Three Dimensional Geometry

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.

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

$L_2:\frac{x+3}{2}=\frac{y-6}{1}=\frac{z-5}{3}$               

Now,

$\overrightarrow{p}*\overrightarrow{q}=\left|\begin{array}{ccc}\hfill \widehat{i}\hfill & \hfill \widehat{j}\hfill & \hfill \widehat{k}\hfill \\ \hfill 2\hfill & \hfill 3\hfill & \hfill -1\hfill \\ \hfill 2\hfill & \hfill 1\hfill & \hfill 3\hfill \end{array}\right|=10\widehat{i}-8\widehat{j}-4\widehat{k}$               

and

  ${\overrightarrow{a}}_2-{\overrightarrow{a}}_1=6\widehat{i}-4\widehat{j}-4\widehat{k}$

  $\therefore S.D=\left|\frac{60+32+16}{\sqrt[]{100+64+16}}\right|=\frac{108}{\sqrt[]{180}}=\frac{18}{\sqrt[]{5}}$      

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

$L_2:\frac{x+3}{2}=\frac{y-6}{1}=\frac{z-5}{3}$               

Now,

$\overrightarrow{p}*\overrightarrow{q}=\left|\begin{array}{ccc}\hfill \widehat{i}\hfill & \hfill \widehat{j}\hfill & \hfill \widehat{k}\hfill \\ \hfill 2\hfill & \hfill 3\hfill & \hfill -1\hfill \\ \hfill 2\hfill & \hfill 1\hfill & \hfill 3\hfill \end{array}\right|=10\widehat{i}-8\widehat{j}-4\widehat{k}$               

and

  ${\overrightarrow{a}}_2-{\overrightarrow{a}}_1=6\widehat{i}-4\widehat{j}-4\widehat{k}$

  $\therefore S.D=\left|\frac{60+32+16}{\sqrt[]{100+64+16}}\right|=\frac{108}{\sqrt[]{180}}=\frac{18}{\sqrt[]{5}}$                           

The equations of the planes are

$\begin{array}{ll}2x - y + 4z = 5 \dots \left(1\right)\hfill & 5x - 2.5y + 10z = 6 \dots \left(2\right)\hfill \end{array}$

It can be seen that,

$\begin{array}{l}\frac{a_1}{a_2}=\frac{2}{5}\\ \frac{b_1}{b_2}=\frac{-1}{-2.5}=\frac{2}{5}\\ \frac{c_1}{c_2}=\frac{4}{10}=\frac{2}{5}\\ \therefore \frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}\end{array}$

Therefore, the given planes are parallel.

Hence, the correct answer is B.

The equations of the planes are

$\begin{array}{lll}2x + 3y + 4z = 4\hfill & 4x + 6y + 8z = 12\hfill & \Rightarrow 2x + 3y + 4z = 6\hfill \end{array}$

It can be seen that the given planes are parallel.

It is known that the distance between two parallel planes,   $ax + by + cz = d_1 and ax + by + cz = d_2,$ is given by,

$\begin{array}{l}D=\left|\frac{d_2-d_1}{}\right|\\ \Rightarrow D=\left|\frac{6-4}{}\right|\\ D=\frac{2}{}\end{array}$

Thus, the distance between the lines is 2/√29 units.

Hence, the correct answer is D.

The equation of a plane having intercepts a,  b,  c with x,  y, and z axes respectively is given by,

$\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1$

The distance (p) of the plane from the origin is given by,

$\begin{array}{l}p=\left|\frac{\frac{0}{a}+\frac{0}{b}+\frac{0}{c}-1}{}\right|\\ \Rightarrow p=\frac{1}{}\\ \Rightarrow p^2=\frac{1}{\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}}\\ \Rightarrow \frac{1}{p^2}=\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\end{array}$

Let the required line be parallel to the vector  $\overrightarrow{b}$  given by,  $\overrightarrow{b}=b_1\widehat{i}+b_2\widehat{j}+b_3\widehat{k}$

The position vector of the point (1, 2, − 4) is  $\overrightarrow{a}=\widehat{i}+2\widehat{j}-4\widehat{k}$

The equation of the line passing through (1, 2, −4) and parallel to vector   $\overrightarrow{b}$  is

$\begin{array}{l}\overrightarrow{r}=\overrightarrow{a}+\lambda \overrightarrow{b}\\ \Rightarrow \overrightarrow{r}=\left(\widehat{i}+2\widehat{j}-4\widehat{k}\right)+\lambda \left(b_1\widehat{i}+b_2\widehat{j}+b_3\widehat{k}\right).......(1)\end{array}$

The equations of the lines are

$\begin{array}{l}\frac{x-8}{3}=\frac{y+19}{-16}=\frac{z-10}{7}........(2)\\ \frac{x-15}{3}=\frac{y-29}{8}=\frac{z-5}{-5}........(3)\end{array}$

Line (1) and line (2) are perpendicular to each other.

$\therefore 3b_1-16b_2+7b_3=0........(4)$

Also, line (1) and line (3) are perpendicular to each other.

$\therefore 3b_1+8b_2-5b_3=0........(5)$

From equations (4) and (5), we obtain

$\begin{array}{l}\frac{b_1}{\left(-16\right)\left(-5\right)-8*7}=\frac{b_2}{7*3-3\left(-5\right)}=\frac{b_3}{3*8-3\left(-16\right)}\\ \Rightarrow \frac{b_1}{24}=\frac{b_2}{36}=\frac{b_3}{72}\\ \Rightarrow \frac{b_1}{2}=\frac{b_2}{3}=\frac{b_3}{6}\end{array}$

Direction ratios of    $\overrightarrow{b}$  are 2, 3, and 6.

$\therefore \overrightarrow{b}=2\widehat{i}+3\widehat{j}+6\widehat{k}$

Substituting  $\overrightarrow{b}=2\widehat{i}+3\widehat{j}+6\widehat{k}$  in equation (1), we obtain

$\overrightarrow{r}=\left(\widehat{i}+2\widehat{j}-4\widehat{k}\right)+\lambda \left(2\widehat{i}+3\widehat{j}+6\widehat{k}\right)$

This is the equation of the required line.