Ncert Solutions Maths class 12th

Let $A\left(-2,-21,29\right),B\left(-1,-16,23\right),P\left(\lambda ,2,1\right),Q\left(4,-2,2\right)$

Given $\overrightarrow{AB}\perp \overrightarrow{PQ}$

$\therefore \overrightarrow{AB}.\overrightarrow{PQ}=0$

$\left(\widehat{i}+5\widehat{j}-6\widehat{k}\right).\left(\left(4-\lambda \right)\widehat{i}-4\widehat{j}+\widehat{k}\right)=0$

$4-\lambda -20-6=0$

$\lambda =-22$

$\therefore {\left(\frac{\lambda }{11}\right)}^2+\left(\frac{-4\lambda }{11}\right)-4=4+8-4=8$

$\sqrt[]{3}co{s}^{2}x=\left(\sqrt[]{3}-1\right)cosx+1$

$\left(\sqrt[]{3}cosx+1\right)\left(cosx-1\right)=0$

$\therefore cosx=-\frac{1}{\sqrt[]{3}}\left(rejected\right)$

Hence, cos x = 1 x = 0 one solution

Information missing. The question was droppedby NTA.

 $y=\left|\left|x-1\right|-2\right|$

Area bounded region = $\frac{1}{2}\times 4\times 2=4$

R = { (P, Q) |P and Q are at the same distance from the origin}.

at $\left(1,-1\right),x^2+y^2={\left(1\right)}^2+{\left(-1\right)}^2=1+1=2$

$P_1:3x+15+21z=9$           

P₂ : x – 3y – z = 5

P₃ : 2x + 10y + 14z = 5

Ratio of the direction cosines of P₁ and P₂

$\frac{3}{2}=\frac{15}{10}=\frac{21}{14}$

Hence, P₁ and P₃ are parallel.

 $\overrightarrow{a}\times \left(\overrightarrow{a}\times \overrightarrow{b}\right)=\left(\overrightarrow{a}.\overrightarrow{b}\right)\overrightarrow{a}-{\left|\overrightarrow{a}\right|}^2\overrightarrow{b}=0-{\left|\overrightarrow{a}\right|}^2\overrightarrow{b}\left(as\overrightarrow{a}.\overrightarrow{b}=0given\right)$

$\overrightarrow{a}\times \left(\overrightarrow{a}\times \left(\overrightarrow{a}\times \overrightarrow{b}\right)\right)=-{\left|\overrightarrow{a}\right|}^2\overrightarrow{a}\times \overrightarrow{b}$

$\overrightarrow{a}\times \left(\overrightarrow{a}\times \left(\overrightarrow{a}\times \left(\overrightarrow{a}\times \overrightarrow{b}\right)\right)\right)=-{\left|\overrightarrow{a}\right|}^2\overrightarrow{a}\times \left(\overrightarrow{a}\times \overrightarrow{b}\right)=-{\left|\overrightarrow{a}\right|}^2\left(-{\left|\overrightarrow{a}\right|}^2\overrightarrow{b}\right)={\left|\overrightarrow{a}\right|}^4\overrightarrow{b}$

Let n be the number of times.

$p=\frac{1}{2},q=\frac{1}{2}$

According to question,

${\Rightarrow }^n{C}_7{=}^n{C}_9\Rightarrow n-7=9\Rightarrow n=16$

$p\left(x=2\right){=}^{16}{C}_2{\left(\frac{1}{2}\right)}^2.{\left(\frac{1}{2}\right)}^{14}=\frac{15}{2^{13}}$

All the points A (1, 5, 35), B (7, 5,5),  $C(1,\lambda ,7),D(2\lambda ,1,2)$ are coplanar. Hence

$\overrightarrow{AB}\times \overrightarrow{AC}.\overrightarrow{AD}=\left|\begin{array}{ccc}\hfill 6\hfill & \hfill 0\hfill & \hfill 30\hfill \\ \hfill 0\hfill & \hfill \lambda -5\hfill & \hfill 28\hfill \\ \hfill 2\lambda -1\hfill & \hfill -4\hfill & \hfill 33\hfill \end{array}\right|=0$

$\Rightarrow 5{\lambda }^2-44\lambda +95=0$

$\Rightarrow sumofroots=\frac{44}{5}$

$\underset{h\to 0}{Lt}\frac{2\times 2\left[\frac{\sqrt[]{3}}{2}sin\left(\frac{\pi }{6}+h\right)-\frac{1}{2}cos\left(\frac{\pi }{6}+h\right)\right]}{2\sqrt[]{3}h\left[\frac{\sqrt[]{3}}{2}cosh-\frac{1}{2}sinh\right]}$

$=\underset{h\to 0}{Lt}4\frac{sin\left(\frac{\pi }{6}+h-\frac{\pi }{6}\right)}{2\sqrt[]{3}hsin\left(\frac{\pi }{3}-h\right)}=\frac{2}{\sqrt[]{3}}\times 1\times \frac{1}{\frac{\sqrt[]{3}}{2}}=\frac{4}{3}$