 # Perpendicular Distance Of A Point From A Plane

### 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,

r.  N = d

In the diagram, 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,

( r- a) . N = 0

In simpler words,

r. N = a. 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’ = d - a.N

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

r. N = 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 =  $\frac{|a^{\rightarrow }.N^{\rightarrow } - D|}{|N^{\rightarrow }|}$

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 = $\frac{|D|}{|N^{\rightarrow }|}$

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 coordinate is given by,

A(x1 , y1 , z1)

You can write the position vector in the following form:

a=  x1 i^ + y1 j^ + z1 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,

N= A i^ + B j^ + C k^

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

d =  $\frac{|a^{\rightarrow }.N^{\rightarrow } - D|}{|N^{\rightarrow }|}$

Substituting this given equation, you get,

d = $\frac{|(x_{1}i^{\Lambda }+y_{1}j^{\Lambda }+z_{1}k^{\Lambda }).(Ai^{\Lambda }+Bj^{\Lambda }+Ck^{\Lambda })-D|}{\sqrt{A^{2}+B^{2}+C^{2}}}$

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 by,

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

d = $\frac{}{\frac{|M_{0}M_{1}\times s|}{\frac{}{s}}}$

### 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.

 d = |2·0 + 4·3 + (-4)·6 -6| = |0 + 12 - 24 - 6| = |-18| √4 + 16 + 16 √36 6

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,

(x - 3) /2 = (y - 1)/1 = (z + 1)/1

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,

M0M1

= {(3 - 0); (1 - 2); (-1 - 3)}

= {3; -1; -4}

 M0M1× s = i j k 3 -1 -4 2 1 2

Solving this you get,

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

Hence,

d = |M0M1 x s|÷|s| =( √22 + ( -14 )2 + 52 ) ÷( √22 + 12 + 22 )

= √225 √9

= 15/3 = 5

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

1. What is the Equation of a Plane Perpendicular to a Vector?

The equation of a plane perpendicular to a vector and which is passing through a point is denoted in the following manner:

The standard form for a plane in R3 is denoted by the equation A(x-x0) + B(y-y0) + C(z-z0) = 0,

where (A,B,C) is the normal vector to the plane and (x0,y0,z0) is the point which lies in the plane.

2. How to Find the Equation of a Plane Passing Through a Point Which is Perpendicular to a Line?

Since the line is perpendicular to the plane, the direction ratios or the cosine ratios of the direction of the plane are same as that of the line, however, since the plane is passing through a specific point, you can put that point in the equation and solve for the value of the constant term. This way you can find the equation.

Consider the following example:

Let the plane be perpendicular to the line aX+bY+cZ+d=0.

You can then denote the equation of the plane like aX+bY+cZ+D=0,

where the a,b, and c, is the same as that of the line and the only quantity not known is D.

Since the line is passing through a point, for example, point P (p,q,r) then it must satisfy the equation of the plane. Thus, when you put the point in the equation you get,

ap+bq+cr+D=0

In this equation, the only quantity which is not known is D. The value of D can be found from the above equation. When you put the value of D in the parent equation, you can find the equation of the plane.