Three Dimensional Geometry

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,  $\overrightarrow{r}.\overrightarrow{N}=d$ is given by,  $D=\frac{\left|\overrightarrow{a}.\overrightarrow{N}-d\right|}{\left|\overrightarrow{N}\right|}$

Here, $\overrightarrow{N}=3\widehat{i}+4\widehat{j}-12\widehat{k}$ and  $d =-13$

Therefore, the distance between the point (1, 1, p) and the given plane is

$\begin{array}{l}{D}_1=\frac{\left|\left(\widehat{i}+\widehat{j}+p\widehat{k}\right).\left(3\widehat{i}+4\widehat{j}-12\widehat{k}\right)+13\right|}{\left|3\widehat{i}+4\widehat{j}-12\widehat{k}\right|}\\ \Rightarrow D_1=\frac{\left|3+4-12p+13\right|}{}\\ \Rightarrow D_1=\frac{\left|20-12p\right|}{13}..........(1)\end{array}$

Similarly, the distance between the point $\left(-3,0,1\right)$ and the given plane is

$\begin{array}{l}{D}_2=\frac{\left|\left(3\widehat{i}+\widehat{k}\right).\left(3\widehat{i}+4\widehat{j}-12\widehat{k}\right)+13\right|}{\left|3\widehat{i}+4\widehat{j}-12\widehat{k}\right|}\\ \Rightarrow D_1=\frac{\left|-9-12+13\right|}{}\\ \Rightarrow D_1=\frac{8}{13}..........(2)\end{array}$

It is given that the distance between the required plane and the points, $\left(1,1, p\right)and\left(-3,0,1\right),$ is equal.

$\therefore D_1 = D_2$

$\begin{array}{l}\Rightarrow \frac{\left|20-12p\right|}{13}=\frac{8}{13}\\ \Rightarrow 20-12p=8,or,-(-20-12p)=8\\ \Rightarrow 12p=12,or,12p=28\\ \Rightarrow p=1,or,p=\frac{7}{3}\end{array}$

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 perpendicular to the plane, $3x +3y + z =0$

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

From equations (2) and (3), we obtain

$\begin{array}{l}\frac{a}{2*1-3*3}=\frac{b}{3*3-1*1}=\frac{c}{1*3-2*3}\\ \Rightarrow \frac{a}{-7}=\frac{b}{8}=\frac{c}{-3}=k(say)\\ \Rightarrow a=-7k,b=8k,c=-3k\end{array}$

Substituting the values of a, b, and c in equation (1), we obtain

$\begin{array}{l}-7k(x+1)+8k(y-3)-3k(z-2)=0\\ \Rightarrow (-7x-7)+(8y-24)-3z+6=0\\ \Rightarrow -7x+8y-3z-25=0\\ \Rightarrow 7x-8y+3z+25=0\end{array}$

This is the required equation of the plane.

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.