Three Dimensional Geometry

${\pi }_1\equiv 2x+ky-5z=1$  

${\pi }_2\equiv 3kx-ky+z=5$                

Given  ${\pi }_1\perp {\pi }_2\Rightarrow 6k-k^2-5=0$  

k = 1, 5

Now required plane is p₁ +   $\lambda {\pi }_2=0$

y intercept = $\frac{1+5\lambda }{1-\lambda }$  

  $\overrightarrow{a}*\overrightarrow{b}=\left|\begin{array}{ccc}\hfill \widehat{i}\hfill & \hfill \widehat{j}\hfill & \hfill \widehat{k}\hfill \\ \hfill \alpha \hfill & \hfill 1\hfill & \hfill \beta \hfill \\ \hfill 3\hfill & \hfill -5\hfill & \hfill 4\hfill \end{array}\right|=\left(4+5\beta \right)\widehat{i}+\left(3\beta -4\alpha \right)\widehat{j}+\left(-5\alpha -3\right)\widehat{k}$                

Compare with  $\overrightarrow{a}*\overrightarrow{b}=-\widehat{i}+9\widehat{j}+12\widehat{k}$  

b = -1, a = -3

Projection of  $\overrightarrow{b}-2\overrightarrow{a}on\overrightarrow{b}+\overrightarrow{a}$  

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

  $\frac{x-a}{3}=\frac{y-b}{-4}=\frac{z-c}{12}=\frac{-2\left(3a-4b+12c+19\right)}{3^2+{\left(-4\right)}^2+12^2}$

$\frac{x-a}{3}=\frac{y-b}{-4}=\frac{z-c}{12}=\frac{-6a+8b-24c-38}{169}$               

$\left(x,y,z\right)\equiv \left(a-6,\beta ,\gamma \right)$               

  $\frac{\left(a-b\right)-a}{3}=\frac{\beta -b}{-4}=\frac{\gamma -c}{12}=\frac{-6a+8b-24c-38}{169}$              

  $\frac{\beta -b}{-4}=-2$              

$\Rightarrow \beta =8+b$               

$\Rightarrow 3a-4b+12c=150$                   ….(i)

a + b + c = 5

$\Rightarrow 3a+3b+3c=15$ ….(ii)

Applying (i) – (ii), we get :

= 56 + 216 + 7b – 9c = 56 + 216 – 135 = 137

 l + m – n = 0 Þ n = l + m

$3l^2+m^2+cnl=0$  

$3l^2+m^2+cl\left(l+m\right)=0$    

$=\left(3+c\right){\left(\frac{l}{m}\right)}^2+c\left(\frac{l}{m}\right)+1=0$    

$?$    Lines are parallel

D = 0

$c=4$     (as c > 0)

 $A\left(-7\widehat{i}+6\widehat{j}\right)C\left(7\widehat{i}+2\widehat{j}+6\widehat{k}\right)$

$\overrightarrow{b}=-6\widehat{i}+7\widehat{j}+\widehat{k}$ $\overrightarrow{d}=-2\widehat{i}+\widehat{j}+\widehat{k}$

$\overrightarrow{b}*\overrightarrow{d}=\left|\begin{array}{ccc}\hfill \widehat{i}\hfill & \hfill \widehat{j}\hfill & \hfill \widehat{k}\hfill \\ \hfill -6\hfill & \hfill 7\hfill & \hfill 1\hfill \\ \hfill -2\hfill & \hfill 1\hfill & \hfill 1\hfill \end{array}\right|=3\widehat{i}+2\widehat{j}+4\widehat{k}$

the shortest distance between the lines

$=$$\left|\frac{42-8+24}{\text{exception 25:}}\right|$$=$$\frac{58}{\text{exception 25:}}$$=$$2$

First plane, P₁ = 2x – 2y + z = 0,

normal vector $\equiv n_1=\left(2,-2,1\right)$

Second plane, P₂ = x – y + 2z = 4,

normal vector $\equiv n_2=\left(1,-1,2\right)$

Plane perpendicular to P₁ and P₂ will have normal vector n₃ where n₃ = (n₁ * n₂)

Hence,

n₃ = (3, 0)

Distance PQ

$=\sqrt[]{21}\Rightarrow {\left(PQ\right)}^2=21$

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

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

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

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

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

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,

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