Ncert Solutions Maths class 12th

The equations of the given planes are

$\begin{array}{l}\overrightarrow{r}.\left(\widehat{i}+2\widehat{j}+3\widehat{k}\right)-4=0.....(1)\\ \overrightarrow{r}.\left(2\widehat{i}+\widehat{j}-\widehat{k}\right)+5=0......(2)\end{array}$

The equation of the plane passing through the line intersection of the plane given in equation (1) and equation (2) is

$\begin{array}{l}\left[\overrightarrow{r}.\left(\widehat{i}+2\widehat{j}+3\widehat{k}\right)-4\right]+\lambda \left[\overrightarrow{r}.\left(2\widehat{i}+\widehat{j}-\widehat{k}\right)+5\right]=0\\ \overrightarrow{r}.\left[\left(2\lambda +1\right)\widehat{i}+\left(\lambda +2\right)\widehat{j}+\left(3-\lambda \right)\widehat{k}\right]+\left(5\lambda -4\right)=0.....(3)\end{array}$

The plane in equation (3) is perpendicular to the plane,  $\overrightarrow{r}.\left(5\widehat{i}+3\widehat{j}-6\widehat{k}\right)+8=0$

$\begin{array}{l}\therefore 5\left(2\lambda +1\right)+3\left(\lambda +2\right)-6\left(3-\lambda \right)=0\\ \Rightarrow 19\lambda -7=0\\ \Rightarrow \lambda =\frac{7}{19}\end{array}$

Substituting λ = 7/19 in equation (3), we obtain

$\begin{array}{l}\Rightarrow \overrightarrow{r}.\left(\frac{13}{19}\widehat{i}+\frac{45}{19}\widehat{j}+\frac{50}{19}\widehat{k}\right)\frac{-41}{19}=0\\ \Rightarrow \overrightarrow{r}.\left(33\widehat{i}+45\widehat{j}+50\widehat{k}\right)-41=0\end{array}$

This is the vector equation o

The coordinates of the points, O and P, are (0, 0) and (1, 2, −3) respectively.

Therefore, the direction ratios of OP are (1 − 0) = 1, (2 − 0) = 2, and (−3 − 0) = −3

It is known that the equation of the plane passing through the point (x1,  y1 z1) is

$a\left(x-x_1\right)+b\left(y-y_1\right)+c\left(z-z_1\right)=0$ where, a,  b, and c 

Equation of one plane is

$\begin{array}{l}\overrightarrow{r}.\left(\widehat{i}+\widehat{j}+\widehat{k}\right)=1\\ \Rightarrow \overrightarrow{r}.\left(\widehat{i}+\widehat{j}+\widehat{k}\right)-1=0\\ \overrightarrow{r}.\left(2\widehat{i}+3\widehat{j}-\widehat{k}\right)+4=0\end{array}$

The equation of any plane passing through the line of intersection of these planes is

$\begin{array}{l}\left[\overrightarrow{r}.\left(\widehat{i}+\widehat{j}+\widehat{k}\right)-1\right]+\lambda \left[\overrightarrow{r}.\left(2\widehat{i}+3\widehat{j}-\widehat{k}\right)+4\right]=0\\ \overrightarrow{r}.\left[\left(2\lambda +1\right)\widehat{i}+\left(3\lambda +1\right)\widehat{j}+\left(1-\lambda \right)\widehat{k}\right]+\left(4\lambda -1\right)=0.....(1)\end{array}$

Its direction ratios are (2λ + 1), (3λ + 1), and (1 − λ).

The required plane is parallel to x-axis. Therefore, its normal is perpendicular to x-axis.

The direction ratios of x-ax

The position vector through the point $\left(1,1, p\right)$ is  $\overrightarrow{a_1}=\widehat{i}+\widehat{j}+p\widehat{k}$

Similarly, the position vector through the point $\left(-3,0,1\right)$ is

$\overrightarrow{a_2}=-4\widehat{i}+\widehat{k}$

The equation of the given plane is  $\overrightarrow{r}.\left(3\widehat{i}+4\widehat{j}-12\widehat{k}\right)+13=0$

It is known that the perpendicular distance between a point whose position vector is  $\overrightarrow{a}$  and the plane, &

The equation of the plane passing through the point $a \left(x +1\right)+ b \left(y -3\right)+ c \left(z -2\right)=0\dots \left(1\right)$ where, a, b, c are the direction ratios of normal to the plane.

It is known that two planes,  $a_{1}x+b_{1}y+c_{1}z+d_1=0\&a_{2}x+b_{2}y+c_{2}z+d_2=0$ are perpendicular, if $a_1{a}_2+b_1{b}_2+c_1{c}_2=0$

Plane (1) is perpendicular to the plane,  $x +2y +3z =5$

$\begin{array}{ll}a.1 + b .2 + c.3 = 0\hfill & a + 2b + 3c = 0 ....\left(2\right)\hfill \end{array}$

Also, plane (1) is perpendicu

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 $\left(3-2,2-4,6\times 2-5\right)i.e.,$&

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,

$5-2k =0$

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

Therefore, the required point is  

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

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

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

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

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

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

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

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

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, equatio