Three Dimensional Geometry

$P_1:3x+15+21z=9$           

P₂ : x – 3y – z = 5

P₃ : 2x + 10y + 14z = 5

Ratio of the direction cosines of P₁ and P₂

$\frac{3}{2}=\frac{15}{10}=\frac{21}{14}$

Hence, P₁ and P₃ are parallel.

All the points A (1, 5, 35), B (7, 5,5),  $C(1,\lambda ,7),D(2\lambda ,1,2)$ are coplanar. Hence

$\overrightarrow{AB}*\overrightarrow{AC}.\overrightarrow{AD}=\left|\begin{array}{ccc}\hfill 6\hfill & \hfill 0\hfill & \hfill 30\hfill \\ \hfill 0\hfill & \hfill \lambda -5\hfill & \hfill 28\hfill \\ \hfill 2\lambda -1\hfill & \hfill -4\hfill & \hfill 33\hfill \end{array}\right|=0$

$\Rightarrow 5{\lambda }^2-44\lambda +95=0$

$\Rightarrow sumofroots=\frac{44}{5}$

${\mathcal{l}}_1:\frac{x-3}{1}=\frac{y+1}{2}=\frac{z-4}{2}=t$

${\mathcal{l}}_2:\frac{x-3}{2}=\frac{y-3}{2}=\frac{z-2}{1}=s$

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

$\mathcal{l}:\overrightarrow{r}=\overrightarrow{0}+\lambda \left(-2\widehat{i}+3\widehat{j}-2\widehat{k}\right)$

$\mathcal{l}\&{\mathcal{l}}_1\to$

Point of intersection $\mathcal{l}\&{\mathcal{l}}_1$ is P (2, 3, 2)

Point Q on ${\mathcal{l}}_2$ is (3 + 2s, 3 + 2s, 2 + s).

  $\frac{a+20}{2}=3+7r$

a + 20 = 6 + 14r     . (i)

b = -2 + 10r           . (ii)

a = 18r – 2              . (iii)

Solving (i) and (iii) we get

20 + 18r – 2 = 6 + 14r

r = -3

$\therefore$   a = 14 + 14 (-3) = -56 and b = -2 -30 = -32

$\left|a+b\right|\left|-56-32\right|=88$

$\overrightarrow{O}P=\left(2,-1,1\right)$

Normal vector to the plane

$=\overrightarrow{A}B*A\overrightarrow{C}=\left|\begin{array}{ccc}\hfill \widehat{i}\hfill & \hfill \widehat{j}\hfill & \hfill \widehat{k}\hfill \\ \hfill 1\hfill & \hfill 1\hfill & \hfill -2\hfill \\ \hfill 1\hfill & \hfill 2\hfill & \hfill -1\hfill \end{array}\right|=\left(3,-1,1\right)=\overrightarrow{n}$

Projection of $\overrightarrow{O}Pon\overrightarrow{n}is\left|\frac{\overrightarrow{O}P.\overrightarrow{n}}{\left|\overrightarrow{n}\right|}\right|$

PN = $\frac{6+1+1}{\sqrt[]{11}}=\frac{8}{\text{exception 25:}}$

$\frac{\text{exception 25:}}{}$Projection of OP on plane = ON =  $\sqrt[]{O{P}^2-P{N}^2}=\sqrt[]{6-\frac{64}{11}}=\sqrt[]{\frac{2}{11}}$

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

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