Class 12th

It is known that the equation of the line through the points, $(x_1, y_1, z_1)and(x_2, y_2, z_2)$ , is

$\frac{x-x_1}{x_2-x_1}=\frac{x-y_1}{y_2-y_1}=\frac{z-z_1}{z_2-z_1}$

Since the line passes through the points, $\left(3,-4,-5\right)and\left(2,-3,1\right)$ , its equation is given by,

$\begin{array}{l}\frac{x-3}{2-3}=\frac{y+4}{-3+4}=\frac{z+5}{1+5}\\ \Rightarrow \frac{x-3}{-1}=\frac{y+4}{1}=\frac{z+5}{6}=k(say)\\ \Rightarrow x=3-k,y=k-4,z=6k-5\end{array}$

Therefore, any point on the line is of the form $\left(3- k, k -4,6k -5\right).$

This point lies on the plane, $2x + y + z =7$

$\begin{array}{lll}\therefore 2 \left(3- k\right) + \left(k -4\right) + \left(6k -5\right) = 7\hfill & 5k - 3 = 7\hfill & k = 2\hfill \end{array}$

Hence, the coordinates of the required point are $\left(3-2,2-4,6*2-5\right)i.e.,$ $\left(1,-2,7\right).$

It is known that the equation of the line passing through the points, $(x_1, y_1, z_1)and(x_2, y_2, z_2),$ is  $\frac{x-x_1}{x_2-x_1}=\frac{x-y_1}{y_2-y_1}=\frac{z-z_1}{z_2-z_1}$

The line passing through the points, $\left(5,1,6\right)and\left(3,4,1\right),$ is given by,

$\begin{array}{l}\frac{x-5}{3-5}=\frac{y-1}{4-1}=\frac{z-6}{1-6}\\ \Rightarrow \frac{x-5}{-2}=\frac{y-1}{3}=\frac{z-6}{-5}=k(say)\\ \Rightarrow x=5-2k,y=3k+1,z=6-5k\end{array}$

Any point on the line is of the form $\left(5-2k,3k +1,6-5k\right).$

Since the line passes through ZX-plane,

$\begin{array}{l}3k+1=0\\ \Rightarrow k=-\frac{1}{3}\\ \Rightarrow 5-2k=5-2\left(-\frac{1}{3}\right)=\frac{17}{3}\\ 6-5k=6-5\left(-\frac{1}{3}\right)=\frac{23}{3}\end{array}$

Therefore, the required point is $\left(\frac{17}{3},0,\frac{23}{3}\right)$

It is known that the equation of the line passing through the points, $(x_1, y_1, z_1)and(x_2, y_2, z_2),$ is  $\frac{x-x_1}{x_2-x_1}=\frac{x-y_1}{y_2-y_1}=\frac{z-z_1}{z_2-z_1}$

The line passing through the points, $\left(5,1,6\right)and\left(3,4,1\right),$ is given by,

$\begin{array}{l}\frac{x-5}{3-5}=\frac{y-1}{4-1}=\frac{z-6}{1-6}\\ \Rightarrow \frac{x-5}{-2}=\frac{y-1}{3}=\frac{z-6}{-5}=k(say)\\ \Rightarrow x=5-2k,y=3k+1,z=6-5k\end{array}$

Any point on the line is of the form $\left(5-2k,3k +1,6-5k\right).$

The equation of $YZ-planeis x =0$

Since the line passes through YZ-plane,

$5-2k =0$

$\begin{array}{l}\Rightarrow k=\frac{5}{2}\\ \Rightarrow 3k+1=3*\frac{5}{2}+1=\frac{17}{2}\\ 6-5k=6-5*\frac{5}{2}=\frac{-13}{2}\end{array}$

Therefore, the required point is  $\left(0,\frac{17}{2},\frac{-13}{2}\right)$ .

The given lines are $\begin{array}{l}\overrightarrow{r}=6\widehat{i}+2\widehat{j}+2\widehat{k}+\lambda \left(\widehat{i}-2\widehat{j}-2\widehat{k}\right)..........(1)\\ \overrightarrow{r}=-4\widehat{i}-\widehat{k}+\mu \left(3\widehat{i}-2\widehat{j}-2\widehat{k}\right)...........(2)\end{array}$ $\begin{array}{l}\overrightarrow{r}=6\widehat{i}+2\widehat{j}+2\widehat{k}+\lambda \left(\widehat{i}-2\widehat{j}-2\widehat{k}\right)..........(1)\\ \overrightarrow{r}=-4\widehat{i}-\widehat{k}+\mu \left(3\widehat{i}-2\widehat{j}-2\widehat{k}\right)...........(2)\end{array}$

It is known that the shortest distance between two lines,  $\overrightarrow{r}=a_1+\lambda b_1\&\overrightarrow{r}=a_2+\lambda b_2$   is given by

$d=\left|\frac{\left(b_1*b_2\right).\left(a_1-a_2\right)}{\left|b_1*b_2\right|}\right|$

Comparing  $\overrightarrow{r}=a_1+\lambda b_1\&\overrightarrow{r}=a_2+\lambda b_2$ to equations (1) and (2), we obtain

$\begin{array}{l}\overrightarrow{a_1}=6\widehat{i}+2\widehat{j}+2\widehat{k}\\ \overrightarrow{b_1}=\widehat{i}-2\widehat{j}-2\widehat{k}\\ \overrightarrow{a_2}=-4\widehat{i}-\widehat{k}\\ \overrightarrow{b_2}=3\widehat{i}-2\widehat{j}-2\widehat{k}\end{array}$

$\Rightarrow \overrightarrow{a_2}-\overrightarrow{a_1}=\left(-4\widehat{i}-\widehat{k}\right)-\left(6\widehat{i}+2\widehat{j}+2\widehat{k}\right)=10\widehat{i}-2\widehat{j}-3\widehat{k}$

$\Rightarrow \overrightarrow{b_1}*\overrightarrow{b_2}=\left|\begin{array}{ccc}\hfill \widehat{i}\hfill & \hfill \widehat{j}\hfill & \hfill \widehat{k}\hfill \\ \hfill 1\hfill & \hfill -2\hfill & \hfill 2\hfill \\ \hfill 3\hfill & \hfill -2\hfill & \hfill -2\hfill \end{array}\right|=(4+4)\widehat{i}-(-2-6)\widehat{j}+(-2+6)\widehat{k}=8\widehat{i}+8\widehat{j}+4\widehat{k}$

$\begin{array}{l}\therefore \left|\overrightarrow{b_1}*\overrightarrow{b_2}\right|==12\\ \left(\overrightarrow{b_1}*\overrightarrow{b_2}\right).\left(\overrightarrow{a_2}-\overrightarrow{a_1}\right)=\left(8\widehat{i}+8\widehat{j}+4\widehat{k}\right).\left(10\widehat{i}-2\widehat{j}-3\widehat{k}\right)=-80-16-12=-108\end{array}$

Substituting all the values in equation (1), we obtain

$d=\left|\frac{-108}{12}\right|=9$

Therefore, the shortest distance between the two given lines is 9 units.

Any plane parallel to the plane,  $\overrightarrow{r}.\left(\widehat{i}+\widehat{j}+\widehat{k}\right)=2$ , is of the form  $\overrightarrow{r}.\left(\widehat{i}+\widehat{j}+\widehat{k}\right)=\lambda .........(1)$           

The plane passes through the point (a, b, c). Therefore, the position vector  $\overrightarrow{r}$  of this point is  $\overrightarrow{r}=a\widehat{i}+b\widehat{j}+c\widehat{k}$

Therefore, equation (1) becomes

$\begin{array}{l}\left(a\widehat{i}+b\widehat{j}+c\widehat{k}\right).\left(\widehat{i}+\widehat{j}+\widehat{k}\right)=\lambda \\ \Rightarrow a+b+c=\lambda \end{array}$

Substituting  $\lambda =a+b+c$ in equation (1), we obtain

$\overrightarrow{r}=\left(\widehat{i}+\widehat{j}+\widehat{k}\right)=a+b+c.........(2)$

This is the vector equation of the required plane.

Substituting  $\overrightarrow{r}=x\widehat{i}+y\widehat{j}+z\widehat{k}$  in equation (2), we obtain

$\begin{array}{l}\left(x\widehat{i}+y\widehat{j}+z\widehat{k}\right).\left(\widehat{i}+\widehat{j}+\widehat{k}\right)=a+b+c\\ \Rightarrow x+y+z=a+b+c\end{array}$

The position vector of the point $\left(1,2,3\right)$ is   $\overrightarrow{r}=\widehat{i}+2\widehat{j}+3\widehat{k}$

The direction ratios of the normal to the plane,   $\overrightarrow{r}=\left(\widehat{i}+2\widehat{j}-5\widehat{k}\right)+9=0$ , are $1,2,and-5$ and the normal vector is  $\overrightarrow{N}=\left(\widehat{i}+2\widehat{j}-5\widehat{k}\right)$

The equation of a line passing through a point and perpendicular to the given plane is given by,

$\overrightarrow{l}=\overrightarrow{r}+\lambda \overrightarrow{N},\lambda \in R\Rightarrow \overrightarrow{l}=\left(\widehat{i}+2\widehat{j}+3\widehat{k}\right)+\lambda \left(\widehat{i}+2\widehat{j}-5\widehat{k}\right)$

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.

The coordinates of $A,B,C,andDare\left(1,2,3\right),\left(4,5,7\right),\left(­-4,3,-6\right),$ and $\left(2,9,2\right)$ respectively.

The direction ratios of $ABare\left(4-1\right)=3,\left(5-2\right)=3,and\left(7-3\right)=4$

The direction ratios of $CDare\left(2-\left(-4\right)\right)=6,\left(9-3\right)=6,and\left(2-\left(-6\right)\right)=8$

It can be seen that,  $\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}=\frac{1}{2}$

Therefore, AB is parallel to CD.

Thus, the angle between $ABandCDiseither0°or18$$0$$°.$

The line parallel to x-axis and passing through the origin is x-axis itself.

Let A be a point on x-axis. Therefore, the coordinates of A are given by $\left(a,0,0\right),where a ?R.$ Direction ratios of $OAare\left(a ?0\right)= a,0,0$

The equation of OA is given by,

$\begin{array}{l}\frac{x?0}{a}=\frac{y?0}{0}=\frac{z?0}{0}\\ ?\frac{x}{1}=\frac{y}{0}=\frac{z}{0}=a\end{array}$

Thus, the equation of line parallel to x-axis and passing through origin is

$\frac{x}{1}=\frac{y}{0}=\frac{z}{0}$

It is given that  $l_1, m_1, n_1 and l_2, m_2, n_2$ are the direction cosines of two mutually perpendicular lines. Therefore,

$\begin{array}{l}{l}_1{l}_2+m_1{m}_2+n_1 n_2=0..........(1)\\ {l_1}^2+{m_1}^2+n_1{ }^2=1..........(2)\\ {l_2}^2+{m_2}^2+n_2{ }^2=1..........(3)\end{array}$

Let  $l, m, n$ be the direction cosines of the line which is perpendicular to the line with direction cosines  $l_1, m_1, n_1 and l_2, m_2, n_2.$

$\begin{array}{l}\therefore l{l}_1+ m{m}_1+ n{n}_1 =0\\ l l_2+m m_2+n n_2=0\\ \therefore \frac{l}{m_1{n}_2-m_2{n}_1}=\frac{m}{n_1{l}_2-n_2{l}_1}=\frac{n}{l_1{m}_2-l_2{m}_1}\\ \Rightarrow \frac{l^2}{{\left(m_1{n}_2-m_2{n}_1\right)}^2}=\frac{m^2}{{\left(n_1{l}_2-n_2{l}_1\right)}^2}=\frac{n^2}{{\left(l_1{m}_2-l_2{m}_1\right)}^2}\\ \Rightarrow \frac{l^2}{{\left(m_1{n}_2-m_2{n}_1\right)}^2}=\frac{m^2}{{\left(n_1{l}_2-n_2{l}_1\right)}^2}=\frac{n^2}{{\left(l_1{m}_2-l_2{m}_2\right)}^2}\\ =\frac{l^2+m^2+n^2}{{\left(m_1{n}_2-m_2{n}_1\right)}^2+{\left(n_1{l}_2-n_2{l}_1\right)}^2+{\left(l_1{m}_2-l_2{m}_2\right)}^2}..........(4)\end{array}$