Three Dimensional Geometry

Let PT perpendicular to QR

$\frac{x+1}{2}=\frac{y+2}{3}=\frac{z-1}{2}=\lambda \Rightarrow T\left(2\lambda -1,3\lambda -2,2\lambda +1\right)$ therefore

$2\left(2\lambda -5\right)+3\left(3\lambda -4\right)+2\left(2\lambda -6\right)=0\Rightarrow \lambda =2$

$T\left(3,4,5\right)\therefore PT=\sqrt[]{1+4+4}=3\therefore QT=\sqrt[]{26-9}=\sqrt[]{17}$

$\therefore \Delta PQR=\frac{1}{2}*2\sqrt[]{17}*3=3\sqrt[]{17}$  

Therefore square of  $ar\left(\Delta PQR\right)$ = 153.

The line x + y – z = 0 = x – 2y + 3z – 5 is parallel to the vector

  $\overrightarrow{b}=\left|\begin{array}{ccc}\hfill \widehat{i}\hfill & \hfill \widehat{j}\hfill & \hfill \widehat{k}\hfill \\ \hfill 1\hfill & \hfill -1\hfill & \hfill 1\hfill \\ \hfill 1\hfill & \hfill -2\hfill & \hfill 3\hfill \end{array}\right|=\left(1,4,-3\right)$ Equation of line through P(1, 2, 4) and parallel to $\overrightarrow{b}\frac{x-1}{1}=\frac{y-2}{-4}=\frac{z-4}{-3}$  

Let  $N\equiv \left(\lambda +1,-4\lambda +2,-3\lambda +4\right)\overline{QN}=\left(\lambda ,-4\lambda +4,-3\lambda -1\right)$  

$\overline{QN}$ is perpendicular to $\overrightarrow{b}\Rightarrow \left(\lambda ,-4\lambda +4,-3\lambda -1\right).\left(1,4,-3\right)=0\Rightarrow \lambda =\frac{1}{2}.$  

Hence  $\overline{QN}\left(\frac{1}{2},2,\frac{-5}{2}\right)and\overline{\left|QN\right|}=\sqrt[]{\frac{21}{2}}$  

$\Delta =\left|\begin{array}{ccc}\hfill 8\hfill & \hfill 1\hfill & \hfill 4\hfill \\ \hfill 1\hfill & \hfill 1\hfill & \hfill 1\hfill \\ \hfill \lambda \hfill & \hfill -3\hfill & \hfill 0\hfill \end{array}\right|=12-3\lambda$  

So for  $\lambda$ = 4, it is having infinitely many solutions. ${\Delta }_x=\left|\begin{array}{ccc}\hfill -2\hfill & \hfill 1\hfill & \hfill 4\hfill \\ \hfill 0\hfill & \hfill 1\hfill & \hfill 1\hfill \\ \hfill \mu \hfill & \hfill -3\hfill & \hfill 0\hfill \end{array}\right|$  = -6 - $3\mu =0\Rightarrow -6-3\mu =0$  

For  $\mu =-2$ distance of $\left(4,-2,-\frac{1}{2}\right)$ from 8x + y + 4z + 2= 0 $\left|\frac{32-2-2+2}{\sqrt[]{64+1+16}}\right|=\frac{10}{3}$  units

Line L is   $\frac{x-1}{1}=\frac{y-2}{2\sqrt[]{2}}=\frac{z}{0}$

this line L makes an angle of 45° with the plane  $\sqrt[]{2}x+y-z=1$

$\therefore$ Required distance PQ is perpendicular distance of plane from P is i.e.,

$PQ=\frac{\left|\sqrt[]{2}+7-\sqrt[]{2}-1\right|}{\sqrt[]{2+1+1}}=3$

Plane containing both line $x=1$

Image of point $(\mathrm{2,1},3)$  in the plane $\frac{\alpha -2}{1}=\frac{\beta -1}{0}=\frac{\gamma -3}{0}=-2\left(1\right)$

$\alpha =0,\beta =1,\gamma =3;\alpha +\beta +\gamma =4$

  $x-cy-bz=0$

$cx-y+az=0$

$bx+ay-z=0$

Equation of any plane passing through the line of intersection of planes (1) and (2)

$x-cy-bz+\lambda (cx-y+az)=0$

$(1+\lambda c)x-(c+\lambda )y+(-b+a\lambda )z=0$

If (3) &  (4) is same plane.

$\frac{1+c\lambda }{b}=\frac{-(c+\lambda )}{a}=\frac{-b+a\lambda }{-1}$

(i)

(ii) (iii)

By (i)  & (ii) $\lambda =-\frac{(a+bc)}{ac+b}$

By (ii) & (iii)

$\lambda =\frac{-(ab+c)}{1-a^2};a^2+b^2+c^2+2abc=1$

$\bar{a}+\bar{b}+\bar{c}+\bar{d}=0$

Magnitude of all vectors will be same as well as angle between these vectors will also be same. $\bar{a}+\bar{b}+\bar{c}=-\bar{d}$

$\begin{array}{c}|\bar{a}{|}^2+|\bar{b}{|}^2+|\bar{c}{|}^2+2\bar{a}\cdot \bar{b}+2\bar{b}\cdot \bar{c}+2\bar{c}\cdot \bar{a}=|d{|}^2\\ \mathrm{}1+1+1+2\mathrm{c}\mathrm{o}\mathrm{s}?\alpha +2\mathrm{c}\mathrm{o}\mathrm{s}?\alpha +2\mathrm{c}\mathrm{o}\mathrm{s}?\alpha =1\\ 6\mathrm{c}\mathrm{o}\mathrm{s}?\alpha =-2\\ \mathrm{c}\mathrm{o}\mathrm{s}?\alpha =-\frac{1}{3}\\ \alpha ={\mathrm{c}\mathrm{o}\mathrm{s}}^{-1}?\left(\frac{-1}{3}\right)=\pi -{\mathrm{c}\mathrm{o}\mathrm{s}}^{-1}?\frac{1}{3}\end{array}$

Volume = |a b c; b c a; c a b| = |3abc – a³ – b³ – c³|
= | (a + b + c) (a² + b² + c² – ab – bc – ca)| = | (a + b + c) (a + b + c)² – 3 (ab + bc + ca)|
= |9 (81-3 * 15)| = 324

  $\left(\lambda ,2\lambda ,3\lambda \right)$

$cos\theta =\frac{6}{\sqrt[]{42}}\Rightarrow sin\theta =\frac{\sqrt[]{6}}{\sqrt[]{42}}$ (where q is the angle between L₁ & L₂)

$Area\Delta OAB=\frac{1}{2}\left(OA\right)\left(OB\right)sin\theta$

$=\frac{1}{2}\left(\sqrt[]{3}\right)\left|\lambda \right|\left(\sqrt[]{14}\right)\frac{\sqrt[]{6}}{\sqrt[]{42}}=\sqrt[]{6}\Rightarrow \lambda =\pm 2$

Kindly consider the following figure