Three Dimensional Geometry

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

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

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

A (1, 4, 3)

2x + my + nz = 4

$M\Rightarrow -4+\frac{7m}{2}+\frac{3n}{2}=4$

7m + 3n = 16

$\overrightarrow{AM}=\left(-1,\frac{-1}{2},\frac{-3}{2}\right)$

$\frac{2}{-1}=\frac{m}{\frac{-1}{2}}=\frac{n}{\frac{-3}{2}}$

m = 1, n = 3

Plane : 2x + y + 3z = 4

$cos\theta =\frac{\left|6-1-12\right|}{\sqrt[]{26}\sqrt[]{14}}=\frac{7}{2\sqrt[]{91}}$

AM = $\frac{\left|-2+4+9-4\right|}{\sqrt[]{14}}=\frac{7}{\sqrt[]{14}}$

$c$$o$$s$$\theta$$=$$\frac{AM}{AB}$$\Rightarrow$$A$$B$$=$$\frac{7/\text{exception 25:}}{7/2\text{exception 25:}}$$=$$\frac{2\text{exception 25:}}{\text{exception 25:}}$$=$$2$$\text{exception 25:}$$=$

$\int {\left(\frac{x}{x\mathrm{s}\mathrm{i}\mathrm{n}?x+\mathrm{c}\mathrm{o}\mathrm{s}?x}\right)}^{2}dx=\int \left(\frac{x}{\mathrm{c}\mathrm{o}\mathrm{s}?x}\right)\cdot \frac{x\mathrm{c}\mathrm{o}\mathrm{s}?xdx}{(x\mathrm{s}\mathrm{i}\mathrm{n}?x+\mathrm{c}\mathrm{o}\mathrm{s}?x{)}^2}$

$\begin{array}{rr}& =\frac{x}{\mathrm{c}\mathrm{o}\mathrm{s}?x}\left(-\frac{1}{x\mathrm{s}\mathrm{i}\mathrm{n}?x+\mathrm{c}\mathrm{o}\mathrm{s}?x}\right)+\int \left(\frac{\mathrm{c}\mathrm{o}\mathrm{s}?x+x\mathrm{s}\mathrm{i}\mathrm{n}?x}{{\mathrm{c}\mathrm{o}\mathrm{s}}^2?x}\right)\left(\frac{1}{x\mathrm{s}\mathrm{i}\mathrm{n}?x+\mathrm{c}\mathrm{o}\mathrm{s}?x}\right)dx\\ & =-\frac{x\mathrm{s}\mathrm{e}\mathrm{c}?x}{x\mathrm{s}\mathrm{i}\mathrm{n}?x+\mathrm{c}\mathrm{o}\mathrm{s}?x}+\int {\mathrm{s}\mathrm{e}\mathrm{c}}^2?xdx\\ & =-\frac{x\mathrm{s}\mathrm{e}\mathrm{c}?x}{x\mathrm{s}\mathrm{i}\mathrm{n}?x+\mathrm{c}\mathrm{o}\mathrm{s}?x}+\mathrm{t}\mathrm{a}\mathrm{n}?x+C\end{array}$

The direction of ratios of the lines,   $\frac{x-3}{-3}=\frac{y-2}{2k}=\frac{z-3}{2}\&\frac{x-1}{3k}=\frac{y-1}{1}=\frac{z-6}{-5}$ , are $-3,2k,2and3k,1,-5$ respectively.

It is known that two lines with direction ratios,   $a_1, b_1, c_1 and a_2, b_2,c_2$ , are perpendicular, if  $a_1{a}_2 + b_1{b}_2 + c_1{c}_2 =0$

$\begin{array}{l}\therefore -3(3k)+2k*1+2(-5)=0\\ \Rightarrow -9k+2k-10=0\\ \Rightarrow 7k=-10\\ \Rightarrow k=\frac{-10}{7}\end{array}$

Therefore, for k= -10/7, the given lines are perpendicular to each other.