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The insitute offers MBA course in the stream of Management. The institute admits students based on national-level entrance exam scores, i.e CAT/MAT/XAT/CMAT or MH-CET. To get admitted to BITM Pune, candidates must meet the eligibility criteria set by the institute. 

Yes, in the stream of Management, Sambalpur University offered an MBA course. It is a management course that runs for a duration of two years. Moreover, the mode of education is full-time. Students willing to pursue a career in Management can pursue this course. Moreover, the university grants MBA admissions based on the scores obtained in the CPET exam.

The top domain that has placed the most number of IAM Kolkata alumni is the Operations domain. The domain-wise IAM Kolkata alumni placements statistics are presented below:

Note: The “Others” category includes the domains of HR, Finance, IT, Administrative, Program and Project Management, Marketing, Arts and Design, Engineering, Community and Social Services and Legal.

The above-mentioned alumni data is sourced from multiple online sources and hence may vary.

The equations of the given planes are  $\overrightarrow{r}.\left(2\widehat{i}+2\widehat{j}-3\widehat{k}\right)=5\&\overrightarrow{r}.\left(3\widehat{i}-3\widehat{j}+5\widehat{k}\right)=3$

It is known that if n1 and n2 are normal to the planes,  $\overrightarrow{r}.\overrightarrow{n_1}=d_1\&\overrightarrow{r}.\overrightarrow{n_2}=d_2$  then the angle between them, Q, is given by,

$cosQ=\left|\frac{\overrightarrow{n_1}.\overrightarrow{n_2}}{\left|\overrightarrow{n_1}\right|\left|\overrightarrow{n_2}\right|}\right|..........(1)$

$\begin{array}{l}Here,\overrightarrow{n_1}=2\widehat{i}+2\widehat{j}-3\widehat{k}\&\overrightarrow{n_2}=3\widehat{i}-3\widehat{j}+5\widehat{k}\\ \therefore \overrightarrow{n_1}.\overrightarrow{n_2}=\left(2\widehat{i}+2\widehat{j}-3\widehat{k}\right)\left(3\widehat{i}-3\widehat{j}+5\widehat{k}\right)=2.3+2.(-3)+(-3).5=-15\\ \left|\overrightarrow{n_1}\right|==\\ \left|\overrightarrow{n_2}\right|==\end{array}$

Substituting the value of $\overrightarrow{n_1}.\overrightarrow{n_2}$ $\left|\overrightarrow{n_1}\right|\&\left|\overrightarrow{n_2}\right|$ in equation (1), we obtain

$\begin{array}{l}cosQ=\left|\frac{-15}{.}\right|\\ \Rightarrow cosQ=\frac{15}{}\\ \Rightarrow cos{Q}^{-1}=\left(\frac{15}{}\right)\end{array}$

BBA Admission at St Anne's Degree College is simple and not too complicated. The admissions starts from the month of June and goes on till October. These are the steps the admissions process generally follows:

1.       Applying for St Anne's Degree College admissions online

2.       Appearing for Counselling online or offline

3.       Do the final document verification at the desired college

During the counselling round candidates have the opportunity to select their desired college and course.

IAM Kolkata provides good placements to its students. It also provides Industrial training and internships. The leading recruiters during IAM Kolkata placements in recent past include Deloitte, Park Group of Hotels, The TAJ and many more.

The equation of the plane through the intersection of the planes, $x+y+z=1and2x+3y+4z=5$ , is

  $\left(x+y+z-1\right) +\lambda \left(2x+3y+4z-5\right)$

$\Rightarrow \left(2\lambda +1\right)x+\left(3\lambda +1\right)y+\left(4\lambda +1\right)z-\left(5\lambda +1\right)=0 ..........\left(1\right)$

The direction ratios,  $a_1, b_1, c_1,$ of this plane are $\left(2\lambda +1\right),\left(3\lambda +1\right),and\left(4\lambda +1\right).$

The plane in equation (1) is perpendicular to $x-y+z=0$

Its direction ratios,  $a_2, b_2, c_2,$ are $1,-1,and1$ .

Since the planes are perpendicular,

$\begin{array}{l}{a}_1{a}_2+b_1{b}_2+c_1{c}_2=0\\ \Rightarrow \left(2\lambda +1\right)-\left(3\lambda +1\right)+\left(4\lambda +1\right)=0\\ \Rightarrow 3\lambda +1=0\\ \Rightarrow \lambda =-\frac{1}{3}\end{array}$

Substituting $\lambda =-\frac{1}{3}$ in equation (1), we obtain

$\begin{array}{l}\frac{1}{3}x-\frac{1}{3}z+\frac{2}{3}=0\\ \Rightarrow x-z+2=0\end{array}$

This is the required equation of the plane.

The equations of the planes are  $\overrightarrow{r}.\left(2\widehat{i}+2\widehat{j}-2\widehat{k}\right)=7,\overrightarrow{r}.\left(2\widehat{i}+5\widehat{j}+3\widehat{k}\right)=9$

$\begin{array}{l}\Rightarrow \overrightarrow{r}.\left(2\widehat{i}+2\widehat{j}-2\widehat{k}\right)-7=0..........(1)\\ \Rightarrow \overrightarrow{r}.\left(2\widehat{i}+5\widehat{j}+3\widehat{k}\right)-9=0..........(2)\end{array}$

The equation of any plane through the intersection of the planes given in equations (1) and (2) is given by,

$\begin{array}{l}\left[\overrightarrow{r}.\left(2\widehat{i}+2\widehat{j}-2\widehat{k}\right)-7\right]+\lambda \left[\overrightarrow{r}.\left(2\widehat{i}+5\widehat{j}+3\widehat{k}\right)-9\right]=0,where,\lambda \in R\\ \overrightarrow{r}.\left[\left(2\widehat{i}+2\widehat{j}-2\widehat{k}\right)+\lambda \left(2\widehat{i}+5\widehat{j}+3\widehat{k}\right)\right]=9\lambda +7\\ \overrightarrow{r}.\left[(2+2\lambda )\widehat{i}+(2+5\lambda )\widehat{j}+(3\lambda -3)\widehat{k}\right]=9\lambda +7..........(3)\end{array}$

The plane passes through the point (2, 1, 3). Therefore, its position vector is given by,

$\overrightarrow{r}=2\widehat{i}+1\widehat{j}+3\widehat{k}$

Substituting in equation (3), we obtain

$\begin{array}{l}\left(2\widehat{i}+\widehat{j}+3\widehat{k}\right).\left[(2+2\lambda )\widehat{i}+(2+5\lambda )\widehat{j}+(3\lambda -3)\widehat{k}\right]=9\lambda +7\\ \Rightarrow 2(2+2\lambda )+1(2+5\lambda )+3(3\lambda -3)=9\lambda +7\\ \Rightarrow 4+4\lambda +2+5\lambda +9\lambda -9=9\lambda +7\\ \Rightarrow 18\lambda -3=9\lambda +7\\ \Rightarrow 9\lambda =10\\ \Rightarrow \lambda =\frac{10}{9}\end{array}$

Substituting $\lambda =\frac{10}{9}$ in equation (3), we obtain

$\begin{array}{l}\overrightarrow{r}.\left(\frac{38}{9}\widehat{i}+\frac{68}{9}\widehat{j}+\frac{3}{9}\widehat{k}\right)=17\\ \Rightarrow \overrightarrow{r}.\left(38\widehat{i}+68\widehat{j}+3\widehat{k}\right)=153\end{array}$