Plane

# Plane

A surface where the straight line joining any two points lies wholly is called a plane surface or simply a plane. The general equation of the first degree in $x$, $y$ and $z$ represents a plane.

The general equation of first degree in $x$, $y$ and $z$ is $Ax+By+Cz+D=0\text{ __(1)}$ where, $A$, $B$, $C$ and $D$ are constants and $A$, $B$, $C$ are not all zero.

et $P(x_1,y_1,z_1)$ and $Q(x_2,y_2,z_2)$ be any two points lying on the plane represented by the equation $\text{(1)}$, then, $Ax_1+By_1+Cz_1+D=9\text{ __(2)}$ $Ax_2+By_2+Cz_2+D=0\text{ __(3)}$ Multiplying equation $\text{(3)}$ by $k$ and adding to equation $\text{(2)},$ $A(kx_2+x_1)+B(ky_2+y_1)+C(kz_2+z_1)+D(k+1)=0$ $A\left(\frac{kx_2+x_1}{k+1}\right)+B\left(\frac{ky_2+y_1}{k+1}\right)+C\left(\frac{kz_2+z_1}{k+1}\right)+D=0$

This shows that the point $\left(\frac{kx_2+x_1}{k+1},\frac{ky_2+y_1}{k+1},\frac{kz_2+z_1}{k+1}\right)$ lies on the locus represented by the equation $\text{(1)}$. Also, this point lies on the line joining the two points $P(x_1,y_1,z_1)$ and $Q(x_2,y_2,z_2)$ and divides in the ratio $k:1$.

Giving different values to $k$ except $-1$, we get all the points on the line $PQ$. Therefore, by definition of the plane, equation $\text{(1)}$ is a plane. Hence, the general equation of first degree in $x$, $y$ and $z$ represents a plane.

## Equation of the plane

### The intercept form

We have to find the equation of the plane which cuts off intercepts $a$, $b$ and $c$ on the coordinate axes.

Let the equation of the plane be $Ax+By+Cz+D=0\text{ __(1)}$ This plane meets the x-axis at the point $(a,0,0)$. Hence, we have $Aa+D=0$ $A=\frac{-D}{a}$ Similarly, it meets the y-axis and z-axis at the point $(0,b,0)$ and $(0,0,z)$ respectively. Hence $B=\frac{-D}{b}\;\text{ and }\;C=\frac{-D}{c}$ Putting the values of $A$, $B$ and $C$ in equation $\text{(1)}$, we get $\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=0$ This is the required equation of the plane.

### The normal form

Let $ABC$ be a plane and let $P(x,y,z)$ be any point on this plane. From origin $O$, draw a perpendicular $ON$ $(=p)$ on the plane. Let $l,m,n$ be the direction cosines of $ON$. Since $ON$ is perpendicular to the plane, it must be perpendicular to every line lying on the plane. Join $NP$, then $\angle ONP=90°$.

Also, $ON$ is the projection of $OP$ on the line $ON$. Hence, $ON$ is equal to the sum of the projections of the segments of $OP$ on the line $ON$. $\therefore ON=lx+my+nz$ Since $ON=p$, we have, $lx+my+nz=p$ This is the required equation of the plane.

#### Reduction of the general equation of the plane to the normal form

The general equation of the plane is $Ax+By+Cz+D=0$ and, the normal form of the plane is $lx+my+nz-p=0$ Since they both represent the same plane, we have $\frac{l}{A}=\frac{m}{B}=\frac{n}{C}=\frac{-p}{D}\text{ __(i)}$

Hence, the direction cosines of the normal to the plane $Ax+By+Cz+D=0$ are proportional to $A,B,C$. Since the axes are rectangular, each of the ratios in $\text{(i)}$ is equal to $\pm\frac{1}{\sqrt{A^2+B^2+C^2}}$.

Here, $p$ is a positive number, if $D$ is also positive, then $\begin{array}{c}p=\frac{D}{\sqrt{A^2+B^2+C^2}},&l=\frac{-A}{\sqrt{A^2+B^2+C^2}}\\m=\frac{-B}{\sqrt{A^2+B^2+C^2}},&n=\frac{-C}{\sqrt{A^2+B^2+C^2}}\end{array}$ If $D$ is negative, then the sign of $\sqrt{A^2+B^2+C^2}$ must be changed.

### Plane through three given points

Let the equation of the plane be $Ax+By+Cz+D=0$ Let any three non collinear points be $(x_1,y_1,z_1)$, $(x_2,y_2,z_2)$ and $(x_3,y_3,z_3)$. Since these points lie on the given plane, we have $Ax_1+By_1+Cz_1+D=0$ $Ax_2+By_2+Cz_2+D=0$ $Ax_3+By_3+Cz_3+D=0$

Eliminating $A$, $B$, $C$ and $D$ from above relations, we get $\left|\begin{array}{c}x&y&z&1\\x_1&y_1&z_1&1\\x_2&y_2&z_2&1\\x_3&y_3&z_3&1\end{array}\right|=0$ This gives the required equation of the plane.

Consider another point $(x_4,y_4,z_4)$. Then, the condition that the fourth point is coplanar is $\left|\begin{array}{c}x_1&y_1&z_1&1\\x_2&y_2&z_2&1\\x_3&y_3&z_3&1\\x_4&y_4&z_4&1\end{array}\right|=0$

### Plane through the intersection of two planes

Let the equations of two given planes be $P=Ax+By+Cz+D=0$ $Q=A’x+B’y+C’z+D=0$ Then, the equation $P+\lambda Q=0$ represents a plane through the intersection of the given planes for all values of $\lambda$.