Three Dimensional Geometry

  $\left(x+3y-z-5\right)+\lambda \left(2x-y+z-3\right)=0$

$\Rightarrow \left(1+2\lambda \right)x+\left(3-\lambda \right)y+\left(\lambda -1\right)z-5-3\lambda =0$                

Point $\left(2,1,-2\right)\Rightarrow \left(1+2\lambda \right)\left(2\right)+\left(3-\lambda \right)\left(1\right)+\left(\lambda -1\right)\left(-2\right)-5-3\lambda =0$  

$-2\lambda +2=0\Rightarrow \lambda =1$                

 P : 3x + 2y + 0.z = 8

X (1, -2, 4)

Y (5, -1, 2)

X + Y = (6, -3, 6)

Y – X = (4, 1, -2)

Let a, b, c be direction ratios of plane containing lines

$\frac{x}{2}=\frac{y}{3}=\frac{z}{5}$

and

$\frac{x}{3}=\frac{y}{7}=\frac{z}{8}$

Equation of plane P is : 1 (x – 3) 1 (y + 4) + 2 (z – 7) = 0

$\Rightarrow x-y+2z-21=0$

Distance from point (2, 5, 11) is

$d=\frac{\left|2+5+22-2\right|}{\sqrt[]{6}}$

$\therefore d^2=\frac{32}{3}$

Radius of circle S touching x-axis and centre  $\left(\alpha ,\beta \right)$ $$ is |. According to given conditions

${\alpha }^2+{\left(\beta -1\right)}^2={\left(\left|\beta \right|+1\right)}^2$

${\alpha }^2=4\beta as\beta >0$

$$ $\therefore$  Required locus is L : x2 = 4y

The area of shaded region = $2{\displaystyle {\int }_0^{4}2\sqrt[]{y}dy}$ ${\displaystyle {}_{}^{}\text{exception 25:}}$

= $4.{\left[\frac{\frac{y^{\frac{3}{2}}}{3}}{2}\right]}_0^4$ ${\frac{\frac{{}^{\frac{}{}}}{}}{}}_{}^{}$

= $\frac{64}{3}$ $\frac{}{}$ square units.

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 + 10l = 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 $\pi \equiv x+2y+3z-2=0$  

$\Rightarrow 7P+3-2P+4-12P+9-2=0\Rightarrow P=2$

${\pi }_1\equiv 2x+ky-5z=1$  

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

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

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