Perpendicular Distance Of A Point From A Plane

In the late Middle English, the word "perpendicular" was introduced. The meaning of this word was "at right angles". In Latin, the word is known as "perpendiculars" or "perpendiculum" which means "plumb line".

On the basis of geometrical point of view, when two objects are kept in a position and the distance between the objects is measured in a particular line, that is considered as perpendicular to one another, then the distance between the two objects is known as perpendicular distance.

Basics of Perpendicularity

When two distinct lines intersect with each other at 90°, or form a right angle, then the lines are said to be perpendicular to each other and are defined as "perpendicular lines".

Criteria that should be satisfied by the lines in order to be perpendicular are as follows: 

• The lines must intersect at right angles.

• If two lines are perpendicular to the same line, then they are said to be parallel to each other and will never intersect and hence will never become perpendicular to each other.

• The adjacent sides of a rectangular and square are always enclosed with right angles and hence they are always perpendicular to each other.

The points by which it can be easily predicted that the lines are not perpendicular and cannot be perpendicular are as follows:

• When two lines are parallel, then even though the lines will be drawn to infinity, still they will remain parallel and will never intersect with each other. This is why they can never be perpendicular to each other.

• In case, two lines are intersecting with each other but do not intersect with right angle, instead, they intersect with acute angles or any other angle, then the lines are also not perpendicular to each other.

• Even though lines intersect with each other but not at 90°, then also, the lines are not perpendicular to each other.

From the overview of the overall statements written above, the conclusion that can be drawn out is:

"All perpendicular lines are intersecting, but all intersecting lines are not compulsorily perpendicular to each other."

Perpendicular Distance of a Point From a Line in 3D

The shortest distance of one point from a plane is called to be along the line that is perpendicular to the plane, or in simpler words, is called as the perpendicular distance of that point from the given plane. Therefore, if you take a normal vector, say for example ň, to the given plane, a line that runs parallel to this vector that meets the point P would give you the shortest distance of that particular point from the plane. If you denote the point of intersection, for example, R, of the line touching the point P, and the plane on which it is said to fall normally, the point R is called as the point on the plane which is the closest to the point P. In this, the distance between the two points P and R would give you the distance of the point P from the plane. In this session, you would study how you can calculate the shortest distance of a point from a plane using two different methods called:

1. the Vector method

2. the Cartesian Method.

Distance of a Point From a Plane Using Vector Form

Consider for example a point A that has a position vector denoted by ȃ and which lies on a plane P. It is given by the equation,

\[\vec{r}.\vec{N}=d\]

Here, N is normal to the plane. If you consider another plane being parallel to the first one and which passes through the point A, you would get the equation of the second plane where N is normal to the plane. This equation can be denoted as,

\[(\vec{r}-\vec{a}).\vec{N}=0\]

In simpler words,

\[\vec{r}.\vec{N}=\vec{a}.\vec{N}\]

If you consider O to be the origin of the coordinates, the distance of the first plane from the origin is denoted as ON. 

Similar to this, the distance of the other plane from O is denoted as ON’. The distance between these two planes can be calculated by ON–ON’. This distance is equal to

\[ON-ON’=d’=\mid d-a.\overbrace{N}\mid \]

You can also find the perpendicular distance of point A on the plane P’ from P. Hence, you can conclude that for a plane which is denoted by the following equation

\[\vec{r}.\vec{N}=\vec{D}\]

and a point A whose position vector is known, you can calculate the perpendicular distance from a point to the plane with the formula given by

\[d=\frac{\mid\vec{r}.\vec{N}-D\mid}{\mid\vec{N}\mid}\]

If you want to calculate the length of the plane from origin O, you need to substitute 0 in the place of the position vector. Doing so, you would get,

\[d=\frac{\mid D\mid }{\mid N\mid }\]

Distance of a Point from a Plane with the Help of Cartesian Form

Consider a plane that is denoted by the Cartesian equation,

Px + Qy + Rz = S

Consider a point that has a position vector ȃ and whose Cartesian coordinates are given by,

A(x1, y1, z1).

You can write the position vector in the following form:

\[\vec{a}=x_1\widehat{i}+y_1\widehat{j}+z_1\widehat{k}\]

For finding the distance of point A from its plane by using the formula given in the vector form, you can find the normal vector to the plane, that is denoted as,

\[\vec{N}=A\widehat{i}+B\widehat{j}+C\widehat{k}\]

When you use the formula, the perpendicular distance of the point A from its given plane is denoted as,

\[d=\frac{\mid\vec{a}.\vec{N}-D\mid}{\mid\vec{N}\mid}\]

Substituting this given equation, you get,

d=\[\frac{|(x_{1}\widehat{i}+y_{1}\widehat{j}+z_{1}\widehat{k}).(A\widehat{i}+B\widehat{j}+C\widehat{k})-D|}{\sqrt{A^{2}+B^{2}+C^{2}}}\]

Reducing the equation further would give you,

d = \[|\frac{Ax_{1}+By_{1}+Cz_{1}-D}{\sqrt{A^{2}+B^{2}+C^{2}}}|\]

This is the equation that would give you the perpendicular distance of a given point from its plane by using the Cartesian method.

The Perpendicular Distance from a Point to a Line 3D

The distance from a point to a line is similar to the perpendicular distance between a line and a point. The perpendicular distance from a point to a line 3d formula is given below.

If M0(x0, y0, z0) are the point coordinates, 

s⎺ = {m; n; p} is the directing vector of line l, M1 (x1, y1, z1) gives you the coordinates of the point on the line l, then you can find the distance between point M0 (x0, y0, z0) and line l using the formula given below:

d = \[\frac{\mid M_{0}M_{1}\times s\mid }{\overline{s}}\]

Solved Examples:

Example 1: Find the distance between the given plane 2x + 4y – 4z – 6 = 0 and the point M(0, 3, 6).

Solution: 

To find the distance from a point to plane, use the perpendicular distance formula.

Solving this, you get =3

Hence, the distance from a point to the plane is equal to 3.

Example 2

Find distance between point M (0, 2, 3) and line in 3d given by,\[\frac{(X-3)}{2}=\frac{(Y-1)}{1}=\frac{(Z+1)}{2}\] .

Solution: 

From the equation of the line, first, find s⎺ = {2; 1; 2} which is the directing vector of line M1(3; 1; –1) which is the coordinates of the point on the line.

Then substituting these in the formula, 

\[\overline{M_0M_1}\] = {(3 − 0); (1 − 2); (−1 − 3)} = {3; −1; −4}

\[\overline{M_0M_1}\times \overline{S}=\begin{bmatrix}i & j & k \\3 & -1 & -4 \\2 & 1 & 2 \end{bmatrix}\]

Solving this you get

i ((−1)·2 − (−4)·1) − j (3·2 − (−4)·2) + k (3·1 −(−1)·2) = {2; −14; 5}

Hence, 

\[d=\frac{\mid\overline{M_0M_1}\times\overline{S}\mid}{\mid\overline{S}\mid}=\frac{\sqrt{ 2^2+(-14)^2+5^2}}{\sqrt {2^2+1^2+2^2}}=\frac{\sqrt {225}}{\sqrt {9}}=\frac{15}{3}=5\]

Therefore, the distance from the point to the line is 5.