Question

Let the plane V be defined by ax + by + cz + d = 0 with at least one of a; b or c different from zero
and d >= 0.
Then the distance between V and the origin is d/(a^2+b^2+c^2)^1/2 : Prove this statement.

Answer 1

Let us consider a point A whose position vector is given by $\vec{a}$ and a plane whose equation is given by

$\vec{r}. \vec{N} = \vec{D}$ , where $\vec{N}$ is the normal to the plane.

If we consider another plane to be parallel to our first plane and let this new plane pass through the point A, then the equation of the second plane, where $\vec{N’}$  is normal to the plane can be written as

$( \vec{r} – \vec{a} ) . \vec{N} = 0$ or, $\vec{r}. \vec{N} = \vec{a}. \vec{N}$

If we consider O to be the origin, then the distance of the first plane from the origin is given by ON. Similarly, the distance of the second plane from the origin is given by ON’. So, we can get the distance between the two planes as ON – ON’. So, the distance equals $|\vec{D} – \vec{a}.\vec{N} |$

So, for a plane whose equation is given by $\vec{r}. \vec{N} = \vec{D}$ , the distance of a point A whose position vector is given by $\vec{a}$ to the plane is given by $d = | \vec{a}. \vec{N} –\vec{D} | / | \vec{N} |$

Now, let us consider a plane V whose Cartesian equation is given by –

Ax + By + Cz = D

Then the position vector $\vec{a}$ of a point whose Cartesian coordinates are given by $(x_1,y_1,z_1)$

Now, the equation of the normal to the plane is –

$\vec{N} = A \hat{i} + B \hat{j} + C \hat{k}$

Thus,

$d = | \vec{a}. \vec{N} – D | / | \vec{N} |$

$= | (x_1 \hat{i} + y_1 \hat{j} + z_1 \hat{k} ) . (A \hat{i} + B \hat{j} + C \hat{k} ) – D | / (A^2 + B^2 + C^2)^{1/2}$

Here point A is the origin itself.

Thus, $(x_1,y_1,z_1)=(0,0,0)$

$\implies d= | D | / (A^2 + B^2 + C^2)^{1/2}$

Hence proved.