Three Dimensional Geometry

$\overrightarrow{r}.\left(\widehat{i}-\widehat{j}+2\widehat{k}\right)=2$

&  $\overrightarrow{r}.\left(2\widehat{i}+\widehat{j}-\widehat{k}\right)=2$ are two planes the direction ratio of the line of intersection of then is collinear to   

$\left|\begin{array}{ccc}\hfill \widehat{i}\hfill & \hfill \widehat{j}\hfill & \hfill \widehat{k}\hfill \\ \hfill 1\hfill & \hfill -1\hfill & \hfill 2\hfill \\ \hfill 2\hfill & \hfill 1\hfill & \hfill -1\hfill \end{array}\right|=-\widehat{i}+5\widehat{j}+3\widehat{k}$

Any point on the line in given by x – y = 2

& 2x + y = 2

$\Rightarrow x=\raisebox{1ex}{$4$}\!\left/ \!\raisebox{-1ex}{$3$}\right.,y=\raisebox{1ex}{$-2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.,z=0$

$\therefore equationoflineL:\frac{x-\raisebox{1ex}{$4$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}{-1}=\frac{y+\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}{5}=\frac{z}{3}=r$

$\therefore pointP\left(\frac{33}{35},\frac{45}{35},\frac{41}{35}\right)\equiv \left(\alpha ,\beta ,\gamma \right)$

$\therefore 35\left(\alpha +\beta +\gamma \right)=119$                                           

$\frac{x-1}{2}=\frac{y-2}{1}=\frac{z-1}{3}.......\left(i\right)$

$\frac{x-2}{1}=\frac{y-\lambda }{2}=\frac{z-3}{4}......\left(ii\right)$

$\overrightarrow{OA}=\left(i+2j+k\right)\overrightarrow{OB}=\left(2\widehat{i}+\lambda \widehat{j}+3\widehat{k}\right)$

$\overrightarrow{a}=2i+j+3k\&\overrightarrow{b}=i+2j+4k$

$\therefore$ shortest distance = $\left|\frac{\left(\overrightarrow{OB}-\overrightarrow{OA}\right).\left(\overrightarrow{a}*\overrightarrow{b}\right)}{\overrightarrow{a}*\overrightarrow{b}}\right|=\frac{1}{\sqrt[]{38}}$

$\Rightarrow \left|14-5\lambda \right|=1$           

$\Rightarrow \lambda =3\&\frac{13}{5}$          

Integer value of $\lambda$ = 3

 Let the smallest angle is C =?

Therefore angle A = 90° -?

And angle B = 90°

i.e. b > a > c

Here, a = 2 R cos? , b = 2R, c = 2R sin?

b² = a² + c²

according to question,

$\begin{array}{l}\frac{1}{c^2}=\frac{1}{a^2}+\frac{1}{b^2}\\ ?\frac{a^2{b}^2}{c^2}=a^2+b^2\end{array}$     

=> $sin?=?\frac{\sqrt[]{5}?1}{2}$

(2, -2) lie in a plane

=>2 + 6 + 4 + β = 0

=> b = -12

Line is perpendicular to normal

α (1) – 5 (3) + 2 (-2) = 0

α = 19

α + β = 7

$\Delta =0\Rightarrow a=3,4$  

 For a = 3, no solution

For a = 4, no solution

n (S₁) = 2, n (S₂) = 0,

Equation of intersection of line

$\frac{x-0}{1}=y=\frac{z-0}{-1}$

Let $\overrightarrow{r}=\widehat{i}-\widehat{k}$

Direction ratio of $\overrightarrow{PQ}$

$\Rightarrow \left(\lambda +1\right)\left(1\right)+\left(-2\right)\left(0\right)+\left(2-\lambda \right)\left(-1\right)=0$

$\lambda =\frac{1}{2}\Rightarrow Q\left(\frac{1}{2},0,\frac{-1}{2}\right)$

$PQ=\frac{\sqrt[]{34}}{2}$

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

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

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

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

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

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

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 =  $\sqrt[]{16+36+9}=\sqrt[]{61}$

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