Three Dimensional Geometry

Given $2\mathcal{l}+2m-n=0........\left(i\right)$

$mn+n\mathcal{l}+\mathcal{l}m=0...........\left(ii\right)$

$\&wehave{\mathcal{l}}^2+m^2+n^2=1..........\left(iii\right)$

$\left(i\right)\Rightarrow 2\left(\mathcal{l}+m\right)=n$  

  $\left(ii\right)\Rightarrow \mathcal{l}m+n\left(\mathcal{l}+m\right)=0$

$2{\mathcal{l}}^2+2m^2+5\mathcal{l}m=0$             

(a) $\frac{\mathcal{l}}{m}=-2$  

(i) $\Rightarrow \frac{2\mathcal{l}}{m}+2-\frac{n}{m}=0$  

$\frac{n}{m}=-2$  

$So,\left(\mathcal{l},m,n\right)=\left(-2m,m,-2m\right)$  

= (-2, 1, -2)

(b)   $\frac{\mathcal{l}}{m}=\frac{-1}{2}givesn=-2\mathcal{l}$

$\left(\mathcal{l},m,n\right)=\left(\mathcal{l},-2\mathcal{l},-2\mathcal{l}\right)=\left(1,-2,-2\right)$

$Now,cos\theta =\frac{-2-2+4}{3*3}=0$

$\Rightarrow \theta =\frac{\pi }{2}$  

${\perp }^{r}planes$

${\overrightarrow{n}}_1,{\overrightarrow{n}}_2=0\Rightarrow \lambda =\frac{3}{4}$

Where required plane P is

$\left(1+\lambda \right)x+\left(2-\lambda \right)y+\left(3-\lambda \right)z+1-6\lambda =0$

$\Rightarrow 2x+y+2z-5=0$

$B\left(1+2\lambda ,3+\lambda ,4+2\lambda \right)$

x – 2y – z = 3 => $\lambda =-6\Rightarrow B\left(-11,-3,-8\right)$

$N\left(1+t,3-2t,4-t\right)$  

$x-2y-z=3\Rightarrow t=2\Rightarrow N\left(3,-1,2\right)$  

Line NB : $\frac{x-3}{7}=\frac{y+1}{1}=\frac{3-2}{5}$

$d^2=P{M}^2=\frac{{\left|P\overrightarrow{Q}*\overrightarrow{R}\right|}^2}{{\left|\overrightarrow{R}\right|}^2}$

${\left|\overrightarrow{R}\right|}^2=75$

$d^2=\frac{1950}{75}=26$  

${\overrightarrow{n}}_1=6\widehat{i}+7\widehat{j}+8\widehat{k}{\overrightarrow{n}}_2=3\widehat{i}+5\widehat{j}+7\widehat{k}$

${\overrightarrow{n}}_1*{\overrightarrow{n}}_2=\left|\begin{array}{ccc}\hfill \widehat{i}\hfill & \hfill \widehat{j}\hfill & \hfill \widehat{k}\hfill \\ \hfill 6\hfill & \hfill 7\hfill & \hfill 8\hfill \\ \hfill 3\hfill & \hfill 5\hfill & \hfill 7\hfill \end{array}\right|=9\widehat{i}-18\widehat{j}+9k$              

$\therefore$ the normal to required plane is $\widehat{i}-2\widehat{j}+\widehat{k}$

Equation of plane 1(x + 1) – 2(y – 1) + (z – 3) = 0

x – 2y + z = 0

P (7, -2, 13)

$\therefore PQ=\left|\frac{7+4+13}{\sqrt[]{1+4+1}}\right|=\frac{24}{\sqrt[]{6}}$

${\left(PQ\right)}^2=\frac{24*24}{6}=96$

Equation of required plane

$\left(x+y+4z-16\right)+\lambda \left(-x+y+z-6\right)=0$ it passes (1, 2, 3)

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

$\Rightarrow \lambda =-\frac{1}{2}$

$\therefore$ Equation of plane

$\left(1-\lambda \right)x+\left(1+\lambda \right)y+\left(4+\lambda \right)z-16-6\lambda =0$              

$\frac{3}{2}x+\frac{1}{2}y+\frac{7}{2}z-13=0$              

$\Rightarrow 3x+y+7z=26$              

  (4, 2, 2) not satisfying the plane

${\mathcal{l}}_1:$

$\overrightarrow{r}=\frac{1}{8}\widehat{i}+\lambda \left(-\frac{1}{8},\frac{1}{4\sqrt[]{2}},0\right)$

$\overrightarrow{r}=-\frac{1}{8}\widehat{i}+\mu \left(\frac{1}{8},0,-\frac{-1}{6\sqrt[]{3}}\right)$

$\overrightarrow{n}=\left|\begin{array}{ccc}\hfill \widehat{i}\hfill & \hfill \widehat{j}\hfill & \hfill \widehat{k}\hfill \\ \hfill \frac{-1}{8}\hfill & \hfill \frac{1}{4\sqrt[]{2}}\hfill & \hfill 0\hfill \\ \hfill \frac{1}{8}\hfill & \hfill 0\hfill & \hfill \frac{-1}{6\sqrt[]{3}}\hfill \end{array}\right|$

$=\left(-\frac{1}{24\sqrt[]{6}},\frac{-1}{48\sqrt[]{3}},-\frac{1}{32\sqrt[]{2}}\right)$

$d=projectionofA\overrightarrow{C}on\overrightarrow{n}=\left(-\frac{2}{8},0,0\right).\frac{\left(4,2\sqrt[]{2},3\sqrt[]{3}\right)}{\sqrt[]{16}+8+27}=\frac{-1}{\sqrt[]{16+8+27}}$

d²⁼ $\frac{1}{51}$                                       

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

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

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 $\pi \equiv x+2y+3z-2=0$  

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