Three Dimensional Geometry

$\frac{h-cos\theta }{2}=\frac{k-sin\theta }{3}=\frac{w-0}{1}$

$=\frac{-1\left(2cos\theta +3sin\theta -6\right)}{14}\Rightarrow h=\frac{cos-2\left(2cos\theta +3sin\theta -6\right)}{14}$

$k=\frac{5sin\theta -6cos\theta +18}{14}$

${\left(5h+6k-12\right)}^2+4{\left(3h+5k-9\right)}^2=1$

$\frac{{\left(x-3\right)}^2}{16}+\frac{{\left(y-4\right)}^2}{9}\le 1,x,y\in N,{\left(x-7\right)}^2+{\left(y-4\right)}^2\ge 36,x,y\in R$        

Total number of common point = 1 + 5 + 7 + 5 + 5 + 3 + 1 = 27

Plane through 4ax – y + 5z – 7a = 0 = 2x – 5y – z – 3

is $x\left(4a+2\lambda \right)+y\left(-1-5\lambda \right)+z\left(5-\lambda \right)=7a+3\lambda$

This plane contains 4, -1, 0

9a + 1 + 10 = 0…… (i)

Plane contains the line $\frac{x-4}{1}=\frac{y+1}{-2}=\frac{z}{1}$

$4a+11\lambda +7=0$ ……. (ii)

From (i) & (ii) a = 1,  $\lambda$ =1

Equation of plane $\pi \equiv x+2y+3z-2=0$

$\Rightarrow$$7$$P$$+$$3$$-$$2$$P$$+$$4$$-$$1$$2$$P$$+$$9$$-$$2$$=$$0$$\Rightarrow$$P$$=2$$\mathrm{}$

Line of shortest distance will be along ${\overline{b}}_1*{\overline{b}}_2$ ${\stackrel{}{}}_{}{\stackrel{}{}}_{}$

where $\begin{array}{l}{\stackrel{}{}}_{}\stackrel{}{}\stackrel{}{}\\ \end{array}$ $\begin{array}{l}{\overline{b}}_1=\widehat{j}+\widehat{k}\\ and\\ {\overline{b}}_2=2\widehat{i}+2\widehat{j}+\widehat{k}\\ {\overline{b}}_1*{\overline{b}}_2=\left|\begin{array}{ccc}\hfill \widehat{i}\hfill & \hfill \widehat{j}\hfill & \hfill \widehat{k}\hfill \\ \hfill 0\hfill & \hfill 1\hfill & \hfill 1\hfill \\ \hfill 2\hfill & \hfill 2\hfill & \hfill 1\hfill \end{array}\right|=-\widehat{i}+2\widehat{j}-2\widehat{k}\end{array}$

Angle between ${\overline{b}}_1*{\overline{b}}_2$ ${\stackrel{}{}}_{}{\stackrel{}{}}_{}$ and plane P,

$\Rightarrow a^2=-\frac{25}{11}\left(notpossible\right)$

$\frac{{\left(x-3\right)}^2}{16}+\frac{{\left(y-4\right)}^2}{9}\le 1,x,y\in N,{\left(x-7\right)}^2+{\left(y-4\right)}^2\ge 36,x,y\in R$

Total number of common point = 1 + 5 + 7 + 5 + 5 + 3 + 1 = 27

$RS\equiv \left(\alpha ,-,1\beta \right)$

$DRofPQ\equiv \left(\frac{56}{17}+2,\frac{43}{17}+1,\frac{111}{17}-1\right)$

$\equiv \left(\frac{90}{17},\frac{60}{17},\frac{94}{17}\right)$

$90\alpha +94\beta =60$

$\begin{array}{l}\beta =-15,\alpha =-15\\ {\alpha }^2+{\beta }^2=450\end{array}$

$\lambda =1+\mu$

$\begin{array}{l}\underset{\_}{2\lambda =3\lambda \mu }\\ \lambda =2\to S\left(2,4,6\right)\end{array}$

$\mu =1$

$\overrightarrow{n}=\left|\begin{array}{ccc}\hfill \widehat{i}\hfill & \hfill \widehat{j}\hfill & \hfill \widehat{k}\hfill \\ \hfill 1\hfill & \hfill 2\hfill & \hfill 3\hfill \\ \hfill 1\hfill & \hfill 1\hfill & \hfill 5\hfill \end{array}\right|=\left(7,-2,-1\right)$

7x – 2y – z + d = 0

(2, 4, 6) Þ d = -14 + 8 + 6

d = 0

7x – 2y – z = 0

7 – 2 = 5

$limPQ:\frac{x-1}{1}=\frac{y-0}{1}=\frac{3-1}{1}=\lambda$

$M\left(1+\lambda ,\lambda ,1+\lambda \right)$

x + y + z = 5

$\Rightarrow \lambda =1$

M (2, 1, 2)

Q (3, 2, 3)

$L:\frac{x-1}{1}=\frac{y+1}{1}=\frac{z+1}{1}=t$

R (3, 1, 1), QR² = 1 + 4 = 5

$\overrightarrow{a}=\left(t,t,t\right)$

$\frac{3t+4t}{5}=7\Rightarrow t=5$

$\overrightarrow{a}=\left(5,5,5\right)$

$\overrightarrow{b}=\mathcal{l}\overrightarrow{a}+m\widehat{i}$

$=\left(5\mathcal{l}+m,5\mathcal{l},5\mathcal{l}\right)$

$\frac{\overrightarrow{b}.\left(3,4,0\right)}{5}=\frac{\pm \frac{5}{\sqrt[]{2}}\left(-6+4+0\right)}{5}=\pm \left(\sqrt[]{2}\right)$

${\overrightarrow{b}}_1*{\overrightarrow{b}}_2=\left|\begin{array}{ccc}\hfill \widehat{i}\hfill & \hfill \widehat{j}\hfill & \hfill \widehat{k}\hfill \\ \hfill 1\hfill & \hfill -a\hfill & \hfill 0\hfill \\ \hfill 1\hfill & \hfill -1\hfill & \hfill 1\hfill \end{array}\right|$ = $-a\widehat{i}-\widehat{j}+\left(a-1\right)\widehat{k}$

${\overrightarrow{a}}_1-{\overrightarrow{a}}_2=-\widehat{i}+\widehat{j}+\widehat{k}$

Shortest distance = $\left|\frac{\left({\overrightarrow{a}}_1-{\overrightarrow{a}}_2\right).\left({\overrightarrow{b}}_1*{\overrightarrow{b}}_2\right)}{\left|{\overrightarrow{b}}_1*{\overrightarrow{b}}_2\right|}\right|$

$=\frac{2\left(a-1\right)}{\sqrt[]{a^2+1+{\left(a+1\right)}^2}}=\sqrt[]{\frac{2}{3}}=\frac{4{\left(a-1\right)}^2}{a^2+1+{\left(a-1\right)}^2}=\frac{2}{3}$

$12{\left(a-1\right)}^2=2\left(a^2+1\right)+2{\left(a-1\right)}^2$

$10{\left(a-1\right)}^2=2\left(a^2+1\right)$

$5a^2-10a+5=a^2+1\Rightarrow 4a^2-10a+4=0$

$2a^2-5a+2=0$

$\left(2a-1\right)\left(a-2\right)=0$

$\therefore a=\frac{1}{2},2$