Ask & Answer: India's Largest Education Community

Once a candidate is selected based on their NMIMS-CET performance and eligibility, the final step is to confirm the seat by paying the tuition fee. For the MBA in Technology Management at NMIMS Mukesh Patel School of Technology, the tuition fee is INR 13.8 Lakh. Timely payment is necessary to secure admission.

While applying for the MBA at NMIMS Mukesh Patel School of Technology, students must upload key documents such as their Class 12 mark sheet, identification proof, passport-sized photographs, and any other documents specified on the application portal. It's essential to follow the official instructions to ensure successful verification and avoid delays.

The selection process for the MBA at NMIMS Mukesh Patel School of Technology is based on merit. Candidates must appear for the NMIMS-CET entrance exam and meet the eligibility criteria. Performance in this exam plays a central role in determining admission, so applicants should prepare thoroughly to secure a competitive score.

GB Shah College for Technology and Management application forms are also available online. Candidates meeting the eligibility criteria can apply for the preferred course. Below are the steps to apply for admission:

Step 1: Visit the official GB Shah College for Technology and Management website.

Step 2: Fill out the form.

Step 3: Complete and submit it.

As per recent data, for the academic year 2024-25, CWRU University has received 37,082 applications, among which 1,619 applicants were enrolled. Every year, around 20% of international students from over 100 countries around the world apply for CWRU admissions. The table below includes the Case Western Reserve University enrollment statistics for undergraduates:

Yes, applications are open at GB Shah College for Technology and Management for various courses. The mode of application is online. Aspirants need to apply at the official website for admission to the various programme. Candidates can fill out the application form on the institute's official website.

In order to get admission at GB Shah College for Technology and Management candidates must keep certain documents ready. Student can check the list of documents below:

• Scanned Class 10 and Class 12 marksheet
• Leaving Certificate
• Aadhaar Card
• Passport-size photograph
• Signature

Note: The list of documents mentioned above is a general documents. For more details, visit the official website.

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

Sol:

$\begin{array}{l}Giventhat\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}=0\\ So,\overrightarrow{a}\times \left(\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}\right)=\overrightarrow{a}\times 0\\ \Rightarrow \overrightarrow{a}\times \overrightarrow{a}+\overrightarrow{a}\times \overrightarrow{b}+\overrightarrow{a}\times \overrightarrow{c}=0\\ \Rightarrow \overrightarrow{0}+\overrightarrow{a}\times \overrightarrow{b}+\overrightarrow{a}\times \overrightarrow{c}=0\left(?\overrightarrow{a}\times \overrightarrow{a}=0\right)\\ \Rightarrow \overrightarrow{a}\times \overrightarrow{b}-\overrightarrow{c}\times \overrightarrow{a}=0\left(\overrightarrow{a}\times \overrightarrow{c}=-\overrightarrow{c}\times \overrightarrow{a}\right)\\ \Rightarrow \overrightarrow{a}\times \overrightarrow{b}=\overrightarrow{c}\times \overrightarrow{a}\dots \left(i\right)\\ Now,\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}=0\\ \Rightarrow \overrightarrow{b}\times \left(\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}\right)=\overrightarrow{b}\times 0\\ \Rightarrow \overrightarrow{b}\times \overrightarrow{a}+\overrightarrow{b}\times \overrightarrow{b}+\overrightarrow{b}\times \overrightarrow{c}=0\\ \Rightarrow \overrightarrow{b}\times \overrightarrow{a}+\overrightarrow{0}+\overrightarrow{b}\times \overrightarrow{c}=0\left(?\overrightarrow{b}\times \overrightarrow{b}=0\right)\\ \Rightarrow -\left(\overrightarrow{a}\times \overrightarrow{b}\right)+\overrightarrow{b}\times \overrightarrow{c}=0\\ \Rightarrow \overrightarrow{b}\times \overrightarrow{c}=\overrightarrow{a}\times \overrightarrow{b}\dots \left(ii\right)\\ Fromeq.\left(i\right)and\left(ii\right)weget\\ \overrightarrow{a}\times \overrightarrow{b}=\overrightarrow{b}\times \overrightarrow{c}=\overrightarrow{c}\times \overrightarrow{a}.Henceproved.\\ GeometricalInterpretation\\ Accordingtofigure,wehaveAreaofparallelogramABCDis\\ \Rightarrow \left|\overrightarrow{a}\times \overrightarrow{b}\right|=\left|\overrightarrow{a}\right|\left|\overrightarrow{b}\right|sin\theta \\ Since,theparallelogramsonthesamebaseandbetweenthesameparallellinesare\\ equalinarea.\\ \therefore \left|\overrightarrow{a}\times \overrightarrow{b}\right|=\left|\overrightarrow{b}\times \overrightarrow{c}\right|=\left|\overrightarrow{c}\times \overrightarrow{a}\right|\\ \Rightarrow \overrightarrow{a}\times \overrightarrow{b}=\overrightarrow{b}\times \overrightarrow{c}=\overrightarrow{c}\times \overrightarrow{a}.\end{array}$

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

Sol:

 $\begin{array}{l}Let\overrightarrow{a}=2\widehat{i}-\widehat{j}+\widehat{k}and\overrightarrow{b}=3\widehat{i}+4\widehat{j}-\widehat{k}\\ andlet\theta betheanglebetween\overrightarrow{a}and\overrightarrow{b}.\\ \therefore cos\theta =\frac{\overrightarrow{a}.\overrightarrow{b}}{\left|\overrightarrow{a}\right|\left|\overrightarrow{b}\right|}=\frac{\left(2\widehat{i}-\widehat{j}+\widehat{k}\right)\left(3\widehat{i}+4\widehat{j}-\widehat{k}\right)}{\sqrt[]{4+1+1}.\sqrt[]{9+16+1}}\\ =\frac{6-4-1}{\sqrt[]{6}\sqrt[]{26}}=\frac{1}{2\sqrt[]{3}.\sqrt[]{13}}=\frac{1}{2\sqrt[]{39}}\\ \therefore \theta =co{s}^{-1}\frac{1}{2\sqrt[]{39}}\Rightarrow co{s}^{-1}\left(\frac{1}{156}\right)\\ Hence,therequiredvalueof\theta isco{s}^{-1}\left(\frac{1}{156}\right).\end{array}$

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

$\begin{array}{l}Weknowthattheequationoflineis\\ \overrightarrow{r}=\overrightarrow{a}+\overrightarrow{b}\lambda \\ Here,\overrightarrow{a}=\widehat{i}-2\widehat{j}+3\widehat{k}and\overrightarrow{b}=3\widehat{i}-2\widehat{j}+6\widehat{k}\\ \therefore Equationoflineis\overrightarrow{r}=\left(\widehat{i}-2\widehat{j}+3\widehat{k}\right)+\lambda \left(3\widehat{i}-2\widehat{j}+6\widehat{k}\right)\\ \Rightarrow \left(x\widehat{i}+y\widehat{j}+z\widehat{k}\right)=\left(\widehat{i}-2\widehat{j}+3\widehat{k}\right)+\lambda \left(3\widehat{i}-2\widehat{j}+6\widehat{k}\right)\\ \Rightarrow \left(x\widehat{i}+y\widehat{j}+z\widehat{k}\right)-\left(\widehat{i}-2\widehat{j}+3\widehat{k}\right)=\lambda \left(3\widehat{i}-2\widehat{j}+6\widehat{k}\right)\\ \Rightarrow \left(x-1\right)\widehat{i}+\left(y+2\right)\widehat{j}+\left(z-3\right)\widehat{k}=\lambda \left(3\widehat{i}-2\widehat{j}+6\widehat{k}\right)\\ Hence,therequiredequationis\left(x-1\right)\widehat{i}+\left(y+2\right)\widehat{j}+\left(z-3\right)\widehat{k}=\lambda \left(3\widehat{i}-2\widehat{j}+6\widehat{k}\right).\end{array}$

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

Sol:

$\begin{array}{l}Let\overrightarrow{a}=2\widehat{i}-\widehat{j}+2\widehat{k}and\overrightarrow{b}=4\widehat{i}-\widehat{j}+3\widehat{k}\\ Weknowthatunitvectorareperpendicularto\overrightarrow{a}and\overrightarrow{b}=\frac{\left(\overrightarrow{a}\times \overrightarrow{b}\right)}{\left|\overrightarrow{a}\times \overrightarrow{b}\right|}\\ \overrightarrow{a}\times \overrightarrow{b}=\left|\begin{array}{ccc}\hfill \widehat{i}\hfill & \hfill \widehat{j}\hfill & \hfill \widehat{k}\hfill \\ \hfill 2\hfill & \hfill -1\hfill & \hfill 2\hfill \\ \hfill 4\hfill & \hfill -1\hfill & \hfill 3\hfill \end{array}\right|\\ =\widehat{i}\left(-3+2\right)-\widehat{j}\left(6-8\right)+\widehat{k}\left(-2+4\right)=-\widehat{i}+2\widehat{j}+2\widehat{k}\\ \therefore \left|\overrightarrow{a}\times \overrightarrow{b}\right|=\sqrt[]{{\left(-1\right)}^2+{\left(2\right)}^2+{\left(2\right)}^2}=\sqrt[]{1+4+4}=\sqrt[]{9}=3\\ So,\frac{\left(\overrightarrow{a}\times \overrightarrow{b}\right)}{\left|\overrightarrow{a}\times \overrightarrow{b}\right|}=\frac{-\widehat{i}+2\widehat{j}+2\widehat{k}}{3}=\frac{1}{3}\left(-\widehat{i}+2\widehat{j}+2\widehat{k}\right)\\ Nowthevectorofmagnitude6=\frac{1}{3}\left(-\widehat{i}+2\widehat{j}+2\widehat{k}\right).6\\ =2\left(-\widehat{i}+2\widehat{j}+2\widehat{k}\right)=-2\widehat{i}+4\widehat{j}+4\widehat{k}\\ Hence,therequiredvectoris-2\widehat{i}+4\widehat{j}+4\widehat{k}.\end{array}$

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

$\begin{array}{l}Let\alpha =60^0,\beta =45^{0}andtheangleinclinedtoOZaxisbe\gamma .\\ Weknowthatco{s}^2\alpha +co{s}^2\beta +co{s}^2\gamma =1\\ \Rightarrow co{s}^{2}60^0+co{s}^{2}45^0+co{s}^2\gamma =1\\ \Rightarrow {\left(\frac{1}{2}\right)}^2+{\left(\frac{1}{\sqrt[]{2}}\right)}^2+co{s}^2\gamma =1\Rightarrow \frac{1}{4}+\frac{1}{2}+co{s}^2\gamma =1\\ \Rightarrow \frac{3}{4}+co{s}^2\gamma =1\Rightarrow co{s}^2\gamma =1-\frac{3}{4}=\frac{1}{4}\\ \therefore cos\gamma =\pm \frac{1}{2}\Rightarrow cos\gamma =\frac{1}{2}\left(Rejectingcos\gamma =-\frac{1}{2},since\gamma <90^0\right)\\ \therefore \overrightarrow{OA}=\left|\overrightarrow{OA}\right|\left(\frac{1}{2}\widehat{i}+\frac{1}{\sqrt[]{2}}\widehat{j}+\frac{1}{2}\widehat{k}\right)=10\left(\frac{1}{2}\widehat{i}+\frac{1}{\sqrt[]{2}}\widehat{j}+\frac{1}{2}\widehat{k}\right)\\ =5\widehat{i}+5\sqrt[]{2}\widehat{j}+5\widehat{k}\\ Hence,thepositionvectorofAis\left(5\widehat{i}+5\sqrt[]{2}\widehat{j}+5\widehat{k}\right).\end{array}$