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This is a Fill in the blanks type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l}{\left|\overrightarrow{a}\times \overrightarrow{b}\right|}^2+{\left(\overrightarrow{a}.\overrightarrow{b}\right)}^2=144\\ \Rightarrow {\left(\left|\overrightarrow{a}\right|\left|\overrightarrow{b}\right|sin\theta \right)}^2+{\left(\left|\overrightarrow{a}\right|\left|\overrightarrow{b}\right|cos\theta \right)}^2=144\\ \Rightarrow {\left|\overrightarrow{a}\right|}^2{\left|\overrightarrow{b}\right|}^{2}si{n}^2\theta +{\left|\overrightarrow{a}\right|}^2{\left|\overrightarrow{b}\right|}^{2}co{s}^2\theta =144\\ \Rightarrow {\left|\overrightarrow{a}\right|}^2{\left|\overrightarrow{b}\right|}^2.\left(si{n}^2\theta +co{s}^2\theta \right)=144\\ \Rightarrow {\left|\overrightarrow{a}\right|}^2{\left|\overrightarrow{b}\right|}^2=144\\ \Rightarrow \left|\overrightarrow{a}\right|\left|\overrightarrow{b}\right|=12\\ \Rightarrow 4.\left|\overrightarrow{b}\right|=12\\ \therefore \left|\overrightarrow{b}\right|=3\\ Hence,thevalueofthefilleris3.\end{array}$

The institution has not released the course-wise placement packages as of yet. Meanwhile the NIRF report has been released. Check out the table below to know more:

NOTE: The above data is obtained from the NIRF report 2025

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

$\begin{array}{l}Equationofplanepassingthroughtwopoints\left(x_1,y_{1,}{z}_1\right)and\left(x_2,y_2,z_2\right)withitsnormald\text{'}ratiosis\\ a\left(x-x_1\right)+b\left(y-y_1\right)+c\left(z-z_1\right)=0\dots \left(i\right)\\ Iftheplaneispassingthroughthegivenpoints\left(2,1,-1\right)and\left(-1,3,4\right)then\\ a\left(x_2-x_1\right)+b\left(y_2-y_1\right)+c\left(z_2-z_1\right)=0\\ \Rightarrow a\left(-1-2\right)+b\left(3-1\right)+c\left(4+1\right)=0\\ \Rightarrow -3a+2b+5c=0\dots \left(ii\right)\\ Since,therequiredplaneisperpendiculartothegivenplanex-2y+4z=10,then\\ 1.a-2.b+4.c=10\dots \left(iii\right)\\ Solving\left(ii\right)and\left(iii\right)weget,\\ \frac{a}{8+10}=\frac{-b}{-12-5}=\frac{c}{6-2}=\lambda \\ a=18\lambda ,b=17\lambda ,c=4\lambda \\ Hence,therequiredplaneis18\lambda \left(x-2\right)+17\lambda \left(y-1\right)+4\lambda \left(z+1\right)=0\\ \Rightarrow 18x-36+17y-17+4z+4=0\\ \Rightarrow 18x+17y+4z-49=0\end{array}$

The selection criteria for the popular courses at ICRI Bangalore are as follows: 

• MBA: Entrance-based
• PG Diploma: Merit-based
• UG Diploma: Merit-based
• BBA: Merit-based

Candidates are required to participate in the admission process, which is based on the selection criteria to secure a seat. 

This is a Fill in the blanks type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l}{\left|\overrightarrow{a}\times \overrightarrow{b}\right|}^2+{\left(\overrightarrow{a}.\overrightarrow{b}\right)}^2={\left(\left|\overrightarrow{a}\right|\left|\overrightarrow{b}\right|sin\theta \right)}^2+{\left(\left|\overrightarrow{a}\right|\left|\overrightarrow{b}\right|cos\theta \right)}^2\\ ={\left|\overrightarrow{a}\right|}^2{\left|\overrightarrow{b}\right|}^{2}si{n}^2\theta +{\left|\overrightarrow{a}\right|}^2{\left|\overrightarrow{b}\right|}^{2}co{s}^2\theta \\ ={\left|\overrightarrow{a}\right|}^2{\left|\overrightarrow{b}\right|}^2.\left(si{n}^2\theta +co{s}^2\theta \right)\\ ={\left|\overrightarrow{a}\right|}^2{\left|\overrightarrow{b}\right|}^2.1={\left|\overrightarrow{a}\right|}^2{\left|\overrightarrow{b}\right|}^2\\ Hence,thevalueofthefilleris{\left|\overrightarrow{a}\right|}^2{\left|\overrightarrow{b}\right|}^2.\end{array}$

The institution has not released the overall average package. The median package offered at Atmiya University placement in the past years are tabulated below:

The average package at Atmiya University Placement  has not been made available yet. However the median package has been released and is given in the table below:

NOTE: The above data is obtained from the NIRF report 2025

This is a Fill in the blanks type Questions as classified in NCERT Exemplar

Sol: 

$\begin{array}{l}Giventhat\left|k\overrightarrow{a}\right|<\left|\overrightarrow{a}\right|andk\overrightarrow{a}+\frac{1}{2}\overrightarrow{a}isparallelto\overrightarrow{a}\\ \therefore \left|k\overrightarrow{a}\right|<\left|\overrightarrow{a}\right|\Rightarrow \left|k\right|\left|\overrightarrow{a}\right|<\left|\overrightarrow{a}\right|\Rightarrow \left|k\right|<1\Rightarrow -1<k<1\\ Nowsincek\overrightarrow{a}+\frac{1}{2}\overrightarrow{a}isparallelto\overrightarrow{a}\\ Here,weseethatk=-\frac{1}{2},k\overrightarrow{a}+\frac{1}{2}\overrightarrow{a}becomenullvectorandthenitwillnotbeparallelto\overrightarrow{a}.\\ \therefore k\overrightarrow{a}+\frac{1}{2}\overrightarrow{a}isparallelto\overrightarrow{a}whenk\in \left(-1,1\right)andk\ne \frac{1}{2}.\\ Hence,therequiredvalueofk\in \left(-1,1\right)andk\ne \frac{1}{2}.\end{array}$

The eligibility criteria for the courses at ICRI Bangalore are as follows: 

Candidates who meet the eligibility criteria for their desired course can participate in the admission process. 

Amity School of Communication offers three MA programmes. Admission to these courses is based on performance in the personal interview round. However, the applicants must also fulfil the course-specific eligibility requirements. The offered course specialisations are :-

• Journalism and Mass Communication 
• Advertising and Marketing Management
• PR and Event Management

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

$\begin{array}{l}Givenpointis\left(3,0,1\right)andtheequationofplanesare\\ x+2y=0\dots \left(i\right)\\ and3y-z=0\dots \left(ii\right)\\ Equationofanylinelpassingthrough\left(3,0,1\right)is\\ l:\frac{x-3}{a}=\frac{y-0}{b}=\frac{z-1}{c}\\ Directionratiosofthenormaltotheplane\left(i\right)and\left(ii\right)are\left(1,2,0\right)and\left(0,3,-1\right)\\ Since,thelineisparalleltoboththeplanes.\\ \therefore 1.a+2.b+0.c=0\Rightarrow a+2b+0c=0\\ and0.a+3.b-1.c=0\Rightarrow 0.a+3b-c=0\\ So,\frac{a}{-2-0}=\frac{-b}{-1-0}=\frac{c}{3-0}=\lambda \\ \therefore a=-2\lambda ,b=\lambda ,c=3\lambda \\ So,equationoflineis\\ \frac{x-3}{-2\lambda }=\frac{y-0}{\lambda }=\frac{z-1}{3\lambda }\\ Hence,therequiredequation\frac{x-3}{-2}+\frac{y-0}{1}+\frac{z-1}{3}\\ orinvectorformis\left(x-3\right)\widehat{i}+y\widehat{j}+\left(z-1\right)\widehat{k}=\lambda \left(-2\widehat{i}+\widehat{j}+3\widehat{k}\right)\end{array}$

This is a Fill in the blanks type Questions as classified in NCERT Exemplar

Sol:

         $\begin{array}{l}Giventhat\overrightarrow{a}=3\widehat{i}-2\widehat{j}+2\widehat{k}\\ and\overrightarrow{b}=-\widehat{i}-2\widehat{k}\\ \therefore \overrightarrow{a}+\overrightarrow{b}=2\widehat{i}-2\widehat{j}and\overrightarrow{a}-\overrightarrow{b}=4\widehat{i}-2\widehat{j}+4\widehat{k}\\ Let\theta betheanglebetweenthetwodiagonalvectors\overrightarrow{a}+\overrightarrow{b}and\overrightarrow{a}-\overrightarrow{b}then\\ cos\theta =\frac{\left(\overrightarrow{a}+\overrightarrow{b}\right).\left(\overrightarrow{a}-\overrightarrow{b}\right)}{\left|\overrightarrow{a}+\overrightarrow{b}\right|\left|\overrightarrow{a}-\overrightarrow{b}\right|}=\frac{\left(2\widehat{i}-2\widehat{j}\right).\left(4\widehat{i}-2\widehat{j}+4\widehat{k}\right)}{\sqrt[]{{\left(2\right)}^2+{\left(-2\right)}^2}\sqrt[]{{\left(4\right)}^2+{\left(-2\right)}^2+{\left(4\right)}^2}}\\ =\frac{8+4}{2\sqrt[]{2}.6}=\frac{12}{2\sqrt[]{2}.6}=\frac{1}{\sqrt[]{2}}\\ \therefore \theta =\frac{\pi }{4}\\ Hence,thevalueofrequiredfilleris\frac{\pi }{4}.\end{array}$