Maths NCERT Exemplar Solutions Class 12th Chapter Ten

$(a+\sqrt[]{2}b\mathrm{c}\mathrm{o}\mathrm{s}?x)(a-\sqrt[]{2}b\mathrm{c}\mathrm{o}\mathrm{s}?y)=a^2-b^2$

$\Rightarrow a^2-\sqrt[]{2}ab\mathrm{c}\mathrm{o}\mathrm{s}?y+\sqrt[]{2}ab\mathrm{c}\mathrm{o}\mathrm{s}?x-2b^2\mathrm{c}\mathrm{o}\mathrm{s}?x\mathrm{c}\mathrm{o}\mathrm{s}?y=a^2-b^2$

Differentiating both sides:

$0-\sqrt[]{2}ab\left(-\mathrm{s}\mathrm{i}\mathrm{n}?y\frac{dy}{dx}\right)+\sqrt[]{2}ab(-\mathrm{s}\mathrm{i}\mathrm{n}?x)-2b^2\left[\mathrm{c}\mathrm{o}\mathrm{s}?x\left(-\mathrm{s}\mathrm{i}\mathrm{n}?y\frac{dy}{dx}\right)+\mathrm{c}\mathrm{o}\mathrm{s}?y(-\mathrm{s}\mathrm{i}\mathrm{n}?x)\right]=0$

At $\left(\frac{\pi }{4},\frac{\pi }{4}\right)$ :

$\begin{array}{rr}& ab\frac{dy}{dx}-ab-2b^2\left(-\frac{1}{2}\frac{dy}{dx}-\frac{1}{2}\right)=0\\ & \Rightarrow \frac{dx}{dy}=\frac{ab+b^2}{ab-b^2}=\frac{a+b}{a-b};a,b>0\end{array}$

$L_1:4x+3y+2=0$  

$L_2:3x-4y-11=0$               

Since circle C touches the line L2 at Q intersection point L1 and L2 is (1, -2)

P lies of L₁

$\therefore P\left(x,-\frac{1}{3}\left(2+4x\right)\right)$               

Now,

PQ = 5 ? (x – 1)² + ${\left(\frac{4x+2}{3}-2\right)}^2=25$  

$\Rightarrow x=4,-2$                     

$?$ The circle lies below the axis

y = -6

p (4, -6)

Now distance of P from 5x – 12 y + 51 = 0

$=\left|\frac{20+72+51}{13}\right|=\frac{143}{13}=11$                

Be the vectors along the diagonals of a parallelogram having are 2.

$\therefore \frac{1}{2}\left|\overrightarrow{a}*\overrightarrow{b}\right|=2\sqrt[]{2}$

$\left|\overrightarrow{a}\right|\left|\overrightarrow{b}\right|sin\theta =4\sqrt[]{2}$

$\Rightarrow \left|\overrightarrow{b}\right|sin\theta =4\sqrt[]{2}$ ……. (i)

And

$\Rightarrow \overrightarrow{c}.\overrightarrow{b}=-2{\left|\overrightarrow{b}\right|}^2=-128......(ii)$

$\Rightarrow \left|\overrightarrow{c}\right|=16\sqrt[]{2}.......(iii)$

From (ii) and (iii)

$\left|\overrightarrow{c}\right|\left|\overrightarrow{b}\right|cos\alpha =-128$

$\Rightarrow cos\alpha =\frac{1}{\sqrt[]{2}}$

$\alpha =\frac{3\pi }{4}$

This is a True or False type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l}If\overrightarrow{a}.\overrightarrow{b}=0then\overrightarrow{a}.\overrightarrow{b}=\left|\overrightarrow{a}\right|\left|\overrightarrow{b}\right|cos90^0\\ Sotheanglebetweentheadjacentsidesoftherhombusshouldbe90^{0}whichisnotpossible.\\ HencethegivenstatementisFalse.\end{array}$

This is a True or False type Questions as classified in NCERT Exemplar

Sol:

       $\begin{array}{l}{\left(\overrightarrow{a}+\overrightarrow{b}\right)}^2=\left(\overrightarrow{a}+\overrightarrow{b}\right).\left(\overrightarrow{a}+\overrightarrow{b}\right)\\ ={\left|\overrightarrow{a}\right|}^2+{\left|\overrightarrow{b}\right|}^2+2\overrightarrow{a}.\overrightarrow{b}\\ HencethegivenstatementisFalse.\end{array}$

This is a True or False type Questions as classified in NCERT Exemplar

Sol: 

$\begin{array}{l}Giventhat\left|\overrightarrow{a}+\overrightarrow{b}\right|=\left|\overrightarrow{a}-\overrightarrow{b}\right|\\ Squaringbothsides,weget\\ {\left|\overrightarrow{a}+\overrightarrow{b}\right|}^2={\left|\overrightarrow{a}-\overrightarrow{b}\right|}^2\\ \Rightarrow {\left|\overrightarrow{a}\right|}^2+{\left|\overrightarrow{b}\right|}^2+2\overrightarrow{a}.\overrightarrow{b}={\left|\overrightarrow{a}\right|}^2+{\left|\overrightarrow{b}\right|}^2-2\overrightarrow{a}.\overrightarrow{b}\\ \Rightarrow 2\overrightarrow{a}.\overrightarrow{b}=-2\overrightarrow{a}.\overrightarrow{b}\\ \Rightarrow \overrightarrow{a}.\overrightarrow{b}=-\overrightarrow{a}.\overrightarrow{b}\\ \Rightarrow 2\overrightarrow{a}.\overrightarrow{b}=0\Rightarrow \overrightarrow{a}.\overrightarrow{b}=0\\ whichimpliesthat\overrightarrow{a}and\overrightarrow{b}areorthogonal.\\ HencethegivenstatementisTrue.\end{array}$

This is a True or False type Questions as classified in NCERT Exemplar

Sol:  $True.$

This is a True or False type Questions as classified in NCERT Exemplar

Sol:         

$\begin{array}{l}If\left|\overrightarrow{a}\right|=\left|\overrightarrow{b}\right|then\overrightarrow{a}=\pm \overrightarrow{b}whichistrue.\\ HencethegivenstatementisTrue.\end{array}$

This is a Fill in the blanks type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l}Let\overrightarrow{a}=a_1\widehat{i}+a_2\widehat{j}+a_3\widehat{k}\\ \therefore \overrightarrow{a}.\widehat{i}=\left(a_1\widehat{i}+a_2\widehat{j}+a_3\widehat{k}\right).\widehat{i}\\ =a_1\\ Similarly,\overrightarrow{a}.\widehat{j}=a_{2}and\overrightarrow{a}.\widehat{k}=a_3\\ \therefore \left(\overrightarrow{a}.\widehat{i}\right).\widehat{i}+\left(\overrightarrow{a}.\widehat{j}\right).\widehat{j}+\left(\overrightarrow{a}.\widehat{k}\right).\widehat{k}=a_1\widehat{i}+a_2\widehat{j}+a_3\widehat{k}=\overrightarrow{a}\\ Hence,thevalueofthefilleris\overrightarrow{a}.\end{array}$