Three Dimensional Geometry

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 of the required plane.

The Cartesian equation of this plane can be obtained by substituting  $\overrightarrow{r}=x\widehat{i}+y\widehat{j}+z\widehat{k}$  in equation (3).

$\begin{array}{l}\left(x\widehat{i}+y\widehat{j}+z\widehat{k}\right).\left(33\widehat{i}+45\widehat{j}+50\widehat{k}\right)-41=0\\ \Rightarrow 33x+45y+50z-41=0\end{array}$

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 are the direction ratios of normal.

Here, the direction ratios of normal are 1, 2, and −3 and the point P is (1, 2, −3).

Thus, the equation of the required plane is

$\begin{array}{l}\mathrm{}\hfill \end{array}$$\begin{array}{ll}1\left(x-1\right)+2\left(y-2\right)-3\left(z+3\right)0\hfill & \Rightarrow x+2y-3z-14=0\hfill \end{array}$

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-axis are 1, 0, and 0.

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

Substituting λ = -1/2 in equation (1), we obtain

$\begin{array}{l}\Rightarrow \overrightarrow{r}.\left[-\frac{1}{2}\widehat{j}+\frac{3}{2}\widehat{k}\right]+\left(-3\right)=0\\ \Rightarrow \overrightarrow{r}\left(\widehat{j}-3\widehat{k}\right)+6=0\end{array}$

Therefore, its Cartesian equation is y − 3z + 6 = 0

This is the equation of the required plane.

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