Three Dimensional Geometry

$L_1:\mathcal{l}x-y+3\left(1-z\right)=1,x+2y-z=2$

plane containing the line P : 3x – 8y + 7z = 4

If $\overrightarrow{n}$ be vector parallel to L.

then $\overrightarrow{n}=\left|\begin{array}{ccc}\hfill \widehat{i}\hfill & \hfill \widehat{j}\hfill & \hfill \widehat{k}\hfill \\ \hfill \mathcal{l}\hfill & \hfill -1\hfill & \hfill 3\left(1-\mathcal{l}\right)\hfill \\ \hfill 1\hfill & \hfill 2\hfill & \hfill -1\hfill \end{array}\right|=\left(6\mathcal{l}-5\right)\widehat{i}+\left(3-2\mathcal{l}\right)\widehat{j}+\left(2\mathcal{l}+1\right)\widehat{k}$ as P containing the line

$\therefore 3\left(6\mathcal{l}-5\right)-8\left(3-2\mathcal{l}\right)+7\left(2\mathcal{l}+1\right)=0$

$\Rightarrow \mathcal{l}=\frac{2}{3}$

If be the acute angle between line L & Y axis then cos = $\frac{5/3}{\sqrt[]{1+\frac{25}{9}+\frac{49}{9}}}=\frac{5}{\sqrt[]{83}}$

$\therefore 415co{s}^2\theta =125$

The normal vector to the plane is $\overline{n_1}*\overline{n_2}=\left|\begin{array}{ccc}\hfill i\hfill & \hfill 3\hfill & \hfill k\hfill \\ \hfill 1\hfill & \hfill a\hfill & \hfill -1\hfill \\ \hfill -1\hfill & \hfill 1\hfill & \hfill -a\hfill \end{array}\right|=\left(1-a\right)\widehat{i}+\widehat{j}+\widehat{k}$

$\therefore equationofplaneis\left(1-a\right)\left(x-1\right)+\left(y-1\right)+z=0$

(1 – a)x + y + z = 2 – a …… (i)

Now distance from (2, 1, 4) = $\sqrt[]{3}$

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

$\Rightarrow a^2+2a-8=0\Rightarrow a=-4,2$ the largest value of a = 2.

he 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 $\stackrel{}{}$$\overline{QN}\left(\frac{1}{2},2,\frac{-5}{2}\right)and\overline{\left|QN\right|}=\text{exception 25:}$

 $\frac{OP}{OA}=tan15°$

$\Rightarrow OA=OPcot15°$

$\frac{OP}{OC}=tan45°\Rightarrow OP=OC$

Now, OP = $\sqrt[]{O{A}^2-8^2}$

$O{P}^2={\left(OP\right)}^{2}co{t}^{2}15°-64$

$\Rightarrow OP=\frac{32}{\sqrt[]{3}}\left(2-\sqrt[]{3}\right)$

Normal of plane P : = $\left|\begin{array}{ccc}\hfill \widehat{i}\hfill & \hfill \widehat{j}\hfill & \hfill \widehat{k}\hfill \\ \hfill -2\hfill & \hfill 1\hfill & \hfill -3\hfill \\ \hfill -1\hfill & \hfill 2\hfill & \hfill -2\hfill \end{array}\right|=4\widehat{i}-j-3\widehat{k}$

Equation of plane P which passes through (2, 2) is 4x – y – 3z – 12 = 0

Now, A (3, 0, 0), B (0, 12 0), C (0, 4)

$\begin{array}{l}\Rightarrow \alpha =3,\beta -12,\gamma =-4\\ \Rightarrow P=\alpha +\beta +\gamma =-13\end{array}$

Now, volume of tetrahedron OABC

$V=\left|\frac{1}{6}\overrightarrow{OA}.\left(\overrightarrow{OB}*\overrightarrow{OC}\right)\right|=24$

(V, P) = (24, 13)

Given,  $L_1:\frac{x-1}{\lambda }=\frac{y-2}{1}=\frac{z-3}{2}$

and $L_2:\frac{x+26}{-2}=\frac{y+18}{3}=\frac{z+28}{\lambda }$

are coplanar

$\Rightarrow \left|\begin{array}{ccc}\hfill 27\hfill & \hfill 20\hfill & \hfill 31\hfill \\ \hfill \lambda \hfill & \hfill 1\hfill & \hfill 2\hfill \\ \hfill 2\hfill & \hfill 3\hfill & \hfill \lambda \hfill \end{array}\right|=0$

$\lambda =3$

Now, normal of plane P, which contains L1 and L2

equation of required plane P : 3x + 13y – 11z + 4 = 0

(0, 4, 5) does not lie on plane P.

$\begin{array}{rr}& D=\left|\begin{array}{ccc}1& 4& -1\\ 3& 1& 5\\ a& b& 6\end{array}\right|=0\Rightarrow 21a-8b-66=0\end{array}$$\begin{array}{r}P:2x-3y+6z=15\end{array}$

so required distance $=\frac{21}{7}=3$

 $P:\left(x+3y-z-6\right)=\lambda \left(-6x+5y-z-7\right)=0$

Passes $\left(2,3,\frac{1}{2}\right)$

$\Rightarrow \left(2+9-\frac{1}{2}-6\right)=\lambda \left(-12+15-\frac{1}{2}-7\right)=0$

$\Rightarrow \lambda =1$

$\frac{{\left|13\overrightarrow{a}\right|}^2}{d^2}=\frac{{\left(13\right)}^2\left(93\right)}{{\left(13\right)}^2}=93$

$\left(5x+8y+13z-29\right)+\lambda \left(8x-7y+z-20\right)=0$  

P₁ passing (2, 1, 3)

(10 + 8 + 39 – 29) + $\lambda \left(16-7+3-20\right)=0$  

$28-8\lambda =0\lambda =7/2$               

$\Rightarrow$ 2X – Y + Z – 6 = 0      ….(i)

For P₂ passes (0, 1, 2)

$\left(8+26-29\right)+\lambda \left(-7+2-20\right)=0$              

$5-25\lambda =0\Rightarrow \lambda =\frac{1}{5}$               

$P_2:x+y+2z-5=0........(ii)$               

Acute angle between the planes

$cos\theta =\frac{2-1+2}{\sqrt[]{4+1+1}\sqrt[]{1+1+4}}=\frac{1}{2}$               

$\Rightarrow \theta =\frac{\pi }{3}$