Maths

$L_1:\mathcal{l}x-y+3\left(1-z\right)=1,x+2y-z=2$

plane containing the line P : 3x – 8y + 7z = 4

If $\overrightarrow{n}$ be vector parallel to L.

then $\overrightarrow{n}=\left|\begin{array}{ccc}\hfill \widehat{i}\hfill & \hfill \widehat{j}\hfill & \hfill \widehat{k}\hfill \\ \hfill \mathcal{l}\hfill & \hfill -1\hfill & \hfill 3\left(1-\mathcal{l}\right)\hfill \\ \hfill 1\hfill & \hfill 2\hfill & \hfill -1\hfill \end{array}\right|=\left(6\mathcal{l}-5\right)\widehat{i}+\left(3-2\mathcal{l}\right)\widehat{j}+\left(2\mathcal{l}+1\right)\widehat{k}$ as P containing the line

$\therefore 3\left(6\mathcal{l}-5\right)-8\left(3-2\mathcal{l}\right)+7\left(2\mathcal{l}+1\right)=0$

$\Rightarrow \mathcal{l}=\frac{2}{3}$

If be the acute angle between line L & Y axis then cos = $\frac{5/3}{\sqrt[]{1+\frac{25}{9}+\frac{49}{9}}}=\frac{5}{\sqrt[]{83}}$

$\therefore 415co{s}^2\theta =125$

 $tan\left(2ta{n}^{-1}\frac{1}{5}+se{c}^{-1}\frac{\sqrt[]{5}}{2}+2ta{n}^{-1}\frac{1}{8}\right)tan\left(2ta{n}^{-1}\frac{\frac{1}{5}+\frac{1}{8}}{1-\frac{1}{5}*\frac{1}{8}}+se{c}^{-1}\frac{\sqrt[]{5}}{2}\right)$

$=tan\left(ta{n}^{-1}\frac{3}{4}+ta{n}^{-1}\frac{1}{2}\right)=tan\left(ta{n}^{-1}\frac{\frac{3}{4}+\frac{1}{2}}{1-\frac{3}{8}}\right)$

$=tan\left(ta{n}^{-1}\frac{\frac{5}{4}}{\frac{5}{8}}\right)=2$

 $\overline{x}=\frac{\sum xi}{40}=30\Rightarrow \sum xi=1200$ ………. (i)

${\alpha }^2=\frac{1}{40}\sum x{i}_i^2-{\left(30\right)}^2=25$

$\Rightarrow \sum x_i^2=37000$

after omitting two wrong observations

$\sum y_i^2=37000-144-100=36756$

$\therefore 38a^2=36756-36158=238$

 $S=\left\{\theta \in \left[0,2\pi \right]:8^{2si{n}^{2}x}+8^{2co{s}^{2}x}=16\right\}$

Now apply AM $\ge GM$ for $\frac{8^{2si{n}^{2}x}+8^{2co{s}^{2}x}}{2}\le {\left(8^{2si{n}^{2}x+2co{s}^{2}x}\right)}^{\frac{1}{2}}\Rightarrow 8^{2si{n}^{2}x}=8^{2co{s}^{2}x}$

$si{n}^2\theta =co{s}^2\theta$ $\therefore \theta =\frac{\pi }{4},\frac{3\pi }{4},\frac{5\pi }{4},\frac{7\pi }{4}$

$=4+\left[cosec\left(\frac{\pi }{2}+\pi \right)+cosec\left(\frac{\pi }{2}+3\pi \right)+cosec\left(\frac{\pi }{2}+5\pi \right)+cosec\left(\frac{\pi }{2}+7\pi \right)\right]$

$=4-2\left(4\right)=-4$

$2si{n}^2\theta =cos2\theta =1-2si{n}^2\theta$

$\Rightarrow 4si{n}^2\theta =1\Rightarrow sin\theta =\pm \frac{1}{2}$

$\frac{\pi }{6},\frac{5\pi }{6},\frac{7\pi }{6},\frac{11\pi }{6}$

$Sum\frac{7\pi }{6}+\frac{11\pi }{6}=3\pi \Rightarrow k=3$

 ${\displaystyle \sum _{k=1}^{10}\frac{k}{k^4+k^2+1}}$

$=\frac{1}{2}{\displaystyle \sum _{k=1}^{10}\left[\frac{1}{k^2-k+1}-\frac{1}{k^2+k+1}\right]}$

$=\frac{1}{2}\left[1-\frac{1}{111}\right]=\frac{110}{222}=\frac{55}{111}=\frac{m}{n}$

$\therefore m+n=166$

Last two digit must be in form

$\begin{array}{l}23,4,516\\ 324\\ 32\\ 52\end{array}\}3*4=12$

Total number of required number = 12 + 18 = 30

 $0\le \frac{2+3p}{6}\le 1\Rightarrow p\in \left[-\frac{2}{3},\frac{4}{3}\right],0\le \frac{2-p}{8}\le 1\Rightarrow p\in \left[-6,2\right]and0\le \frac{1-p}{2}\le 1\Rightarrow p\in \left[-1,1\right]$

$0<P\left(E_1\right)+P\left(E_2\right)+P\left(E_3\right)\le 10\le \frac{13}{12}-\frac{p}{8}\le 1\Rightarrow p\in \left[\frac{2}{3},\frac{26}{3}\right]$ Taking intersection to all $p\in \left[\frac{2}{3},1\right]$

$\therefore p_1+p_2=\frac{5}{3}$

The normal vector to the plane is $\overline{n_1}*\overline{n_2}=\left|\begin{array}{ccc}\hfill i\hfill & \hfill 3\hfill & \hfill k\hfill \\ \hfill 1\hfill & \hfill a\hfill & \hfill -1\hfill \\ \hfill -1\hfill & \hfill 1\hfill & \hfill -a\hfill \end{array}\right|=\left(1-a\right)\widehat{i}+\widehat{j}+\widehat{k}$

$\therefore equationofplaneis\left(1-a\right)\left(x-1\right)+\left(y-1\right)+z=0$

(1 – a)x + y + z = 2 – a …… (i)

Now distance from (2, 1, 4) = $\sqrt[]{3}$

$\Rightarrow \sqrt[]{3}=\left|\frac{2\left(1-a\right)+1+4-\left(2-a\right)}{\sqrt[]{{\left(1-a\right)}^2+1+1}}\right|$

$\Rightarrow a^2+2a-8=0\Rightarrow a=-4,2$ the largest value of a = 2.