The intercept equation of a plane is
$$\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1$$
where $x,y,z$ are the co-ordinates of an arbitrary point (say,$P$) on the plane and $a,b,c$ are the $x$-axis, $y$-axis and $z$-axis intercepts of the plane respectively
I was wondering the relation between the intercepts and the plane's distance from the origin. I tried in the following manner:
The general equation of the plane in Cartesian form is as follows:
$Ax+By+Cz+D=0$
Where $A,B,C$ are the direction ratios of the normal vector to the plane from the origin and $x,y,z$ are the same as above.
When we compare these two equations, we get
$A=\frac{1}{a}$, $B=\frac{1}{b}$ and $C=\frac{1}{c}$ and $D=-1$
Clearly, we can state that the vector, $$\frac{1}{a}î+\frac{1}{b}\hat{j}+\frac{1}{c}\hat{k}$$ is perpendicular to the plane from the origin
But if that is the case, then it must satisfy the identity
$\vec{r}.\hat{n}=d$
Where $\vec{r}$ is the position vector of the poin $P$ and $\hat{n}$ is the unit normal vector to the plane from the origin and $d$ is the distance to the plane from the origin
To convert $$\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1$$ of the form $\vec{r}.\hat{n}=d$, we state
$$\frac{1}{\sqrt{a^2+b^2+c^2}}(\frac{x}{a}+\frac{y}{b}+\frac{z}{c})=\frac{1}{\sqrt{a^2+b^2+c^2}}$$
Where the distance to the plane from the origin is $\frac{1}{\sqrt{a^2+b^2+c^2}}$
This is the general expression I have deduced of the distance of the plane. Is this correct? If not, please tell the correct relationship between the intercepts and the distance to the plane. I think I am mistaken in the first step itself.
Best Answer
Given a plane whose equation is $$ Ax + By + Cz + D = 0, \tag{0} $$ the distance of this plane from the point $P_0\left( x_0, y_0, z_0 \right)$ is $$ \frac{ \left\lvert Ax_0 + By_0 + Cz_0 + D \right\rvert }{\sqrt{A^2 + B^2 + C^2 }}. \tag{1} $$ So if we put $x_0 = y_0 = z_0 = 0$ into (1), then the distance of the plane given by (0) from the origin is $$ \frac{ \lvert D \rvert }{ \sqrt{ A^2 + B^2 + C^2 }}. \tag{2}. $$ Now the equation of your plane can be rewritten as $$ \frac{x}{a} + \frac{y}{b} + \frac{z}{c} -1 = 0. \tag{3} $$ Therefore the distance of the plane given by (3) from the origin is $$ \frac{ \lvert -1 \rvert }{ \sqrt{ \frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2} } } = \frac{ 1 }{ \sqrt{ \frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2} } }. $$