Just like how in plane two-dimensional geometry we talk of parallel lines that are equidistant from each other everywhere, in the same way in three-dimensional space geometry, we talk of equidistant planes. Such planes can also be called parallel planes. Think of the floor and ceiling of your house. They are parallel planes or equidistant planes. The distance between them at any point in the room would be same. That is why they are called equidistant planes.

Shown above is a picture of the Great Wall of China. Note the equidistant planes in there.

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Consider two planes in three-dimensional space. If the perpendicular distance between the two planes remains constant everywhere even if the planes are extended infinitely in all directions, then the two planes are said to be equidistant. They can also be called parallel planes. Another way of defining equidistant planes would be: If two planes are such that they share the same normal, then the planes are said to be parallel or equidistant.

See the above figure, there are two planes $l_1$ and $l$. There is one normal $\overleftrightarrow{DC}$, that is normal to both the planes. Thus both the planes can have a common normal. We can say that the planes are equidistant or parallel.

Normals of intersecting planes would intersect in exactly one point as shown in the figure below:

If the position vector of a point on a plane is $r_0$ ($x_0$, $y_0$, $z_0$) and the normal vector to the plane is $n$($a$, $b$, $c$), then the equation of the plane can be given by the vector equation:

$n∙(r\ -\ r_0)$ = $0$

If we replace the dot product with the components of the respective vectors the equation would look like this:

$a(x\ -\ x_0)\ +\ b(y\ -\ y_0)\ +\ c(z\ -\ z_0)$ = $0$

This can also be called the scalar equation.

Now if we have another plane that is equidistant from this plane. In other words, the new plane is parallel to this plane, then its normal vector would be the same. That is: n(a, b, c). If the position vector of a point on this new plane is $r_1$ ($x_1$, $y_1$, $z_1$) then the vector equation of this plane would be:

$n\ ∙\ (r\ -\ r_1)$ = $0$

On replacing the vectors with the components and doing the dot product, we get the scalar equation as:

$a(x\ -\ x_1)\ +\ b(y\ -\ y_1)\ +\ c(z\ -\ z_1)$ = $0$

Consider two equidistant planes with the equations:

$ax\ +\ by\ +\ cz\ +\ d_1$ = $0$

And

$ax\ +\ by\ +\ cz\ +\ d_2$ = $0$

Note that since the two planes are parallel, the normal vector of both the planes

would be $n\ (a,\ b,\ c)$. Now if we have a point $(x_1,\ y_1,\ z_1)$ in the first plane, then the perpendicular distance between this point and the other plane would be the distance between the two planes. The formula for distance between a point and a plane is:

$D$ = $\frac{|ax_1\ +\ by_1\ +\ cz_1\ +\ d_1|}{\sqrt{a^2\ +\ b^2\ +\ c^2}}$

Using this formula we can find the required distance.

Equidistant projection is a map projection. The distance between a particular point on a map and every other point on the same map bear a constant ratio with the corresponding distances on the sphere. This is called equidistant projection. So basically, there is a constant scale factor involved that is used to convert the distances on a sphere to those on the map. The particular point with respect to which the distances are measured on the map can also be two particular points instead. If this specific point is at the center of the map, then it is called the Azimuthal equidistant projection. The logo of the united nations is a classic example of this projection:Now we shall learn how to find the equation of a plane that is equidistant from two points. Suppose the two points between which we wish to find the equidistant plane are $(x_1,\ y_1,\ z_1)$ and $(x_2,\ y_2,\ z_2)$. Then the mid-point of the line segment joining these two points would be:

Mid point = $(\frac{x_1\ +\ x_2}{2},\ \frac{y_1\ +\ y_2}{2},\ \frac{z_1\ +\ z_2}{2})$

This midpoint would lie on the plane of which we wish to find the equation of.

The direction vector between the two point would also be a normal vector to the required plane. Thus,

$n$ = $(x_1\ -\ x_2,\ y_1\ -\ y_2,\ z_1\ -\ z_2)$

So for our plane, we have the normal vector n and we have $r(x_0,\ y_0,\ z_0)$ = $(\frac{x_1\ +\ x_2}{2},\ \frac{y_1\ +\ y_2}{2},\ \frac{z_1\ +\ z_2}{2})$. Thus using the formula for vector equation of a plane we have:

$n\ ∙\ (r\ -\ r_0)$ = $0$

Substituting the values we have:

$(x_1\ -\ x_2,\ y_1\ -\ y_2,\ z_1\ -\ z_2 )\ ∙$ $(x\ -\ \frac{x_1\ +\ x_2}{2},\ y\ -\ \frac{y_1\ +\ y_2}{2},\ z\ -\ \frac{z_1\ +\ z_2}{2})$ = $0$

This dot product can be evaluated to get the required equation of the plane equidistant from the two given points $(x_1,\ y_1,\ z_1)$ and $(x_2,\ y_2,\ z_2)$.

Let us now take a look at how to find the equation of a set of points that are equidistant from two given planes. Suppose the equations of the planes given to us are $v_1\ (x,\ y,\ z)$ = $a$ and $v_2\ (x,\ y,\ z)$ = $b$. Now if the two normal vectors, v_1 and v_2 of these two planes are proportional to each other, then that means that the two planes are parallel. In that case the set of points equidistant from these two planes would be a third plane in between these two planes that is parallel to those planes. The equation of such a plane would be:

$v_1\ (x,\ y,\ z)$ = $\frac{a\ +\ b}{2}$

In another scenario, where the two planes are not parallel to each other, then obviously the two planes intersect. In this case, the planes that bisect the angle between the two given planes would contain all the points that are equidistant from the two given planes. To find the equation of these planes, we first find the normal vectors to these planes and then use them to get the required equation of the bisecting planes.

$z$ = $2x\ +\ 5y\ +\ 1$ and $6x\ +\ 10y\ -\ 3z$ = $5$

To find a point (any point) that lies on the first plane we plug in $x$ = $0$ and $y$ = $0$. So,

$z$ = $2\ (0)\ +\ 5\ (0)\ +\ 1$ = $1$

Thus the point $(0,\ 0,\ 1)$ lies on the first plane. The distance between this point and the second plane is given by the formula:

$D$ = $\frac{|\ ax_1\ +\ by_1\ +\ cz_1\ +\ d_1\ |}{\sqrt{a^2\ +\ b^2\ +\ c^2}}$

Substituting the values $(x_1,\ y_1,\ z_1)$ = $(0,\ 0,\ 1)$ and $(a,\ b,\ c)$ = $(6,\ 10,\ -3)$. For the second plane, $d_1$ = $-5$. Thus,

$D$ = $\frac{|6\ (0)\ +\ 10\ (0)\ -\ 3(1)\ -\ 5\ |}{\sqrt{6^2\ +\ 10^2\ +\ (-3)^2)}}$

$D$ = $\frac{|-8|}{\sqrt{145}}$

$D$ = $\frac{8}{\sqrt{145}}$ $\leftarrow$Answer!

The mid-point of the line joining these two points would be:

$(\frac{1\ -\ 1}{2},\ \frac{2\ +\ 1}{2},\ \frac{0\ +\ 1}{2})$

= $(0,\ 1.5,\ 0.5)$

The direction vector of the line joining these two points would be:

$((-1\ -1),\ (1\ -\ 2),\ (1\ -\ 0))$

= $(-2,\ -1,\ 1)$

So for this problem,

$n$ = $(-2,\ -1,\ 1)$

And

$r(x_0,\ y_0,\ z_0)$ = $(0,\ 1.5,\ 0.5)$

Thus the equation of the required plane would be:

$n\ ∙\ (x\ -\ x_0,\ y\ -\ y_0,\ z\ -\ z_0)$ = $0$

$(-2,\ -1,\ 1)\ ∙\ (x\ -\ 0,\ y\ -\ 1.5,\ z\ -\ 0.5)$ = $0$

$-2(x)\ -1\ (y\ -\ 1.5)\ +\ 1\ (z\ -\ 0.5)$ = $0$

$-2x\ -\ y\ +\ z\ +\ 1$ = $0$ $\leftarrow$ Answer!

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