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BPharm applicants need to keep essential documents ready, including their Class 10 and 12 mark sheets, transfer certificate, passport-sized photographs, ID proof, and caste certificate (if applicable). All documents must be scanned and uploaded as per the NIILM university's instructions during the application process. Timely submission of documents is important to avoid delays in the admission process.

The equations of the given planes are

$\begin{array}{l}\overrightarrow{r}.\left(\widehat{i}+2\widehat{j}+3\widehat{k}\right)-4=0.....(1)\\ \overrightarrow{r}.\left(2\widehat{i}+\widehat{j}-\widehat{k}\right)+5=0......(2)\end{array}$

The equation of the plane passing through the line intersection of the plane given in equation (1) and equation (2) is

$\begin{array}{l}\left[\overrightarrow{r}.\left(\widehat{i}+2\widehat{j}+3\widehat{k}\right)-4\right]+\lambda \left[\overrightarrow{r}.\left(2\widehat{i}+\widehat{j}-\widehat{k}\right)+5\right]=0\\ \overrightarrow{r}.\left[\left(2\lambda +1\right)\widehat{i}+\left(\lambda +2\right)\widehat{j}+\left(3-\lambda \right)\widehat{k}\right]+\left(5\lambda -4\right)=0.....(3)\end{array}$

The plane in equation (3) is perpendicular to the plane,  $\overrightarrow{r}.\left(5\widehat{i}+3\widehat{j}-6\widehat{k}\right)+8=0$

$\begin{array}{l}\therefore 5\left(2\lambda +1\right)+3\left(\lambda +2\right)-6\left(3-\lambda \right)=0\\ \Rightarrow 19\lambda -7=0\\ \Rightarrow \lambda =\frac{7}{19}\end{array}$

Substituting λ = 7/19 in equation (3), we obtain

$\begin{array}{l}\Rightarrow \overrightarrow{r}.\left(\frac{13}{19}\widehat{i}+\frac{45}{19}\widehat{j}+\frac{50}{19}\widehat{k}\right)\frac{-41}{19}=0\\ \Rightarrow \overrightarrow{r}.\left(33\widehat{i}+45\widehat{j}+50\widehat{k}\right)-41=0\end{array}$

This is the vector equation of the required plane.

The Cartesian equation of this plane can be obtained by substituting  $\overrightarrow{r}=x\widehat{i}+y\widehat{j}+z\widehat{k}$  in equation (3).

$\begin{array}{l}\left(x\widehat{i}+y\widehat{j}+z\widehat{k}\right).\left(33\widehat{i}+45\widehat{j}+50\widehat{k}\right)-41=0\\ \Rightarrow 33x+45y+50z-41=0\end{array}$

34.$=\frac{sin5x+sin3x}{cos5x+cos3x.}$

$=\frac{2sin\left(\frac{5x+3x}{2}\right)cos\left(\frac{5x-3x}{2}\right)}{2cos\left(\frac{5x+3x}{2}\right)cos\left(\frac{5x-3x}{2}\right)}$

$=\frac{sin\frac{8x}{2}cos\frac{2x}{2}}{cos\frac{8x}{2}cos\frac{2x}{2}}$

$=\frac{sin4x}{cos4x}=tan4x=$R.H.S

Yes, the application form for Sambalpur University MBA admissions is available online. Students who wish to apply for the university's MBA course must apply via the SAMS official website. Check out below to learn how to access and fill out the form.

1. Visit the online admission portal of SAMS Odisha.

2. Click on the 'New User' button.

3. Register to make a new student account.

5. Log in to the account and fill out the application form.

6. Upload scanned documents and pay the Sambalpur University MBA application fees 2025.

7. Submit the form.

The coordinates of the points, O and P, are (0, 0) and (1, 2, −3) respectively.

Therefore, the direction ratios of OP are (1 − 0) = 1, (2 − 0) = 2, and (−3 − 0) = −3

It is known that the equation of the plane passing through the point (x1,  y1 z1) is

$a\left(x-x_1\right)+b\left(y-y_1\right)+c\left(z-z_1\right)=0$ where, a,  b, and c are the direction ratios of normal.

Here, the direction ratios of normal are 1, 2, and −3 and the point P is (1, 2, −3).

Thus, the equation of the required plane is

$\begin{array}{l}\mathrm{}\hfill \end{array}$$\begin{array}{ll}1\left(x-1\right)+2\left(y-2\right)-3\left(z+3\right)0\hfill & \Rightarrow x+2y-3z-14=0\hfill \end{array}$

To apply for the BPharm programme at NIILM University, students must visit the official admissions portal and complete the online application form. They should select the desired course, fill in accurate academic and personal details, upload the required documents, and pay the application fee. Regularly checking the website for updates is strongly recommended.

Candidates wishing to get admission to the courses at Somaiya School of Civilisation Studies must participate in the entrance based admission process. Candidates who do clear the admission process are required to pay the course fee to guarantee their admission. 

Candidates can check the official website of the institute to learn more about the fee structure. 

Equation of one plane is

$\begin{array}{l}\overrightarrow{r}.\left(\widehat{i}+\widehat{j}+\widehat{k}\right)=1\\ \Rightarrow \overrightarrow{r}.\left(\widehat{i}+\widehat{j}+\widehat{k}\right)-1=0\\ \overrightarrow{r}.\left(2\widehat{i}+3\widehat{j}-\widehat{k}\right)+4=0\end{array}$

The equation of any plane passing through the line of intersection of these planes is

$\begin{array}{l}\left[\overrightarrow{r}.\left(\widehat{i}+\widehat{j}+\widehat{k}\right)-1\right]+\lambda \left[\overrightarrow{r}.\left(2\widehat{i}+3\widehat{j}-\widehat{k}\right)+4\right]=0\\ \overrightarrow{r}.\left[\left(2\lambda +1\right)\widehat{i}+\left(3\lambda +1\right)\widehat{j}+\left(1-\lambda \right)\widehat{k}\right]+\left(4\lambda -1\right)=0.....(1)\end{array}$

Its direction ratios are (2λ + 1), (3λ + 1), and (1 − λ).

The required plane is parallel to x-axis. Therefore, its normal is perpendicular to x-axis.

The direction ratios of x-axis are 1, 0, and 0.

$\begin{array}{l}\therefore 1.\left(2\lambda +1\right)+0\left(3\lambda +1\right)+0\left(1-\lambda \right)=0\\ \Rightarrow 2\lambda +1=0\\ \Rightarrow \lambda =-\frac{1}{2}\end{array}$

Substituting λ = -1/2 in equation (1), we obtain

$\begin{array}{l}\Rightarrow \overrightarrow{r}.\left[-\frac{1}{2}\widehat{j}+\frac{3}{2}\widehat{k}\right]+\left(-3\right)=0\\ \Rightarrow \overrightarrow{r}\left(\widehat{j}-3\widehat{k}\right)+6=0\end{array}$

Therefore, its Cartesian equation is y − 3z + 6 = 0

This is the equation of the required plane.

33.$=\frac{cos9x-cos5x}{sin17x-sin3x}$

$=\frac{-2sin\left(\frac{9x+5x}{2}\right)sin\left(\frac{9x-5x}{2}\right)}{2cos\left(\frac{17x+3x}{2}\right)sin\left(\frac{17x-3x}{2}\right)}$

$=\frac{-sin\frac{14x}{2}sin\frac{4x}{2}}{cos\frac{20x}{2}sin\frac{14x}{2}}=\frac{-sin7xsin2x}{cos10xsin7x}=\frac{-sin2x}{cos10x}$

= R.H.S

The CPET for the session 2025 has already been conducted. The exam was held from May 3, 2035, to May 15, 2025. For admission to the Sambalpur University MBA course, students are required to present a valid non-zero score in the CPET exam. Those who have appeared for the exam must participate in the subsequent counselling process.

The position vector through the point $\left(1,1, p\right)$ is  $\overrightarrow{a_1}=\widehat{i}+\widehat{j}+p\widehat{k}$

Similarly, the position vector through the point $\left(-3,0,1\right)$ is

$\overrightarrow{a_2}=-4\widehat{i}+\widehat{k}$

The equation of the given plane is  $\overrightarrow{r}.\left(3\widehat{i}+4\widehat{j}-12\widehat{k}\right)+13=0$

It is known that the perpendicular distance between a point whose position vector is  $\overrightarrow{a}$  and the plane,  $\overrightarrow{r}.\overrightarrow{N}=d$ is given by,  $D=\frac{\left|\overrightarrow{a}.\overrightarrow{N}-d\right|}{\left|\overrightarrow{N}\right|}$

Here, $\overrightarrow{N}=3\widehat{i}+4\widehat{j}-12\widehat{k}$ and  $d =-13$

Therefore, the distance between the point (1, 1, p) and the given plane is

$\begin{array}{l}{D}_1=\frac{\left|\left(\widehat{i}+\widehat{j}+p\widehat{k}\right).\left(3\widehat{i}+4\widehat{j}-12\widehat{k}\right)+13\right|}{\left|3\widehat{i}+4\widehat{j}-12\widehat{k}\right|}\\ \Rightarrow D_1=\frac{\left|3+4-12p+13\right|}{}\\ \Rightarrow D_1=\frac{\left|20-12p\right|}{13}..........(1)\end{array}$

Similarly, the distance between the point $\left(-3,0,1\right)$ and the given plane is

$\begin{array}{l}{D}_2=\frac{\left|\left(3\widehat{i}+\widehat{k}\right).\left(3\widehat{i}+4\widehat{j}-12\widehat{k}\right)+13\right|}{\left|3\widehat{i}+4\widehat{j}-12\widehat{k}\right|}\\ \Rightarrow D_1=\frac{\left|-9-12+13\right|}{}\\ \Rightarrow D_1=\frac{8}{13}..........(2)\end{array}$

It is given that the distance between the required plane and the points, $\left(1,1, p\right)and\left(-3,0,1\right),$ is equal.

$\therefore D_1 = D_2$

$\begin{array}{l}\Rightarrow \frac{\left|20-12p\right|}{13}=\frac{8}{13}\\ \Rightarrow 20-12p=8,or,-(-20-12p)=8\\ \Rightarrow 12p=12,or,12p=28\\ \Rightarrow p=1,or,p=\frac{7}{3}\end{array}$

The equation of the plane passing through the point $a \left(x +1\right)+ b \left(y -3\right)+ c \left(z -2\right)=0\dots \left(1\right)$ where, a, b, c are the direction ratios of normal to the plane.

It is known that two planes,  $a_{1}x+b_{1}y+c_{1}z+d_1=0\&a_{2}x+b_{2}y+c_{2}z+d_2=0$ are perpendicular, if $a_1{a}_2+b_1{b}_2+c_1{c}_2=0$

Plane (1) is perpendicular to the plane,  $x +2y +3z =5$

$\begin{array}{ll}a.1 + b .2 + c.3 = 0\hfill & a + 2b + 3c = 0 ....\left(2\right)\hfill \end{array}$

Also, plane (1) is perpendicular to the plane, $3x +3y + z =0$

$\begin{array}{ll}a .3 + b.3 + c.1 = 0\hfill & 3a + 3b + c = 0 .....\left(3\right)\hfill \end{array}$

From equations (2) and (3), we obtain

$\begin{array}{l}\frac{a}{2\times 1-3\times 3}=\frac{b}{3\times 3-1\times 1}=\frac{c}{1\times 3-2\times 3}\\ \Rightarrow \frac{a}{-7}=\frac{b}{8}=\frac{c}{-3}=k(say)\\ \Rightarrow a=-7k,b=8k,c=-3k\end{array}$

Substituting the values of a, b, and c in equation (1), we obtain

$\begin{array}{l}-7k(x+1)+8k(y-3)-3k(z-2)=0\\ \Rightarrow (-7x-7)+(8y-24)-3z+6=0\\ \Rightarrow -7x+8y-3z-25=0\\ \Rightarrow 7x-8y+3z+25=0\end{array}$

This is the required equation of the plane.

32. L.H.S$=cot4x(sin5x+sin3x)$

$=cot4x\left[2sin\frac{8x}{2}cos\frac{2x}{2}\right]$

$=\frac{cos4x}{4x}\times 2sin4xcosx$

$=2·cos4x·cosx$

R.H.S$=cotx[sin5x-sin3x]$

$=cotx\left[2cos\frac{8x}{2}·sin\frac{2x}{2}\right]$

$=\frac{cosx}{sinx}·2cos4xsinx$

$=2·cos4x·cosx.$

Hence, L.H.S. = R.H.S.

It is known that the equation of the line through the points, $(x_1, y_1, z_1)and(x_2, y_2, z_2)$ , is

$\frac{x-x_1}{x_2-x_1}=\frac{x-y_1}{y_2-y_1}=\frac{z-z_1}{z_2-z_1}$

Since the line passes through the points, $\left(3,-4,-5\right)and\left(2,-3,1\right)$ , its equation is given by,

$\begin{array}{l}\frac{x-3}{2-3}=\frac{y+4}{-3+4}=\frac{z+5}{1+5}\\ \Rightarrow \frac{x-3}{-1}=\frac{y+4}{1}=\frac{z+5}{6}=k(say)\\ \Rightarrow x=3-k,y=k-4,z=6k-5\end{array}$

Therefore, any point on the line is of the form $\left(3- k, k -4,6k -5\right).$

This point lies on the plane, $2x + y + z =7$

$\begin{array}{lll}\therefore 2 \left(3- k\right) + \left(k -4\right) + \left(6k -5\right) = 7\hfill & 5k - 3 = 7\hfill & k = 2\hfill \end{array}$

Hence, the coordinates of the required point are $\left(3-2,2-4,6\times 2-5\right)i.e.,$ $\left(1,-2,7\right).$

NALSAR University of Law, Hyderabad, has a horizontal reservation for women students with Persons with Benchmark Disabilities (PWD). For NALSAR Hyderabad, the Round 1 CLAT cutoff 2025 for women PWD students ranged from 5524 to 6528 within the General category. Candidates can refer to the table below to view the CLAT cutoff 2025 for NALSAR Hyderabad under the horizontal reservation for PWD female (w) students. 

 * indicates that the seat is allotted based on the horizontal reservation

The three colleges that are compared have similar BSc tuition fees. Candidates can have a look at the table below to compare there BBA tuition fees: 

It is known that the equation of the line passing through the points, $(x_1, y_1, z_1)and(x_2, y_2, z_2),$ is  $\frac{x-x_1}{x_2-x_1}=\frac{x-y_1}{y_2-y_1}=\frac{z-z_1}{z_2-z_1}$