Three Dimensional Geometry

 Normal vector to the given plane be

$2\widehat{i}-\widehat{j}+3\widehat{k}so$                   

Equation of line QS :

$\frac{x-1}{2}=\frac{y-3}{-1}=\frac{z-4}{1}=\lambda$

So let P $\left(2\lambda +1,-\lambda +3,\lambda +4\right)$  

Now P lies on given plane so

$4\lambda +2+\lambda -3+8\lambda +4+3=0$  

So, S (-3, 5, 2)

also given R lies on given plane so

6 – 5 + $\gamma$ + 3 = 0 so   $\gamma$ = -4

So, R (3, 5, -4)

SR² = 72

Required equation of plane will be (x – y – z – 1) + $\lambda$ (2x + y – 3z + 4) = 0

Given ${\perp }^r$ distance of (i) from origin = $\sqrt[]{\frac{2}{21}}$  

$\frac{\left|4\lambda -1\right|}{\sqrt[]{{\left(2\lambda +1\right)}^2+{\left(\lambda -1\right)}^2+{\left(3\lambda +1\right)}^2}}=\sqrt[]{\frac{2}{21}}$   

$\Rightarrow \lambda =\frac{1}{2}or\frac{15}{154}$

So plane be 4x – y – 5z + 2 = 0 for $\lambda =\frac{1}{2}$

Equation of line PQ :

$\frac{x-1}{2}=\frac{y+2}{3}=\frac{z-3}{-6}=k$

So, let Q (2k + 1, 3k – 2, -6k + 3) Q lies on given plane so

2k + 1 – 3k + 2 – 6k + 3 = 5

$-7k=-1ork=\frac{1}{7}$

So, $Q\left(\frac{9}{7},\frac{-11}{7},\frac{15}{7}\right)$

Now required distance = PQ = $\sqrt[]{\frac{4}{49}+\frac{9}{49}+\frac{36}{49}}=1$

Equation of the plane will be $\left\{\overrightarrow{r}.\left(\widehat{i}+\widehat{j}+\widehat{k}\right)-1\right\}+\lambda \left\{\overrightarrow{r}.\left(2\widehat{i}+3\widehat{j}-\widehat{k}\right)+4\right\}=0$   

$\overrightarrow{r}.\left\{\left(1+2\lambda \right)i+\left(1+3\lambda \right)\widehat{j}+\left(1-\lambda \right)\widehat{k}\right\}+\left(4\lambda -1\right)=0$               

-> $\left(1+2\lambda \right)x+\left(1+3\lambda \right)y+\left(1-\lambda \right)z+\left(4\lambda -1\right)=0.........\left(i\right)$

(i) is parallel to x-axis so its d.r.s will be (1, 0, 0)

-> $1+2\lambda =0so\lambda =\frac{-1}{2}$

Hence required equation will be

$\overrightarrow{r}\left\{-\frac{1}{2}\widehat{j}.+\frac{3}{2}\widehat{k}\right\}+\left(-3\right)=0$

$\overrightarrow{r}.\left(\widehat{j}-3\widehat{k}\right)+6=0$              

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⁼ $\frac{1}{51}$