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This is an Objective type Questions as classified in NCERT Exemplar

Sol:    

            $\begin{array}{l}Theprojectionvectorof\overrightarrow{a}on\overrightarrow{b}=\left(\frac{\overrightarrow{a}.\overrightarrow{b}}{\left|\overrightarrow{b}\right|}\right).\overrightarrow{b}\\ Hence,thecorrectoptionis\left(a\right).\end{array}$

The application form fee for ICRI Bangalore is INR 1,300. Candidates are required to visit the official website to get the application form. The form must be filled with accurate details related to their academic and personal records. Candidates must upload the asked for documents. 

For an OBC rank of 11,137 in JEE Main 2025, NIT Hamirpur's CSE programme is a strong contender in a Northern state. Also you have option like NIT Srinagar, NIT jalandhar they may also fell in these rank

This is a short answer type question as classified in NCERT Exemplar

$\begin{array}{l}WehaveA\left(a,b,c\right)andO\left(0,0,0\right)\\ \therefore directionratiosofOA=a-0,b-0,c-0\\ \therefore directioncosinesoflineOA=\frac{a}{\sqrt[]{a^2+b^2+c^2}},\frac{b}{\sqrt[]{a^2+b^2+c^2}},\frac{c}{\sqrt[]{a^2+b^2+c^2}}\\ Now,directionratiosofthenormaltotheplaneare\left(a,b,c\right).\\ \therefore EquationoftheplanepassingthroughthepointA\left(a,b,c\right)is\\ a\left(x-a\right)+b\left(y-b\right)+c\left(z-c\right)=0\\ \Rightarrow ax-a^2+by-b^2+cz-c^2=0\\ \Rightarrow ax+by+cz=a^2+b^2+c^2\\ Hence,therequiredequationisax+by+cz=a^2+b^2+c^2.\end{array}$

Hindu College is the top-ranked college of India according to NIRF, and students of the college experience some of the most vibrant on-campus lives. The college excels in academics and provides quality higher education, with a leaning towards the Humanities. Further, the large number of clubs and societies (Music, Dance, Exam Prep, etc.) provide students with an all-round character development.

The Ajay Kumar Garg Engineering College has had successful placements over the years and is completely worth joining. For more information on AKGEC placements during 2024 and 2025, refer to the table below:

For the registration process of Amity School of Communication admissions 2025, the candidate needs scanned copies of:

• Class 10 and 12 marksheets (UG applicants)
• Graduation marksheet (PG applicants)
• Recent passport-sized photograph
• ID proof (Aadhaar/PAN)

This is an Objective type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l}Giventhat\left|\overrightarrow{a}\right|=\left|\overrightarrow{b}\right|=\left|\overrightarrow{c}\right|=1\\ and\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}=\overrightarrow{0}\\ \therefore \left(\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}\right).\left(\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}\right)=\overrightarrow{0}.\overrightarrow{0}=0\\ {\left|\overrightarrow{a}\right|}^2+\overrightarrow{a}.\overrightarrow{b}+\overrightarrow{a}.\overrightarrow{c}+\overrightarrow{b}.\overrightarrow{a}+{\left|\overrightarrow{b}\right|}^2+\overrightarrow{b}.\overrightarrow{c}+\overrightarrow{c}.\overrightarrow{a}+\overrightarrow{c}.\overrightarrow{b}+{\left|\overrightarrow{c}\right|}^2=0\\ \Rightarrow {\left|\overrightarrow{a}\right|}^2+{\left|\overrightarrow{b}\right|}^2+{\left|\overrightarrow{c}\right|}^2+2\overrightarrow{a}.\overrightarrow{b}+2\overrightarrow{b}.\overrightarrow{c}+2\overrightarrow{c}.\overrightarrow{a}=0\\ \Rightarrow 1+1+1+2\left(\overrightarrow{a}.\overrightarrow{b}+\overrightarrow{b}.\overrightarrow{c}+\overrightarrow{c}.\overrightarrow{a}\right)=0\\ \Rightarrow 2\left(\overrightarrow{a}.\overrightarrow{b}+\overrightarrow{b}.\overrightarrow{c}+\overrightarrow{c}.\overrightarrow{a}\right)=-3\\ \Rightarrow \overrightarrow{a}.\overrightarrow{b}+\overrightarrow{b}.\overrightarrow{c}+\overrightarrow{c}.\overrightarrow{a}=\frac{-3}{2}\\ Hence,thecorrectoptionis\left(c\right).\end{array}$

MAAC South Extension was the first 3D Animation & Visual Effects Institute in Delhi, established in 2001. Maya Academy of Advanced Cinematics, South Extension, located in Delhi, is a renowned educational institution.

MAAC Institute offers a wide range of certificate and diploma programmes across various fields, including Animation, Arts (Fine/ Visual/ Performing), Mass Communication & Media and Design. Some of the most popular programme offered at the institute include 3D Animation and VFX and Gaming.

The application fee for Amity School of Communication admissions 2025 is INR 1,500. This is same for all the courses and specialisations and is mandatory to be paid in order to submit the application form and complete the registration process successfully.

There is no provision to submit XAT application form online. The process to register for XAT exam is completely online. All the activities are conducted online at the xatonline.in website such as registration, form filling, fee payment. Candidates can fill the XAT application form on a desktop/laptop/tablet or any device with internet. Fee payment is to be done through debit/credit card or net banking.

The Ajay Kumar Garg Engineering College has had a strong record when it comes to placements. Refer to the table below for more placement details:

Yes, candidates have to fill the admission forms of XAT accepting colleges separately as there is no option to choose multiple colleges in the XAT form. Candidates can only choose XLRI in the XAT application form, for a fee of INR 200.

The application process for Amity School of Communication is conducted both online and offline. For online, the form needs to submitted directly on the portal and a standard fee of INR 1,500 needs to be paid, while for offline the candidates can download the form and submit it to the admission cell.

Hindu College admissions are held on the basis of students' performance in the CUET entrance exam. Earlier, students could get direct admissions via the HSC/10+2 merit-based cutoff system. Now, students must clear their 12th standard exams and appear for the prescribed CUET/CUET PG test for admissions into Hindu College UG and PG courses, respectively.

The Ajay Kumar Garg Engineering College is considered as good when it comes to placement packages. The key information on the AKGEC placement packages are presented below:

There is no provision of form correction in XAT. Thus, candidates must make sure they fill correct details in their XAT application forms or if they have made a mistake in any field, then must rectify it before pay the exam fee. Once the exam fee is paid, the form will get submitted and there will be no option to make changes.

This is a short answer type question as classified in NCERT Exemplar

$\begin{array}{l}Thegiventhatl,m,nandl+\delta l,m+\delta m,n+\delta narethedirectioncosinesofavariable\\ lineintwopositions\\ \therefore l^2+m^2+n^2=1\dots \left(i\right)\\ and{\left(l+\delta l\right)}^2+{\left(m+\delta m\right)}^2+{\left(n+\delta n\right)}^2=1\dots \left(ii\right)\\ \Rightarrow l^2+\delta l^2+2l.\delta l+m^2+\delta m^2+2m.\delta m+n^2+\delta n^2+2n.\delta n=1\\ \Rightarrow \left(l^2+m^2+n^2\right)+\left(\delta l^2+\delta m^2+\delta n^2\right)+2\left(l.\delta l+m.\delta m+n.\delta n\right)=1\\ \Rightarrow 1+\left(\delta l^2+\delta m^2+\delta n^2\right)+2\left(l.\delta l+m.\delta m+n.\delta n\right)=1\\ \Rightarrow l.\delta l+m.\delta m+n.\delta n=-\frac{1}{2}\left(\delta l^2+\delta m^2+\delta n^2\right)\\ Let\overrightarrow{a}and\overrightarrow{b}betheunitvectorsalongalinewith{d}^{\text{'}}cosinesl,m,nand\left(l+\delta l\right),\left(m+\delta m\right),\left(n+\delta n\right)\\ \therefore \overrightarrow{a}=l\widehat{i}+m\widehat{j}+n\widehat{k}and\overrightarrow{b}=\left(l+\delta l\right)\widehat{i}+\left(m+\delta m\right)\widehat{j}+\left(n+\delta n\right)\widehat{k}\\ \therefore cos\delta \theta =\frac{\overrightarrow{a}.\overrightarrow{b}}{\left|\overrightarrow{a}\right|.\left|\overrightarrow{b}\right|}\\ \Rightarrow cos\delta \theta =\frac{\left(l\widehat{i}+m\widehat{j}+n\widehat{k}\right)\left[\left(l+\delta l\right)\widehat{i}+\left(m+\delta m\right)\widehat{j}+\left(n+\delta n\right)\widehat{k}\right]}{1.1}\left[?\left|\overrightarrow{a}\right|=\left|\overrightarrow{b}\right|=1\right]\\ \Rightarrow cos\delta \theta =l\left(l+\delta l\right)+m\left(m+\delta m\right)+n\left(n+\delta n\right)\\ \Rightarrow cos\delta \theta =l^2+l.\delta l+m^2+m.\delta m+n^2+n.\delta n\\ \Rightarrow cos\delta \theta =\left(l^2+m^2+n^2\right)+\left(l.\delta l+m.\delta m+n.\delta n\right)\\ \Rightarrow cos\delta \theta =1-\frac{1}{2}\left(\delta l^2+\delta m^2+\delta n^2\right)\\ \Rightarrow 1-cos\delta \theta =\frac{1}{2}\left(\delta l^2+\delta m^2+\delta n^2\right)\\ \Rightarrow 2si{n}^2\frac{\delta \theta }{2}=\frac{1}{2}\left(\delta l^2+\delta m^2+\delta n^2\right)\\ \Rightarrow 4si{n}^2\frac{\delta \theta }{2}=\delta l^2+\delta m^2+\delta n^2\\ \Rightarrow 4{\left(\frac{\delta \theta }{2}\right)}^2=\delta l^2+\delta m^2+\delta n^2\left[?\frac{\delta \theta }{2}isverysmallso,sin\frac{\delta \theta }{2}=\frac{\delta \theta }{2}\right]\\ \Rightarrow {\left(\delta \theta \right)}^2=\delta l^2+\delta m^2+\delta n^{2}Henceproved.\end{array}$