Class 11th

Mid point of BC is $\frac{1}{2}\left(5\widehat{i}+\left(\alpha -2\right)\widehat{j}+9\widehat{k}\right)$ $\frac{}{}\stackrel{}{}\stackrel{}{}\stackrel{}{}$

$\overline{AB}=\widehat{i}+\left(\alpha -4\right)\widehat{j}+\widehat{k}$

$\overline{AC}=\widehat{i}+\left(-2-\alpha \right)\widehat{j}+\widehat{k}$

For = 1, $\overline{AB}$ $\stackrel{}{}$ and $\overline{AC}$ $\stackrel{}{}$ will be collinear. So for non collinearity

= 2

Equation of perpendicular bisector of AB is

$y-\frac{3}{2}=-\frac{1}{5}\left(x-\frac{5}{2}\right)\Rightarrow x+5y=10$

Solving it with equation of given circle

${\left(x-5\right)}^2+{\left(\frac{10-x}{5}-1\right)}^2=\frac{13}{2}$

$\Rightarrow x-5=\pm \frac{5}{2}\Rightarrow x=\frac{5}{2}or\frac{15}{2}$

But $x\ne \frac{5}{2}$

because AB is not the diameter.

So, centre will be

$\left(\frac{15}{2},\frac{1}{2}\right)$

Now,

$r^2={\left(\frac{15}{2}-2\right)}^2+{\left(\frac{1}{2}+1\right)}^2=\frac{65}{2}$

${\left(x-\frac{1}{\sqrt[]{2}}\right)}^2+{\left(y-\frac{1}{\sqrt[]{2}}\right)}^2=1$

here AB = $\sqrt[]{2}$ $\text{exception 25:}$ , BC = 2, AC = $\sqrt[]{2}$ $\text{exception 25:}$

area = $\frac{1}{2}*\sqrt[]{2}*\sqrt[]{2}=1$

It does not contains 2^nd most abundant element by weight in earth crust because that is Si Calgon -> Na₂ [Na₄ (PO₃)₆]

$\underset{}{\overset{Watersoluble}{\to }}2N{a}^{+}{\left[N{a}_4{\left(P{O}_3\right)}_6\right]}^{2-}$

$\underset{}{\overset{C{a}^{2+}}{\to }}2N{a}^{+}{\left[N{a}_{2}Ca{\left(P{O}_3\right)}_6\right]}^{2-}$

Tangents making angle $\frac{\pi }{4}$ $\frac{}{}$ with y = 3x + 5.

$tan\frac{\pi }{4}=\left|\frac{m-3}{1+3m}\right|\Rightarrow m=-2,\frac{1}{2}$

So, these tangents are $\perp$ $$ . So ASB is a focal chord.

RM = $\left|\frac{3+7-5}{\sqrt[]{2}}\right|=\frac{5}{\sqrt[]{2}}$

$\mathcal{l}sin60°=\frac{5}{\sqrt[]{2}}\Rightarrow \mathcal{l}=5\sqrt[]{\frac{2}{3}}$

$\therefore Areaof\Delta PQR=\frac{\sqrt[]{3}}{4}{\mathcal{l}}^2=\frac{25}{2\sqrt[]{3}}$

Tritium is radioactive and it decays into He³ during emission of b-radiation

Let S = 1 $+\frac{5}{6}+\frac{12}{6^2}+\frac{22}{6^3}+....$ $\frac{}{}\frac{}{{}^{}}\frac{}{{}^{}}$

$\underset{\_}{-\frac{S}{6}=\frac{1}{6}+\frac{5}{6^2}+....}$

$\frac{5S}{6}=1+\frac{4}{6}+\frac{7}{6^2}+\frac{10}{6^3}+....$

$\underset{\_}{-\frac{5S}{36}=\frac{1}{6}+\frac{4}{6^2}+\frac{7}{6^3}+.....}$

$\frac{25S}{36}=1+\frac{3}{6}+\frac{3}{6^2}+\frac{3}{6^3}+......$

$\frac{25S}{36}=1+\frac{3/6}{1-1/6}$

$\frac{25S}{36}=1+\frac{3/5}{1}$

$S=\frac{288}{125}$