 2019 |

(4 Days) | (1 Day) | (1 Day) | (1 Day) | (9 Days) | |

6 | Transfer of Non-Reporting & Non-Joining Vacant seat to Deemed/ Central Universities/ National Institutes for Stray Vacancy Round on 09^{th} September, 2019 |

1^{ST} ROUND | |||||

Sl. No | Registration/ Payment & Choice Filling | Choice Filling/ Locking | Processing of Seat Allotment | Result | Reporting |

1 | 19^{th} June to 24^{th} June, 2019 up to 05:00 PM of 24^{th} June, 2019 only as per Server Time *Payment facility will be available till 25^{th} June, 2019 up to 2:00 PM as per Server Time | 25^{th} June, 2019 (10:00am to till 05:00 PM) as per Server Time | 26^{th} June, 2019 | 27^{th} June, 2019 | 28^{th} June, 2019 to 3^{rd} July, 2019. |

(7-Days) | (1-Day) | (1-Day) | (1-Day) | (6-Days) | |

2^{ND} ROUND | |||||

2 | 6^{th} July, 2018 to 8^{th} July, 2019 till 05:00 PM as per Server Time *Payment facility will be available till 9^{th} July, 2019 till 12:00 NOON as per Server Time | 9^{th} July, 2019 (10:00 to till 05:00 PM) as per Server Time | 10^{th} July, 2019 to 11^{th} July, 2019 | 12^{th} July, 2019 | 13^{th} July 2019 to 22^{nd} July, 2019 |

(3-Days) | (1-Day) | (2-Days) | (1-Day) | (9-Days) | |

Transfer of Non-Reporting & Non-Joining Vacant seats to State Quota 23^{rd} July, 2019 (Only 15% All India Quota) | |||||

MOP-UP ROUND for Central/ Deemed Uni./ ESIC only | |||||

3 | 13^{th} August, 2019 to 15^{th} August, 2019 05:00 PM as per Server Time *Payment facility will be available till 16^{th} August, 2018 2:00 PM as per Server Time | 16^{th} August, 2019 (10:00 to till 05:00 PM) as per Server Time | 17^{th} August, 2019 | 18^{th} August, 2019 | 20^{th} August, 2019 to 26^{th} August, 2019 |

(3-Days) | (1-Day) | (1-Day) | (1-Day) | (7-Days) | |

Transfer of Non-Reporting & Non-Joining Vacant seat to Deemed/ Central Universities/ ESIC for Mop-up on 27^{th} August, 2019 |

The admissions in the State of Tamil Nadu are as per the provisions of the Government of Tamil Nadu and in compliance to the amendment to section 10(D) of the Board of Governors in Supersession of Indian Medical Council of India, under Indian Medical Council of India Act, 1956 and section 10(D) of Dental Council Act, 1948 and the orders of the Hon’ble Supreme Court of India, the eligibility for admission to MBBS/ BDS Degree Courses within the respective categories shall be based solely on marks obtained in the National Eligibility Cum Entrance Test (NEET UG – 2019).

**Eligibility Criteria:**

**Age Limit:**

Candidates should have
completed the age of 17 years at the time of admission or should complete that
age on or before 31^{st} December of that year.

**Nationality:**

Candidates should be a Citizen of India.

Overseas Citizens of India, who are registered under Section 7A of the Citizenship Act, 1955 (Central Act 57 of 1955) are eligible to apply for MBBS/ BDS Degree Courses subject to the production of proof of such registration. However, OCI candidates will not be eligible for any kind of reservation and will be treated only as Open Category.

**Educational Qualification:**

Admission to MBBS/ BDS Degree Courses shall be based solely on marks obtained in the National Eligibility Cum Entrance Test (NEET – UG 2019).

**Nativity:**

(a) Candidates should be a Native of Tamil Nadu.

(b) Candidates belonging to
Native of Tamil Nadu and having studied from Standard VI to Standard XII in
schools of Tamil Nadu need not submit their “Nativity Certificate”. However,
such candidates should produce the true copies of their parent’s certificates
such as Birth Certificate, Ration Card SSLC/ 10^{th}/ 12^{th}/
Degree/ Diploma/ Professional course. In case, parents are not literate, then
No Graduation Certificate for parents, from the Revenue Authority of competent
jurisdiction, to substantiate their parent’s place of birth in Tamil Nadu and
also “Parent’s Community Certificate to claim the communal reservation”. If the
candidates do not submit the above-mentioned certificates, then the candidates
will be considered under Open Category only.

(c) Candidates belonging to other States who are residing at Tamil Nadu cannot claim nativity of Tamil Nadu and they will be considered under Open Category.

(d) The other State candidates who are not native of Tamil Nadu and have studied from Standard VI to XII in Tamil Nadu will be considered under Open Category.

(e) Permanent Residence Certificate in lieu of Nativity Certificate will not be accepted.

(f) Candidates who are Native
of Tamil Nadu, but studied from VI Standard to XII Standard outside Tamil Nadu
either partly or completely in one or more States should produce the true
copies of their parent’s certificates such as Birth Certificate, Ration Card,
SSLC/ 10^{th}/ 12^{th}/ Degree/ Diploma/ Professional course.
In case, parents are not literate, then No Graduation Certificate for parents,
from the Revenue Authority of competent jurisdiction, to substantiate their
parent’s place of birth in Tamil Nadu and also “Parent’s Community Certificate
to claim the communal reservation”. If the candidates do not submit the above-mentioned
certificates, then the candidate’s application will be summarily rejected.

(g) Nativity Certificate obtained prior to the last date of receipt of filled in application alone will be accepted and sending the Nativity Certificate separately will not be entertained.

(h) If false Nativity certificate is submitted and at a later date, if it is found out, then the candidate will be expelled from the course and also criminal proceedings will be initiated against the candidate as well as his/ her parents as per provision of the law.

**Mode of Application:** Online

**Application Fee:** Rs. 500/-

**Mode of Payment:** Online/ Offline

For Applying: https://tnmedicalonline.xyz/Default.aspx?ID=PRyxQzzugjUK7KqMgT4fbCI/L/NABDwijg7NwIclT8U=

**Important Dates:**

Date of Notification | 07/06/2019 |

Date of Commencement of online application | 07/06/2019 10:00 AM |

Last Date for online submission of application | 20/06/2019 upto 5:00 PM |

Last date for receipt of filled in online application | 21/06/2019 upto 5:00 PM |

Address to which the filled in online application along with enclosures are to be sent | The SECRETARY, SELECTION COMMITTEE, 162, PERIYAR E.V.R. HIGH ROAD, KILPAUK, CHENNAI – 600010. |

Tentative date of declaration of Rank list | 02/07/2019 |

Special Category and wards of I.R.T Employees 1 ^{st} Phase of Counselling2 ^{nd} Phase of Counselling | 04/07/2019 05/07/2019 to 12/07/2019 (Excluding Sunday) Depends on the schedule of All India Quota Counselling |

Mop-Up Counselling | (If time permits) |

Commencement of Courses | 01/08/2019 |

Closure of admission | 18/08/2019 |

As per regulations framed under the Indian Medical Council Act, 1956 and the Dentists Act-1948, admissions in 100% seats of M.B.B.S/ B.D.S will be done through NATIONAL ELIGIBILITY CUM ENTRANCE TEST- NEET in Medical/Dental Colleges run with the approval of Medical Council of India/ Dental Council of India except for the institutions established through an Act of Parliament i.e. AIIMS and JIPMER Puducherry.

Medical Counselling Committee in the Directorate General of Health Services does the Counselling for admission to the M.B.B.S/ B.D.S seats under the following Quotas:

i. All India Quota Seats (Filled by DGHS) 15% seats surrendered by Government Medical/ Dental Colleges run as per the approval of MCI/ DCI are filled through the counselling conducted by DGHS are known as All India Quota seats.

ii. Central Institutions/Universities (Filled by DGHS)

iii. Seats in Deemed Universities (Filled by DGHS)

**Eligibility Criteria:**

**M.B.B.S & B.D.S:**

Candidates seeking admission to the M.B.B.S/ B.D.S, Degree courses must have passed H. Sc. (Academic) examination conducted by the Board of Higher Secondary Examination of Tamil Nadu or any other equivalent thereto, with a minimum of 50% marks (40% of marks for OBC/ BCM/ MBC/ EBC/ BT/ SC/ ST candidates) in aggregate, in the prescribed subjects of Physics, Chemistry and Biology or Bio-technology, Botany and Zoology. Further they should have studied English as one of the subjects and they must have passed the above four subjects individually.

**B.A.M.S:**

(a) Candidates seeking admission for BAMS course must have passed in the subjects of Physics, Chemistry, Biology & English individually and must have obtained a minimum mark taken together in Physics, Chemistry and Biology at the qualifying examination with a minimum of 50% marks for General and 45% of marks for OBC/ BCM/ MBC/ EBC/ OPH (General &OBC) and 40% of marks for SC/ ST/ BT& OPH.

(b) The Higher Secondary Examination or the Indian School Certificate Examination which is equivalent to10 +2 Higher Secondary Examination after a period of 12 years study, the last two years of study comprising of Physics, Chemistry, Biology and Mathematics or any other elective subjects with English at a level not less than core course of English as prescribed by the National Council of Educational Research and Training after the introduction of 10+2+3 years educational structure as recommended by the National Committee on education one of the subjects and they must have passed the above four subjects individually.

**Bachelor of Veterinary Science & Animal Husbandry [B.V. Sc & A.H]:**

Candidates seeking admission to the B.V.Sc & A.H Degree courses must have passed H. Sc. (Academic) examination conducted by the Board of Higher Secondary Examination of Tamil Nadu or any other equivalent thereto, with a minimum of 50% marks (40% of marks for OBC/ BCM/ MBC/ EBC/ BT/ SC/ ST candidates) in aggregate in Biology/ Botany & Zoology, Physics and Chemistry

**RESERVATION:**

(i) As per UT Govt. norms for all Government quota seats, only Candidates who are resident of the UT of Puducherry are eligible.

**Minority Quota Seats in Private Medical/ Dental Colleges:**

(ii) For Minority Quota seats in Minority Medical/ Dental Colleges candidates who are residents of the UT of Puducherry and belonging to respective minorities are only eligible.

i.e., for the minority quota seats in PIMS, candidates who are resident of UT of Puducherry and belonging to Christian religious minority alone are eligible. For the minority quota seats in SVMC&RI candidates who are resident of UT of Puducherry and belonging to Telugu linguistic minority alone are eligible.

(iii) UT of Puducherry candidates claiming seats under Christian Minority Quota should produce any valid documentary proof like Transfer Certificate or similar document to prove his/ her religion is CHRISTIANITY.

(iv) UT of Puducherry candidates claiming seats under Telugu Minority Quota should produce any valid documentary proof like Transfer Certificate or School admission certificate to prove his/her mother tongue is TELUGU.

(v) Conversion of Minority Quota seats into General Management Quota seats: In case these minority quota seats are not filled in the TWO rounds of Counselling, then these unfilled seats will be converted to general management quota seats and filled on All India basis.

(vi) For Management Quota seats in Private Colleges candidates are eligible on All India basis without any restriction to residence in the UT of Puducherry.

**Mode of Application:** Online

**Application Fee:**

Type of seat applied for | Others | SC/ ST |

Govt. Quota | 1000 | 500 |

Management Quota/ SS – (Other States candidates) | 2000 | 1000 |

NRI/ Foreign Nationals | 2000 | – |

**SELECTION PROCEDURE:**

Admission to the M.B.B.S, B.D.S, B.A.M.S & B. V. Sc & AH courses is based on the NEET Rank for the current year of the candidates.

Merit List will be based on NEET Rank. Allotment will be based on the eligibility, Category of Seats and NEET Rank.

**Important Dates:**

Submission of Online Application: Started

Last date for Applying to Courses: 16^{th} June 2019

When an external force is applied to a body then at each cross section of the body an internal restoring force is setup which tends to restore the body to its original state. The restoring setup inside the body per unit area is known as stress. The normal stresses can be either tensile or compressive whether the stresses act out of the area or into the area.

Basically, three different types of stresses can be identified. These are related to the nature of the deforming force applied on the body. That is, whether they are tensile, compressive or shearing.

**What is Tensile Stress?**

If the deforming force or applied force results in the increase in the object’s length then the resulting stress is termed as tensile stress. For example: When a rod or wire is stretched by pulling it with equal and opposite forces at both ends.

Consider a uniform bar of cross-sectional area A subjected to an axial tensile force P. The stress at any section x – x normal to the line of action of the tensile force P is specifically called tensile stress (P_{t}). Since internal resistance R at x – x is equal to the applied force P, we have:

\(Tensile\,\,Stress{{\left( P
\right)}_{t}}=\frac{Internal\,\,resis\tan
ce\,\,at\,\,x-x}{\operatorname{Re}sisting\,\,area\,\,at\,\,x-x}=\frac{R}{A}=\frac{Applie\,\,Force(P)}{Cross\,\,\operatorname{Sec}tional\,\,Area(A)}\)**.**

Under tensile stress the bas suffers stretching or elongation. Tensile stress causes stress corrosion cracking, which is the combined influence of tensile stress and a corrosive environment. The required tensile stresses may be in the form of directly applied stresses or in the form of residual stresses.

]]>When an external force is applied to a body then at each cross section of the body an internal restoring force is setup which tends to restore the body to its original state. The restoring setup inside the body per unit area is known as stress. The normal stresses can be either tensile or compressive whether the stresses act out of the area or into the area.

There are three ways to deform a solid with the help of external force. Compressive stress is the one among three ways.

**What is Compressive Stress?**

Compressive stress is defined as the stress experienced by a material which leads to a smaller volume. In order to increase the compressive stress, compressive strength must be reached. It can also define as the measure of force that one needs to break a material.

Examples in which compressive stresses are developed are connecting rods, link of structure and column. The figure below shows the compressive stress developed under the application of the external load.

**Formula of
Compressive Stress: **

The compressive stress that develops when an external compressive force is applied on an object is given by:

Stress (σ) = Force (F)/ Area (A)

Unit of compressive stress is N/ m². Higher the compressive force, higher the compressive stress. Compressive stress results in the shortening of the solid and it is due to the application of external compressive force. Compressive strength is the measure of force that one needs to break a material.

]]>The vector triple product of three vectors \(\overrightarrow{a}\), \(\overrightarrow{b}\) and \(\overrightarrow{c}\) is the vector \(\overrightarrow{a}\times \left( \overrightarrow{b}\times \overrightarrow{c} \right)=\left( \overrightarrow{a}.\overrightarrow{c} \right)\overrightarrow{b}-\left( \overrightarrow{a}.\overrightarrow{b} \right)\overrightarrow{c}\).

Also, \(\left( \overrightarrow{a}\times \overrightarrow{b} \right)\times \overrightarrow{c}=\left( \overrightarrow{a}.\overrightarrow{c} \right)\overrightarrow{b}-\left( \overrightarrow{b}.\overrightarrow{c} \right)\overrightarrow{a}\).

In general, \(\overrightarrow{a}\times \left( \overrightarrow{b}\times \overrightarrow{c} \right)\ne \left( \overrightarrow{a}\times \overrightarrow{b} \right)\times \overrightarrow{c}\).

If \(\overrightarrow{a}\times \left( \overrightarrow{b}\times \overrightarrow{c} \right)=\left( \overrightarrow{a}\times \overrightarrow{b} \right)\times \overrightarrow{c}\), then the vectors \(\overrightarrow{a}\) and \(\overrightarrow{c}\) are collinear.

\(\overrightarrow{p}=\overrightarrow{a}\times \left( \overrightarrow{b}\times \overrightarrow{c} \right)\) is a vector perpendicular to \(\overrightarrow{a}\) and \(\overrightarrow{b}\times \overrightarrow{c}\) but \(\overrightarrow{b}\times \overrightarrow{c}\) is a vector perpendicular to the plane of \(\overrightarrow{b}\) and \(\overrightarrow{c}\).

Hence, vector \(\overrightarrow{p}\) must lie in the plane of \(\overrightarrow{b}\) and \(\overrightarrow{c}\).

\(\overrightarrow{p}=\overrightarrow{a}\times \left( \overrightarrow{b}\times \overrightarrow{c} \right)=x.\overrightarrow{b}+y.\overrightarrow{c}\) … (i)

Multiplying (i) scalarly by \(\overrightarrow{a}\), we have

\(\overrightarrow{p}.\overrightarrow{a}=x\left( \overrightarrow{a}.\overrightarrow{b} \right)+y\left( \overrightarrow{a}.\overrightarrow{c} \right)\) … (ii)

But \(\overrightarrow{p}\bot \overrightarrow{a}\) ⇒ \(\overrightarrow{p}.\overrightarrow{a}=0\). Therefore,

\(x\left( \overrightarrow{a}.\overrightarrow{b} \right)=-y\left( \overrightarrow{c}.\overrightarrow{a} \right)\), i.e, \(\frac{x}{\overrightarrow{c}.\overrightarrow{a}}=\frac{-y}{\overrightarrow{a}.\overrightarrow{b}}=\lambda \).

∴ \(x=\lambda \left( \overrightarrow{c}.\overrightarrow{a} \right)\), \(y=-\lambda \left( \overrightarrow{a}.\overrightarrow{b} \right)\) … (iii)

Substituting x and y from (iii) in (i), \(\overrightarrow{a}\times \left( \overrightarrow{b}\times \overrightarrow{c} \right)=\lambda \left[ \left( \overrightarrow{c}.\overrightarrow{a} \right)\overrightarrow{b}-\left( \overrightarrow{a}.\overrightarrow{b} \right)\overrightarrow{c} \right]\) … (iv)

The simplest way to determine λ is by taking specific vectors \(\overrightarrow{a}=\hat{i}\), \(\overrightarrow{b}=\hat{i}\), \(\overrightarrow{c}=\hat{j}\).

We have from (iv), \(\hat{i}\times \left( \hat{i}\times \hat{j} \right)=\lambda \left[ \left( \hat{i}.\hat{j} \right)\hat{i}-\left( \hat{i}.\hat{i} \right)\hat{j} \right]\), i.e, \(\hat{i}\times \hat{k}=\lambda \left[ 0\hat{i}-1\hat{j} \right]\), i.e, \(-\hat{j}=-\lambda \hat{j}\).

∴ λ = 1

Substituting λ in (iv),

\(\overrightarrow{a}\times \left( \overrightarrow{b}\times \overrightarrow{c} \right)=\left( \overrightarrow{a}.\overrightarrow{c} \right)\overrightarrow{b}-\left( \overrightarrow{a}.\overrightarrow{b} \right)\overrightarrow{c}\).

**Lagrange’s Identity:**

\(\left( \overrightarrow{a}\times \overrightarrow{b} \right).\left( \overrightarrow{c}\times \overrightarrow{d} \right)=\overrightarrow{a}\left[ \overrightarrow{b}\times \left( \overrightarrow{c}\times \overrightarrow{d} \right) \right]\),

\(=\overrightarrow{a}.\left[ \left( \overrightarrow{b}.\overrightarrow{d} \right)\overrightarrow{c}-\left( \overrightarrow{b}.\overrightarrow{c} \right)\overrightarrow{d} \right]\),

\(=\left( \overrightarrow{a}.\overrightarrow{c} \right)\left( \overrightarrow{b}.\overrightarrow{d} \right)-\left( \overrightarrow{a}.\overrightarrow{d} \right)\left( \overrightarrow{b}.\overrightarrow{c} \right)\),

\(=\left| \begin{matrix} \overrightarrow{a}.\overrightarrow{c} & \overrightarrow{a}.\overrightarrow{d} \\ \overrightarrow{b}.\overrightarrow{c} & \overrightarrow{b}.\overrightarrow{d} \\\end{matrix} \right|\).** **

This is called Lagrange’s Identity.

]]>Electric field lines used to describe the basic nature of the electric field in two dimensional forms. What if we want to describe electric field in a three-dimensional form, this can be done with the help of electric flux. It is the measure of flow of the electric field through a given area. It is proportional to the number of electric field lines going through a normally perpendicular surface.

**What is Electric
Flux?**

Electric flux is defined as a number of electric filed lines, passing per unit area. It is a physical quantity to measure the strength of electric filed and frame the basics of electrostatics. It is also defined as the product of electric field and surface area projected in a direction perpendicular to the electric field.

Mathematically, φ_{E} = E.A

When the same plane is tilted at an angle θ, the projected area is given as A cosθ and the total flux through this surface is:

Electric Flux (φ_{E}) = E.A cos θ; Where,

E = Magnitude of the Electric Field,

A = surface area,

θ = Angle made by the plane and the axis parallel to the direction of flow of the electric field.

**Applications of
Electric Flux:**

Electric flux forms the basic foundations of electrostatics and it has application like:

- Electric flux simplifies evaluation of electric field in complex figures.
- Electric flux helps in calculation of electric field.

Finding Angle between Two Vectors

If \(\overrightarrow{a}={{a}_{1}}\hat{i}+{{a}_{2}}\hat{j}+{{a}_{3}}\hat{k}\) and \(\overrightarrow{b}={{b}_{1}}\hat{i}+{{b}_{2}}\hat{j}+{{b}_{3}}\hat{k}\) are non-zero vectors, then the angle between them is given by \(\cos \theta =\frac{\overrightarrow{a.}\overrightarrow{b}}{\left| \overrightarrow{a} \right|\left| \overrightarrow{b} \right|}=\frac{{{a}_{1}}{{b}_{1}}+{{a}_{2}}{{b}_{2}}+{{a}_{3}}{{b}_{3}}}{\sqrt{a_{1}^{2}+a_{2}^{2}+a_{3}^{2}}\sqrt{b_{1}^{2}+b_{2}^{2}+b_{3}^{2}}}\).

Also

\(\frac{{{\left( {{a}_{1}}{{b}_{1}}+{{a}_{2}}{{b}_{2}}+{{a}_{3}}{{b}_{3}} \right)}^{2}}}{\left( a_{1}^{2}+a_{2}^{2}+a_{3}^{2} \right)\left( b_{1}^{2}+b_{2}^{2}+b_{3}^{2} \right)}={{\cos }^{2}}\theta \le 1\).

Or \({{\left( {{a}_{1}}{{b}_{1}}+{{a}_{2}}{{b}_{2}}+{{a}_{3}}{{b}_{3}} \right)}^{2}}\le \left( a_{1}^{2}+a_{2}^{2}+a_{3}^{2} \right)\left( b_{1}^{2}+b_{2}^{2}+b_{3}^{2} \right)\).

**Cosine Rule using Dot Product:**

Using vector method, prove that in a triangle a² = b² + c² – 2bc cos A (Cosine law).

In ΔABC,

Let \(\overrightarrow{AB}=\overrightarrow{c}\), \(\overrightarrow{BC}=\overrightarrow{a}\), \(\overrightarrow{CA}=\overrightarrow{b}\).

Since \(\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}=0\), we have \(\overrightarrow{a}=-\left( \overrightarrow{b}+\overrightarrow{c} \right)\),

∴ \(\left| \overrightarrow{a} \right|=\left| -\left( \overrightarrow{b}+\overrightarrow{c} \right) \right|\),

or \({{\left| \overrightarrow{a} \right|}^{2}}={{\left| \overrightarrow{b}+\overrightarrow{c} \right|}^{2}}\),

or \({{\left| \overrightarrow{a} \right|}^{2}}={{\left| \overrightarrow{b} \right|}^{2}}+{{\left| \overrightarrow{c} \right|}^{2}}+2\,\overrightarrow{b}.\overrightarrow{c}\),

or \({{\left| \overrightarrow{a} \right|}^{2}}={{\left| \overrightarrow{b} \right|}^{2}}+{{\left| \overrightarrow{c} \right|}^{2}}+2\left| \overrightarrow{b} \right|\left| \overrightarrow{c} \right|\cos \left( \pi -A \right)\) (∵ angle between \(\overrightarrow{b}\) and \(\overrightarrow{c}\) = angle between CA produced and AB)

or a² = b² + c² – 2bc cos A.

]]>**1. Communitive property **

\(\overrightarrow{a}+\overrightarrow{b}=\overrightarrow{b}+\overrightarrow{a}\).

**2. Associative property**

\(\left( \overrightarrow{a}+\overrightarrow{b} \right)+\overrightarrow{c}=\overrightarrow{a}+\left( \overrightarrow{b}+\overrightarrow{c} \right)\).

**3. Additive identity**

\(\overrightarrow{a}+\overrightarrow{0}=\overrightarrow{a}\).

**4. Additive inverse**

\(\overrightarrow{a}+\left( -\overrightarrow{a} \right)=\overrightarrow{0}\).

5. \(\left| \overrightarrow{a}+\overrightarrow{b} \right|\le \left| \overrightarrow{a} \right|+\left| \overrightarrow{b} \right|\)and \(\left| \overrightarrow{a}-\overrightarrow{b} \right|\ge \left| \overrightarrow{a} \right|-\left| \overrightarrow{b} \right|\).

**Examples 1: **If vector \(\overrightarrow{a}+\overrightarrow{b}\) bisects
the angle between \(\overrightarrow{a}\) and \(\overrightarrow{b}\), then prove that \(\left| \overrightarrow{a} \right|=\left|
\overrightarrow{b} \right|\).

**Solution: **We know that vector \(\overrightarrow{a}+\overrightarrow{b}\) is along
the diagonal of the parallelogram whose adjacent sides are vectors \(\overrightarrow{a}\) and \(\overrightarrow{b}\).

Now if \(\overrightarrow{a}+\overrightarrow{b}\) bisects the angle between vector \(\overrightarrow{a}\) and \(\overrightarrow{b}\), then the parallelogram must be a rhombus

Hence \(\left| \overrightarrow{a} \right|=\left| \overrightarrow{b} \right|\).

**Example 2:** \(\overrightarrow{AO}+\overrightarrow{OB}=\overrightarrow{BO}+\overrightarrow{OC}\),
then prove that B is the midpoint of AC.

**Solution: **Given that \(\overrightarrow{AO}+\overrightarrow{OB}=\overrightarrow{BO}+\overrightarrow{OC}\),

\(\overrightarrow{AB}=\overrightarrow{BC}\),

Thus, vectors \(\overrightarrow{AB}\) and \(\overrightarrow{BC}\) are collinear

Point A, B, C are collinear

\(\overrightarrow{AB}=\overrightarrow{BC}\),

B is the midpoint of AC.

**Example 3: **ABCDE is a pentagon prove that the
resultant of forces \(\overrightarrow{AB},\ \overrightarrow{AE},\
\overrightarrow{BC},\ \overrightarrow{DC},\ \overrightarrow{ED}\) and \(\overrightarrow{AC}\) is \(3\overrightarrow{AC}\).

**Solution: **Given that \(\overrightarrow{AB},\ \overrightarrow{AE},\
\overrightarrow{BC},\ \overrightarrow{DC},\ \overrightarrow{ED}\),

\(R=\overrightarrow{AB}+\ \overrightarrow{AE}+\overrightarrow{BC}+\ \overrightarrow{DC}+\ \overrightarrow{ED}+\overrightarrow{AC}\),

\(=\left( \overrightarrow{AB}+\overrightarrow{BC} \right)+\left( \overrightarrow{AE}+\overrightarrow{BD}+\overrightarrow{DC} \right)+\left( \overrightarrow{AC} \right)\),

\(\overrightarrow{AC}+\overrightarrow{AC}+\overrightarrow{AC}=3\overrightarrow{AC}\).

]]>