Ask & Answer: India's Largest Education Community

This is a Long Answer type Questions as classified in NCERT Exemplar

$\begin{array}{l}Since,\overrightarrow{a},\overrightarrow{b}and\overrightarrow{c}aretheverticesof\Delta ABC\\ \therefore \overrightarrow{AB}=\overrightarrow{b}-\overrightarrow{a},\overrightarrow{BC}=\overrightarrow{c}-\overrightarrow{b}and\overrightarrow{Ac}=\overrightarrow{c}-\overrightarrow{a}\\ \therefore Areaof\Delta ABC=\frac{1}{2}\left|\overrightarrow{AB}\times \overrightarrow{AC}\right|\\ =\frac{1}{2}\left|\left(\overrightarrow{b}-\overrightarrow{a}\right)\times \left(\overrightarrow{c}-\overrightarrow{b}\right)\right|=\frac{1}{2}\left|\overrightarrow{b}\times \overrightarrow{c}-\overrightarrow{b}\times \overrightarrow{a}-\overrightarrow{a}\times \overrightarrow{c}+\overrightarrow{a}\times \overrightarrow{a}\right|\\ =\frac{1}{2}\left|\overrightarrow{b}\times \overrightarrow{c}+\overrightarrow{a}\times \overrightarrow{b}+\overrightarrow{c}\times \overrightarrow{a}\right|\left[\begin{array}{l}?\overrightarrow{a}\times \overrightarrow{b}=-\overrightarrow{b}\times \overrightarrow{a}\\ \overrightarrow{c}\times \overrightarrow{a}=-\overrightarrow{a}\times \overrightarrow{c}\\ \overrightarrow{a}\times \overrightarrow{a}=\overrightarrow{0}\end{array}\right]\\ Forthreevectorsarecollinear,areaof\Delta ABC=0\\ \therefore \frac{1}{2}\left|\overrightarrow{b}\times \overrightarrow{c}+\overrightarrow{a}\times \overrightarrow{b}+\overrightarrow{c}\times \overrightarrow{a}\right|=0\\ \left|\overrightarrow{a}\times \overrightarrow{b}+\overrightarrow{b}\times \overrightarrow{c}+\overrightarrow{c}\times \overrightarrow{a}\right|=0\\ whichistheconditionofcollinearityof\overrightarrow{a},\overrightarrow{b}and\overrightarrow{c}.\\ Let\widehat{n}betheunitvectornormaltotheplaneofthe\Delta ABC\\ \therefore \widehat{n}=\frac{\overrightarrow{AB}\times \overrightarrow{AC}}{\left|\overrightarrow{AB}\times \overrightarrow{AC}\right|}=\frac{\overrightarrow{a}\times \overrightarrow{b}+\overrightarrow{b}\times \overrightarrow{c}+\overrightarrow{c}\times \overrightarrow{a}}{\left|\overrightarrow{a}\times \overrightarrow{b}+\overrightarrow{b}\times \overrightarrow{c}+\overrightarrow{c}\times \overrightarrow{a}\right|}\end{array}$

Yes, BCA course is available at GB Shah College for Technology and Management. The institute offers BCA courses at the UG level. Candidate must meet the eligibility criteria to enrol for BCA course admission set by the college. Aspirants must pass Class 12. GB Shah College for Technology and Management's admission process is merit-based.

The Case Western Reserve University application fee varies by the level of the programme that applicants choose. The application fee as per the programme is mentioned below:

• For UG: USD 75 (INR 6.4 K)
• For PG (including MBA): USD 100 (INR 8.5 K)

(USD 1 = INR 85.48)

Candidates have to apply online to secure BA admission at Christ University (Bannerghatta Campus). For this, candidates must visit the university's official website once the application portal opens. The application dates vary from year to year. Based on the previous year trends, the application process usually starts in Decemeber of the year preceding the admission year. It usually continues till May or June. For exact admission dates, candidates must visit the university's official website.

This is a Long Answer type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l}Here,inthegivenfigure,thecomponentsofcareccosAandandcsinA.\\ \therefore \overrightarrow{CD}=b-ccosA\\ In\Delta BDC,\\ a^2=C{D}^2+B{D}^2\\ \Rightarrow a^2={\left(b-ccosA\right)}^2+{\left(csinA\right)}^2\\ \Rightarrow a^2=b^2+c^{2}co{s}^{2}A-2bccosA+c^{2}si{n}^{2}A\\ \Rightarrow a^2=b^2+c^2\left(co{s}^{2}A+si{n}^{2}A\right)-2bccosA\\ \Rightarrow a^2=b^2+c^2-2bccosA\\ \Rightarrow 2bccosA=b^2+c^2-a^2\\ \therefore cosA=\frac{b^2+c^2-a^2}{2bc}\\ Henceproved.\end{array}$

The IELTS exam fee is different in each country. Here are some IELTS test fees for different countries:

Yes, IELTS is accepted worldwide. More than 12,500+ universities, companies, and government offices in over 140 countries accept IELTS scores. So you can use it in many places across the world.

Christ University Bannerghatta Campus BA admissions 2026 is currently not open. Candidates can go to the Christ Univeristy website to stay updated on any news regarding BA admission dates. This will help to ensure that candidates can apply for their preferred BA course before the scheduled Christ University Bannerghatta BA deadline. 

To secure a seat in the MBA programme at NMIMS Mukesh Patel School of Technology Shirpur, candidates must apply online, meet the eligibility criteria, and appear for the NMIMS-CET entrance exam. Admission is granted based on merit and exam performance. Timely application and preparation are essential to improve your chances of selection.

This is a Short Answer type Questions as classified in NCERT Exemplar

 $\begin{array}{l}LetABCDandABFEbetwoparallelogramsonthesamebaseABandbetweensame\\ parallellinesABandDF.\\ Let\overrightarrow{AB}=\overrightarrow{a}and\overrightarrow{AD}=\overrightarrow{b}\\ \therefore AreaofparallelogramABCD=\left|\overrightarrow{a}\times \overrightarrow{b}\right|\\ NowAreaofparallelogramABFE=\left|\overrightarrow{AB}\times \overrightarrow{AE}\right|\\ =\left|\overrightarrow{a}\times \left(\overrightarrow{AD}+\overrightarrow{DE}\right)\right|=\left|\overrightarrow{a}\times \left(\overrightarrow{b}+K\overrightarrow{a}\right)\right|\\ =\left|\left(\overrightarrow{a}\times \overrightarrow{b}\right)+K\left(\overrightarrow{a}\times \overrightarrow{a}\right)\right|=\left|\overrightarrow{a}\times \overrightarrow{b}\right|+0\left[?\overrightarrow{a}\times \overrightarrow{a}=0\right]\\ =\left|\overrightarrow{a}\times \overrightarrow{b}\right|\\ Henceproved.\end{array}$

IELTS stands for International English Language Testing System. It is an exam that shows how good your English is, for university admission, job, or immigration abroad.

The total seat intake for the integrated BTech in Computer Engineering + MBA in Technology Management at NMIMS Mukesh Patel School of Technology Shirpur is 120. These seats are filled annually based on merit through the NMIMS-CET entrance exam and are subject to reservation policies and eligibility requirements.

The IELTS exam is an English proficiency test which is conducted to check your English language skills. It tests you in Listening, Reading, Writing, and Speaking. If you want to study abroad, get a job, or move to an English-speaking country, you need this test. The IELTS in India is administered and conducted by IDP.

This is a Short Answer type Questions as classified in NCERT Exemplar

Sol:

 $\begin{array}{l}GiventhatA\left(1,2,3\right),B\left(2,-1,4\right)andC\left(4,5,-1\right)\\ \overrightarrow{AB}=\left(2-1\right)\widehat{i}+\left(-1-2\right)\widehat{j}+\left(4-3\right)\widehat{k}\\ \overrightarrow{AB}=\widehat{i}-3\widehat{j}+\widehat{k}\\ \overrightarrow{AC}=\left(4-1\right)\widehat{i}+\left(5-2\right)\widehat{j}+\left(-1-3\right)\widehat{k}=3\widehat{i}+3\widehat{j}-4\widehat{k}\\ Areaof\Delta ABC=\frac{1}{2}\left|\overrightarrow{AB}\times \overrightarrow{AC}\right|=\frac{1}{2}\left|\begin{array}{ccc}\hfill \widehat{i}\hfill & \hfill \widehat{j}\hfill & \hfill \widehat{k}\hfill \\ \hfill 1\hfill & \hfill -3\hfill & \hfill 1\hfill \\ \hfill 3\hfill & \hfill 3\hfill & \hfill -4\hfill \end{array}\right|\\ =\frac{1}{2}\left[\widehat{i}\left(12-3\right)-\widehat{j}\left(-4-3\right)+\widehat{k}\left(3+9\right)\right]=\frac{1}{2}\left|9\widehat{i}+7\widehat{j}+12\widehat{k}\right|\\ =\frac{1}{2}\sqrt[]{{\left(9\right)}^2+{\left(7\right)}^2+{\left(12\right)}^2}=\frac{1}{2}\sqrt[]{81+49+144}=\frac{1}{2}\sqrt[]{274}\\ Hence,therequiredareais\frac{\sqrt[]{274}}{2}.\end{array}$

GB Shah College for Technology and Management offers courses at UG level. The college offers a three/four-year BBA and BCA programme at UG level, respectively. Admission to the offered programmes is both merit-based.

This is a Short Answer type Questions as classified in NCERT Exemplar

 $\begin{array}{l}Here,PositionvectorofA=\widehat{i}+\widehat{j}-\widehat{k}\\ PositionvectorofB=2\widehat{i}-\widehat{j}+3\widehat{k}\\ PositionvectorofC=2\widehat{i}-3\widehat{k}\\ PositionvectorofD=3\widehat{i}-2\widehat{j}+\widehat{k}\\ \overrightarrow{AB}=P.VofB-P.VofA\\ =\left(2\widehat{i}-\widehat{j}+3\widehat{k}\right)-\left(\widehat{i}+\widehat{j}-\widehat{k}\right)=\widehat{i}-2\widehat{j}+4\widehat{k}\\ \overrightarrow{CD}=P.VofD-P.VofC\\ =\left(3\widehat{i}-2\widehat{j}+\widehat{k}\right)-\left(2\widehat{i}-3\widehat{k}\right)=\widehat{i}-2\widehat{j}+4\widehat{k}\\ Projectionof\overrightarrow{AB}on\overrightarrow{CD}=\frac{\overline{AB}\overline{.CD}}{\left|\overline{CD}\right|}=\frac{\left(\widehat{i}-2\widehat{j}+4\widehat{k}\right)\left(\widehat{i}-2\widehat{j}+4\widehat{k}\right)}{\sqrt[]{{\left(1\right)}^2+{\left(-2\right)}^2+{\left(4\right)}^2}}\\ =\frac{1+4+16}{\sqrt[]{1+4+16}}=\frac{21}{\sqrt[]{21}}=\sqrt[]{21}\\ Hence,therequiredprojection=\sqrt[]{21}.\end{array}$

The BCA full form is Bachelor of Computer Applications. It is a 3-year-long undergraduate degree programme primarily focused on computer applications and software development. It serves as a gateway for students who wish to pursue a career in the field of Information Technology (IT). Students can check the list of career options available after completing BCA from GB Shah College for Technology and Management:

This is a Short Answer type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l}Giventhat\overrightarrow{a}=3\widehat{i}+\widehat{j}+2\widehat{k}and\overrightarrow{b}=2\widehat{i}-2\widehat{j}+4\widehat{k}\\ Weknowthat\left|\overrightarrow{a}\times \overrightarrow{b}\right|=\left|\overrightarrow{a}\right|\left|\overrightarrow{b}\right|sin\theta \\ \therefore \overrightarrow{a}\times \overrightarrow{b}=\left|\begin{array}{ccc}\hfill \widehat{i}\hfill & \hfill \widehat{j}\hfill & \hfill \widehat{k}\hfill \\ \hfill 3\hfill & \hfill 1\hfill & \hfill 2\hfill \\ \hfill 2\hfill & \hfill -2\hfill & \hfill 4\hfill \end{array}\right|\\ =\widehat{i}\left(4+4\right)-\widehat{j}\left(12-4\right)+\widehat{k}\left(-6-2\right)\\ =8\widehat{i}-8\widehat{j}-8\widehat{k}\\ \therefore \left|\overrightarrow{a}\times \overrightarrow{b}\right|=\sqrt[]{{\left(8\right)}^2+{\left(-8\right)}^2+{\left(-8\right)}^2}\\ =\sqrt[]{64+64+64}=\sqrt[]{192}=\sqrt[]{64\times 3}=8\sqrt[]{3}\\ \Rightarrow \left|\overrightarrow{a}\right|=\sqrt[]{{\left(3\right)}^2+{\left(1\right)}^2+{\left(2\right)}^2}=\sqrt[]{9+1+4}=\sqrt[]{14}\\ \Rightarrow \left|\overrightarrow{b}\right|=\sqrt[]{{\left(2\right)}^2+{\left(-2\right)}^2+{\left(4\right)}^2}=\sqrt[]{4+4+16}=\sqrt[]{24}=2\sqrt[]{6}\\ \therefore sin\theta =\frac{\left|\overrightarrow{a}\times \overrightarrow{b}\right|}{\left|\overrightarrow{a}\right|\left|\overrightarrow{b}\right|}=\frac{8\sqrt[]{3}}{\sqrt[]{14}.2\sqrt[]{6}}=\frac{4\sqrt[]{3}}{\sqrt[]{84}}=\frac{4\sqrt[]{3}}{2\sqrt[]{21}}=\frac{2}{\sqrt[]{7}}\\ Hence,sin\theta =\frac{2}{\sqrt[]{7}}.\end{array}$