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No,  SRMJEEE is not required for admisson to the BA course admisson at SRM University Sonepat. Admission to the BA course is subject to a valid score in the last qualifying exam. Candidates must have completed Class 12 with at least 45% and must be inclusive of subjects related to specialisations candidates want to apply to. Candidates must have completed Class 12 with a minimum of 50% aggregate in the final examination. 

Let AB be the line joining the points $\left(1,-1,2\right)$ and $\left(3,4,-2\right)$ and CD be the line joining the point $\left(0,3,2\right)$ and $\left(3,5,6\right)$ .

The direction ratios, a, b, c of AB are $\begin{array}{l}\left(3-1\right),\left(4-\left(-1\right)\right),\left(-2-2\right)\\ =\left(2,5,-4\right)\end{array}$

The direction ratios $a_2,b_2,c_2$ of CD are $\begin{array}{l}\left(3-0\right),\left(5-3\right),\left(6-2\right)\\ =\left(3,2,4\right)\end{array}$

AB and CD will be perpendicular to each other, if $a_1{a}_2+b_1{b}_2+c_1{c}_2=0$

$\begin{array}{l}=2\times 3+5\times 2+\left(-4\right)\times 4\\ =6+10-16\\ =0\end{array}$

$\therefore$ Therefore, AB and CD are perpendicular to each other.

18. $$

$\begin{array}{l}{\left(\frac{1}{2}\right)}^2+{\left(\frac{1}{2}\right)}^2-1^2\\ =\frac{1}{4}+\frac{1}{4}-1\\ =\frac{1+1+4}{4}\\ =\frac{-2}{4}\\ =\frac{-1}{2}\end{array}$

=R.H.S

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Two lines with direction cosines l, m, n and l₂, m₂, n₂ are perpendicular to each other, if $l_1{l}_2+m_1{m}_2+n_1{n}_2=0$ .

Now, for the 3 lines with direction cosine,

$\begin{array}{l}\frac{12}{13},\frac{-3}{13},\frac{-4}{13}and\frac{4}{13},\frac{12}{13},\frac{3}{13}\\ l_1{l}_2+m_1{m}_2+n_1{n}_2=\frac{12}{13}\times \frac{4}{13}+\frac{\left(-3\right)}{13}\times \frac{12}{13}+\frac{\left(-4\right)}{13}\times \frac{3}{13}\\ =\frac{48}{169}-\frac{36}{169}-\frac{12}{169}\\ =0\end{array}$

Hence, the lines are perpendicular.

For lines with direction cosines,

$\begin{array}{l}\frac{4}{13},\frac{12}{13},\frac{3}{13}and\frac{3}{13},\frac{-4}{13},\frac{12}{13}\\ l_1{l}_2+m_1{m}_2+n_1{n}_2=\frac{4}{13}\times \frac{3}{13}+\frac{12}{13}\times \frac{\left(-4\right)}{13}+\frac{3}{13}\times \frac{12}{13}\\ =\frac{12}{169}-\frac{48}{169}+\frac{36}{169}\\ =0\end{array}$

Hence, these lines are perpendicular.

For the lines with direction cosines,

$\begin{array}{l}\frac{3}{13},\frac{-4}{13},\frac{12}{13}and\frac{12}{13},\frac{-3}{13},\frac{-4}{13}\\ l_1{l}_2+m_1{m}_2+n_1{n}_2=\frac{3}{13}\times \frac{12}{13}+\frac{\left(-4\right)}{13}\times \frac{\left(-3\right)}{13}+\frac{12}{13}\times \frac{\left(-4\right)}{13}\\ =\frac{36+12-48}{169}\\ =0\end{array}$

Hence, these lines are perpendicular.

Therefore, all the lines are perpendicular.

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The vertices of ABC are A (3,5, -4), B (-1,1,2) and C (-5, -5, -2)

Direction ratio of side AB $\begin{array}{l}=\left(-1-3\right)\left(1-5\right)\left(2-(-4)\right)\\ =\left(-4,-4,6\right)\end{array}$

 

Direction cosine of AB,

Direction ratios of BC= $\begin{array}{l}\left(-5-\left(-1\right)\right),\left(-5-1\right),\left(-2-2\right)\\ =\left(-4,-6,-4\right)\end{array}$

Direction cosine of BC =

 

 

 

Direction of CA= $\begin{array}{l}\left(-5-3\right)\left(-5-5\right)\left(-2-\left(-4\right)\right)\\ =\left(-8,-10,2\right)\end{array}$

Direction cosine of CA =

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Given,

A (2,3,4), B (-1, -2,1), C (5,8,7)

Direction ratio of AB= $\begin{array}{l}\left(-1-2\right),\left(-2-3\right),\left(1-4\right)\\ =\left(3,-5,-3\right)\end{array}$

Where, a₁=3, b₁=-5, c₁=-3

Direction ratio of BC= $\begin{array}{l}\left(5-\left(-1\right)\right),\left(8-\left(-2\right)\right),\left(7-1\right)\\ =\left(6,10,6\right)\end{array}$

Where, a₂=6, b₂=10, c₂=6

Now,

$\begin{array}{l}\frac{a_2}{a_1}=\frac{6}{-3}=-2\\ \frac{b_2}{b_1}=\frac{10}{-5}=-2\\ \frac{c_2}{c_1}=\frac{6}{-3}=-2\end{array}$

Here, direction ratio of two-line segments are proportional.

So, A, B, C are collinear.

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Direction cosine are 

$\begin{array}{l}\frac{-18}{},\frac{+12}{},\frac{-4}{}\\ =\frac{-18}{22},\frac{12}{22},\frac{-4}{22}\\ =\frac{-9}{11},\frac{6}{11},\frac{-2}{11}\end{array}$

26. Given f (x) = $\{\begin{array}{ll}k{x}^2\hfill & if x\le 2.\hfill \\ 3\hfill & if x>2.\hfill \end{array}$

For continuous at x = 2,

f (2) = k (2)2 = 4x.

L.H.L. = $\underset{x\to 2^{-}}{lim}f(x)=\underset{x\to 2^{-}}{lim}{x}^2=4x$

R.H.L. = $\underset{x\to 2^{+}}{lim}f(x)=\underset{x\to 2^{+}}{lim}3=3$

Then, L.H.L = R.H.L. = f (2)

i e, 4x = 3

$\Rightarrow x=\frac{3}{4}.$

Let the angles be α, β, r which are equal  

Let the direction cosines of the line be l, m, n.

$\begin{array}{l}l=cos\alpha ,m=cos\beta ,n=cosr\\ \alpha =\beta =r\\ l^2+m^2+n^2=1\\ co{s}^2\alpha +co{s}^2\beta +co{s}^{2}r=1\\ co{s}^2\alpha +co{s}^2\alpha +co{s}^2\alpha =1\\ 3co{s}^2\alpha =1\\ co{s}^2\alpha =\frac{1}{3}\\ cos\alpha =\frac{1}{}\end{array}$

16. Kindly go through the solution

No, JCAT is not compulsory to get admissions into the BA course at Jindal School of Design and Architecture. Because the institute also accepts the valid scores from other entrance exams too, such as UCEED/ SAT/ ACT/ NID/ NIFT. Therefore, JCAT is not compulsory to get admission into BA courses. However, it is one of the accepted entrance exams for admissions to BA courses. 

Let the direction cosine of the line be l, m, n.

Then,

$\begin{array}{l}l=90^{?}=cos90^{?}=0\\ m=cos135=\frac{-1}{}\\ n=cos45^{?}=\frac{1}{}\end{array}$