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Yes, admissions for the MBA in Technology Management at NMIMS Mukesh Patel School of Technology are currently open. Interested candidates must apply through the official NMIMS portal and ensure they register for the NMIMS-CET entrance exam within the specified deadlines to be considered for the 2025 intake.

Assam Science and Technology University offers a wide range of courses at the undergraduate and postgraduate levels. Students need to satisfy the admission criteria of the desired programme before applying for the same. Check out the list below to know about the ASTU courses:

• MSc
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• B.Arch
• M.Arch
• MPlan
• MPharma

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

Sol:

$\begin{array}{l}LetthegivenpointsareA\left(k,-10,3\right),B\left(1,-1,3\right)andC\left(3,5,3\right)\\ \overrightarrow{AB}=\left(1-k\right)\widehat{i}+\left(-1+10\right)\widehat{j}+\left(3-3\right)\widehat{k}=\left(1-k\right)\widehat{i}+9\widehat{j}+0\widehat{k}\\ \therefore \left|\overrightarrow{AB}\right|=\sqrt[]{{\left(1-k\right)}^2+{\left(9\right)}^2}=\sqrt[]{{\left(1-k\right)}^2+81}\\ \overrightarrow{BC}=\left(3-1\right)\widehat{i}+\left(5+1\right)\widehat{j}+\left(3-3\right)\widehat{k}=2\widehat{i}+6\widehat{j}+0\widehat{k}\\ \therefore \left|\overrightarrow{BC}\right|=\sqrt[]{{\left(2\right)}^2+{\left(6\right)}^2}=\sqrt[]{4+36}=\sqrt[]{40}=2\sqrt[]{10}\\ \overrightarrow{AC}=\left(3-k\right)\widehat{i}+\left(5+10\right)\widehat{j}+\left(3-3\right)\widehat{k}=\left(3-k\right)\widehat{i}+6\widehat{j}+0\widehat{k}\\ \therefore \left|\overrightarrow{AC}\right|=\sqrt[]{{\left(3-k\right)}^2+{\left(15\right)}^2}=\sqrt[]{{\left(3-k\right)}^2+225}\\ IfA,B,andCarecollinear,then\\ \left|\overrightarrow{AB}\right|+\left|\overrightarrow{BC}\right|=\left|\overrightarrow{AC}\right|\\ \sqrt[]{{\left(1-k\right)}^2+81}+2\sqrt[]{10}=\sqrt[]{{\left(3-k\right)}^2+225}\\ Squaringbothsides,wehave\\ {\left[\sqrt[]{{\left(1-k\right)}^2+81}+\sqrt[]{40}\right]}^2={\left[\sqrt[]{{\left(3-k\right)}^2+225}\right]}^2\\ \Rightarrow {\left(1-k\right)}^2+81+40+2\sqrt[]{40}\sqrt[]{{\left(1-k\right)}^2+81}={\left(3-k\right)}^2+225\\ \Rightarrow 1+k^2-2k+121+2\sqrt[]{40}\sqrt[]{1+k^2-2k+81}=9+k^2-6k+225\\ \Rightarrow 122-2k+2\sqrt[]{40}\sqrt[]{k^2-2k+82}=234-6k\\ Dividingby2,weget\\ \Rightarrow 61-k+\sqrt[]{40}\sqrt[]{k^2-2k+82}=117-3k\\ \Rightarrow \sqrt[]{40}\sqrt[]{k^2-2k+82}=117-61-3k+k\\ \Rightarrow \sqrt[]{40}\sqrt[]{k^2-2k+82}=56-2k\\ \Rightarrow 2\sqrt[]{10}\sqrt[]{k^2-2k+82}=56-2k\\ \Rightarrow \sqrt[]{10}\sqrt[]{k^2-2k+82}=28-k\left(Dividingby2\right)\\ Squaringbothsides,weget\\ \Rightarrow 10\left(k^2-2k+82\right)=784+k^2-56k\\ \Rightarrow 10k^2-20k+820=784+k^2-56k\\ \Rightarrow 10k^2-k^2-20k+56k+820-784=0\\ \Rightarrow 9k^2+36k+36=0\Rightarrow k^2+4k+4=0\\ \Rightarrow {\left(k+2\right)}^2=0\Rightarrow k+2=0\Rightarrow k=-2\\ Hence,therequiredvalueisk=-2.\end{array}$

Lady Shri Ram College for Women offered the highest placement package of INR 10.20 LPA and the prominent recruiters participated include McKinsey & Co., Accenture, KPMG, Boston Consulting Group, Bain & Company, etc. 

Listed below are the top Arts colleges in Delhi NCR along with their average placement package and the top recruiters:

Disclaimer: This information is sourced from the official website and may vary.

The following Practical Skills Module is available for the Level 3 CFA candidates. They need to complete the PSM in order to view their CFA result:

• Analyst Skills
• Financial modelling
• Portfolio Development and Construction
• Python, data Science and AI
• Python Programming Fundamentals. 

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

$\begin{array}{l}GiventhatBC=1.5BA\\ \Rightarrow \frac{BC}{BA}=1.5=\frac{3}{2}\\ \Rightarrow \frac{\overrightarrow{c}-\overrightarrow{b}}{\overrightarrow{a}-\overrightarrow{b}}=\frac{3}{2}\\ \Rightarrow 2\overrightarrow{c}-2\overrightarrow{b}=3\overrightarrow{a}-3\overrightarrow{b}\Rightarrow 2\overrightarrow{c}=3\overrightarrow{a}-3\overrightarrow{b}+2\overrightarrow{b}\\ \Rightarrow 2\overrightarrow{c}=3\overrightarrow{a}-\overrightarrow{b}\\ \therefore \overrightarrow{c}=\frac{3\overrightarrow{a}-\overrightarrow{b}}{2}\\ Hence,therequiredvectoris\overrightarrow{c}=\frac{3\overrightarrow{a}-\overrightarrow{b}}{2}.\end{array}$

The Students can apply for admissions at Amity School of Communication via the application portal website. Aspirants seeking admission are required the following: -

• Fill in their personal details
• Course preferences
• Academic qualification
• Complete other formalities

The table below shows the top government Arts colleges in Delhi NCR along with their tuition fees and India Today ranking 2024:

Disclaimer: This information is sourced from the official website and may vary.

The following PSM is availabel for the Level 2 CFA exam:

• Analyst Skills
• Python, Data Science and AI
• Python Programming Fundamentals

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

Sol:

 $\begin{array}{l}GivencoordinatesareP\left(5,0,8\right)andQ\left(3,3,2\right)\\ \therefore \overrightarrow{PQ}=\left(3-5\right)\widehat{i}+\left(3-0\right)\widehat{j}+\left(2-8\right)\widehat{k}=-2\widehat{i}+3\widehat{j}-6\widehat{k}\\ \therefore Unitvectorinthedirectionof\overrightarrow{PQ}=\frac{\overrightarrow{PQ}}{\left|\overrightarrow{PQ}\right|}\\ =\frac{-2\widehat{i}+3\widehat{j}-6\widehat{k}}{\sqrt[]{{\left(-2\right)}^2+{\left(3\right)}^2+{\left(-6\right)}^2}}=\frac{-2\widehat{i}+3\widehat{j}-6\widehat{k}}{\sqrt[]{4+9+36}}\\ =\frac{-2\widehat{i}+3\widehat{j}-6\widehat{k}}{\sqrt[]{49}}=\frac{-2\widehat{i}+3\widehat{j}-6\widehat{k}}{7}\\ =\frac{1}{7}\left(-2\widehat{i}+3\widehat{j}-6\widehat{k}\right)\\ Hence,therequiredunitvectoris\frac{1}{7}\left(-2\widehat{i}+3\widehat{j}-6\widehat{k}\right).\end{array}$

MBA programme at NMIMS Mukesh Patel School of Technology is offered in a single specialisation, Technology Management. This integrated course combines engineering and management education, focusing specifically on developing leadership skills in tech-driven industries.

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

Sol:

$\begin{array}{l}Giventhat\overrightarrow{a}=\widehat{i}+\widehat{j}+2\widehat{k}and\overrightarrow{b}=2\widehat{i}+\widehat{j}-2\widehat{k}\\ \left(i\right)6\overrightarrow{b}=6\left(2\widehat{i}+\widehat{j}-2\widehat{k}\right)=12\widehat{i}+6\widehat{j}-12\widehat{k}\\ \therefore Unitvectorinthedirectionof6\overrightarrow{b}=\frac{6\overrightarrow{b}}{\left|6\overrightarrow{b}\right|}\\ =\frac{12\widehat{i}+6\widehat{j}-12\widehat{k}}{\sqrt[]{{\left(12\right)}^2+{\left(6\right)}^2+{\left(-12\right)}^2}}=\frac{12\widehat{i}+6\widehat{j}-12\widehat{k}}{\sqrt[]{144+36+144}}\\ =\frac{12\widehat{i}+6\widehat{j}-12\widehat{k}}{\sqrt[]{324}}=\frac{12\widehat{i}+6\widehat{j}-12\widehat{k}}{18}\\ =\frac{6}{18}\left(2\widehat{i}+\widehat{j}-2\widehat{k}\right)=\frac{1}{3}\left(2\widehat{i}+\widehat{j}-2\widehat{k}\right)\\ Hence,therequiredunitvectoris\frac{1}{3}\left(2\widehat{i}+\widehat{j}-2\widehat{k}\right).\\ \left(ii\right)2\overrightarrow{a}-\overrightarrow{b}=2\left(\widehat{i}+\widehat{j}+2\widehat{k}\right)-\left(2\widehat{i}+\widehat{j}-2\widehat{k}\right)\\ =2\widehat{i}+2\widehat{j}+4\widehat{k}-2\widehat{i}-\widehat{j}+2\widehat{k}=\widehat{j}+6\widehat{k}\\ \therefore Unitvectorinthedirectionof2\overrightarrow{a}-\overrightarrow{b}=\frac{2\overrightarrow{a}-\overrightarrow{b}}{\left|2\overrightarrow{a}-\overrightarrow{b}\right|}\\ =\frac{\widehat{j}+6\widehat{k}}{\sqrt[]{{\left(1\right)}^2+{\left(6\right)}^2}}=\frac{\widehat{j}+6\widehat{k}}{\sqrt[]{1+36}}\\ =\frac{\widehat{j}+6\widehat{k}}{\sqrt[]{37}}=\frac{1}{\sqrt[]{37}}\left(\widehat{j}+6\widehat{k}\right)\\ Hence,therequiredunitvectoris\frac{1}{\sqrt[]{37}}\left(\widehat{j}+6\widehat{k}\right).\end{array}$