Maths NCERT Exemplar Solutions Class 12th Chapter Three

ax + by + cz = d

2a + 3b – 5c = d               2a + 3b – 5c = d …………….(v)

Putting (i) & (iv) in (v) we get a = -9d.

In conditions

$\left(2\widehat{i}+\widehat{j}-5\widehat{k}\right),\left(a\widehat{i}+b\widehat{j}+c\widehat{k}\right)=0$

2b = d ………….(i) d > 0

2a + b – 5c = 0                  2a + 3b – 5c = d   $\left|a\right|,\left|b\right|,\left|c\right|,d\to g.c.d$           

=   $\frac{\alpha }{2}$

$\therefore \frac{\alpha }{2}=1\Rightarrow \alpha =2$

$\left(3\widehat{i}+5\widehat{j}-7\widehat{k}\right).\left(a\widehat{i}+b\widehat{j}+c\widehat{k}\right)=0$

3a + 5b – 7c = 0) *   $\frac{4}{7}...................\left(iii\right)$

$\therefore a=-18,b=1$

c = -7, d = 2

(iii)…(ii) ® 2a + 3b   $\frac{-33c}{7}=0$

2a + 3b =  $\frac{33}{7}c$ $c=\frac{-7}{2}d.................\left(iv\right)$

Putting values

a + 7b + c + 20d = 22

$a=\left|\frac{4.2-3\left(-1\right)-21}{5}\right|$

$=\left|\frac{8+3-21}{5}\right|=2$

$\therefore L=4a=8$

$\overrightarrow{a}=\alpha \widehat{i}+2\widehat{j}-\widehat{k}\&\overrightarrow{b}=-2\widehat{i}+\alpha \widehat{j}+\widehat{k}$

$\left|\overrightarrow{a}*\overrightarrow{b}\right|=\sqrt[]{15\left(a^2+4\right)}$

$2{\left|\overrightarrow{a}\right|}^2+\left(\overrightarrow{a}.\overrightarrow{b}\right){\left|\overrightarrow{b}\right|}^2=?$

$\overrightarrow{a}.\overrightarrow{b}=\left|\overrightarrow{a}\right|\left|\overrightarrow{b}\right|cos\theta$

$cos\theta =\frac{-1}{\left(a^2+5\right)}$ $\therefore \left|sin\theta \right|=\frac{\sqrt[]{{\left(a^2+5\right)}^2-1}}{\left(a^2+5\right)}$

$\left|\overrightarrow{a}*\overrightarrow{b}\right|=\left|\stackrel{?}{a}\right|*\left|\overrightarrow{b}\right|*\left|sin\theta \right|$

$=\left(a^5+5\right)*\frac{\sqrt[]{{\left(a^2+5\right)}^2-1}}{\left(a^2+5\right)}=\sqrt[]{15\left(a^2+4\right)}$

${\left(a^2+5\right)}^2-1=15\left(a^2+4\right)$

$\Rightarrow \left(\alpha =\pm 3\right)$

$2{\left|\overrightarrow{a}\right|}^2+\left(\overrightarrow{a}.\overrightarrow{b}\right){\left|\overrightarrow{b}\right|}^2$

= $2\left(a^2+5\right)-\left(a^2+5\right)$

$\left(a^2+5\right)=14$

$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$

$e=\sqrt[]{1+\frac{b^2}{a^2}}$ $e\text{'}=\sqrt[]{1+\frac{a^2}{b^2}}$

$\mathcal{l}=\frac{2b^2}{a}$ $\mathcal{l}\text{'}=\frac{2a^2}{b}$

$\left(1+\frac{b^2}{a^2}\right)=\frac{11}{14}*\frac{2b^2}{a}$ $\left(1+\frac{a^2}{b^2}\right)=\frac{11}{8}*\frac{2a^2}{b}$

$7*\left\{{\left(\frac{7b}{4}\right)}^2+b^2\right\}=11*b^2*\frac{7b}{4}\Rightarrow a*77=65$

65b² = 44b³

65 = b * 44

$\therefore 77a+44b=65*2=130$

$\frac{dy}{dx}+2ytanx=2sinx$  

  $I.F.=e^{2{\displaystyle \int tanxdx}}=se{c}^{2}x$             

$\therefore$  Solution y sec²x =   $2{\displaystyle \int sinxse{c}^{2}xdx=2{\displaystyle \int secxtanxdx}}$

y sec² x = 2sec x + c

y = 2 cos x + c cos²x passes $\left(\frac{\pi }{4},0\right)=B$     C =   $-2\sqrt[]{2}$

y = 2 cos x  $2\sqrt[]{2}co{s}^{2}x$ $\therefore {\displaystyle \underset{0}{\overset{\pi /2}{\int }}ydx}=2-\frac{\pi }{\sqrt[]{2}}$                              

$2y{e}^{x/y^2}dx+\left(y^2-4x{e}^{x/y^2}\right)dy=0$

  $\Rightarrow 2e^{2/y^2}\left(ydx-2xdy\right)+y^{2}dy=0$            

$2e^{x/y^2}d\left(\frac{x}{y^2}\right)+\frac{dy}{y}=0$

Integrating   $2e^{x/y^2}+Iny=c$

y = 1, n = 0  c = 2

  ^{$2e^{x/y^2}+Iny=2$}  

$\therefore y=e\Rightarrow 2e^{x/e^2}+1=2$

x = -e² In2

  $y=3?\left|x?\frac{1}{2}\right|=\left|x+1\right|$

Graph

Area =   $\left(\frac{1}{2}*\frac{3}{2}*\frac{3}{4}\right)*2+\frac{3}{2}*\frac{3}{2}=\frac{3}{2}*\frac{3}{2}*\frac{3}{2}=\frac{27}{8}$