Ncert Solutions Maths class 12th

$\overrightarrow{a}*\left[\left(\overrightarrow{r}-\overrightarrow{b}\right)*\overrightarrow{a}\right]+\overrightarrow{b}*\left[\left(\overrightarrow{r}-\overrightarrow{c}\right)*\overrightarrow{b}\right]+\overrightarrow{c}*\left[\left(\overrightarrow{r}-\overrightarrow{a}\right)*\overrightarrow{c}\right]=\overrightarrow{0}$

As ${\left|\overrightarrow{a}\right|}^2={\left|\overrightarrow{b}\right|}^2={\left|\overrightarrow{c}\right|}^2$

$={\left|\overrightarrow{a}\right|}^2\left(3\overrightarrow{r}-\left(\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}\right)\right)-\left(\left(\overrightarrow{a}.\overrightarrow{r}\right)\overrightarrow{a}+\left(\overrightarrow{a}.\overrightarrow{r}\right)\overrightarrow{b}+\left(\overrightarrow{c}.\overrightarrow{r}\right)\overrightarrow{c}\right)$

Let $\overrightarrow{r}=\frac{\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}}{2}$

line: 3y – 2z – 1 = 0, 3x – z + 4 = 0

$\Rightarrow apointP\left(\frac{t-4}{3},\frac{2t+1}{3},t\right)$  on the line, Q (2, -1, 6)

$P{Q}^2=\frac{2}{9}\left(7t^2-56t+220\right)$

${\left(PQ\right)}_{min}=2\sqrt[]{6}$  

$dist.=\frac{{\overrightarrow{A}}_1{A}_2.\left({\overrightarrow{b}}_1*{\overrightarrow{b}}_2\right)}{\left|{\overrightarrow{b}}_1*{\overrightarrow{b}}_2\right|}$

$=\frac{\left(-4-\alpha ,-2,-3\right).\left(2,2,1\right)}{3}$

$\left|-5-\frac{2\alpha }{3}\right|=9\Rightarrow -5-\frac{2\alpha }{3}=\pm 9$

$\Rightarrow \frac{2\alpha }{3}=4,-14,$

but a > 0

a = 6

For the plane P,

$\overrightarrow{n}=\frac{1}{2}\left(-2\widehat{j}\right)*\left(-\widehat{i}+\widehat{j}-3\widehat{k}\right)=\widehat{j}*\widehat{i}+3\widehat{j}*\widehat{k}=-\widehat{k}+3\widehat{i}$

$\overrightarrow{a}.\overrightarrow{n}=0\Rightarrow 3\alpha -\gamma =0.......\left(i\right)$

$\overrightarrow{a}.\left(1,2,3\right)=0\Rightarrow \alpha +2\beta +3\gamma =0..........\left(ii\right)$

$\overrightarrow{a}.\left(1,1,2\right)=2\Rightarrow \alpha +\beta +2\gamma =2...........\left(iii\right)$

(i), (ii) & (iii) Þ a = 1, β = -5, $\gamma =$ 3

$T_{r+1}{=}^{120}{C}_r{4}^{\frac{120-r}{4}}.5^{r/6}$

For to be rational

r should be multiple of 6

Number of such terms = 21

$\left|\overrightarrow{a}\right|=\left|\overrightarrow{b}\right|=\left|\overrightarrow{c}\right|=\mathcal{l}$

$\overrightarrow{a}.\overrightarrow{b}=\overrightarrow{b}.\overrightarrow{c}=\overrightarrow{c}.\overrightarrow{a}=0$

${\left|\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}\right|}^2=3{\mathcal{l}}^2$

$\Rightarrow {\mathcal{l}}^2=\sqrt[]{3}{\mathcal{l}}^{2}cos\theta \Rightarrow cos\theta =\frac{1}{\sqrt[]{3}}$

36 cos² 2q = 36 ${\left(\frac{2}{3}-1\right)}^2=4$

P (E₁)= 0.9 $P\left({\overline{E}}_t\right)=0.1andP\left(E_2\right)=0.8,P\left({\overline{E}}_2\right)=0.2$

$\therefore RequiredprobabilityP=\frac{0.8*0.1}{0.8*0.1+0.2*0.1+0.9*0.2}=\frac{0.8}{2.8}=\frac{2}{7}$

$\therefore 98P=28$

Let zc be the centre of the circle

$\frac{z_c-2}{z_c+2}=i\Rightarrow z_c=2i$

Radius of the circle =  $\sqrt[]{4+4}=2\sqrt[]{2}$

Square off minimum distance AB =  ${\left(7\sqrt[]{2}\right)}^2=98$

$?\frac{x-1}{2}=\frac{y-2}{3}=\frac{z+1}{6}=r\left(say\right)$

Let P (1 + 2r, 2 + 3r, 1 + 6r) lies on the plane 2x – y + z = 6

$\therefore r=1\Rightarrow P\left(3,5,5\right)$

Distance between P and Q is PQ =  $\sqrt[]{16+36+9}=\sqrt[]{61}$

$\therefore P{Q}^2=61$

$f\left(x\right)=x^2+ax+1\Rightarrow f\text{'}\left(x\right)=2x+a$

for increasing   $f\text{'}\left(x\right)\ge 0$

$\therefore 2x+a\ge 0\Rightarrow a\ge -2x\Rightarrow a\ge -2*2\Rightarrow a\ge -4\therefore R=-4$

And for decreasing   $f\text{'}\left(x\right)\le 0\Rightarrow a\le -2x\Rightarrow a\le -2*1\therefore S=-2$

|R – S| = 2