CBSE Class 11 » CBSE Class 11 Study Materials » Mathematics » Perpendicular Distance of a Point

Perpendicular Distance of a Point

The distance between two points is an important concept of physics. This article contains information about the perpendicular distance of a point from a line in 3d.

Distance of a point from a line can be considered as the shortest distance between two points on a line. The measured length of the line segment joining the point nearest point on the given line as the shortest length is called the perpendicular distance of a point in line. Knowing the distance between point to line can be beneficial in day to day life. In this article, we will learn about the perpendicular distance of a point from a line in 3d and the distance of a point from a line in vector form. 

Derivation Of Perpendicular Distance of a Point From a Line in 3D Formula

Using Vector Formula

For deriving the perpendicular distance of a point from a line in 3d formula, let’s assume a point A that has a position vector that lies on the plane P. This formula is given by the equation 

→ →

  1. N = d

N is normal to the plane. And if you assume another plane parallel and passing through the first one. Then you would get an equation of the second plane, where N is the normal to the plane. This equation can also be represented as

→→.  →

(r-a ). N =0

This could also be written as 

→ →  →→

  1.   N = a.  N

If we take O to be origin of the coordinates, the distance of first plane just like this the other plane distance can also be formed and it is given by 

ON= ON’= |d-A. Ñ|

You can also calculate the perpendicular distance from the above equation which is given by the below-mentioned value. 

d= |r.N-D| / N

And the length of the plane is given by

d= N

Using Cartesian Form

Assume that a plane is defined by the Cartesian equation,

Px + Qy + Rz = S

You can take a point that has a vector â and the Cartesian coordinates of this equation can be given as 

A (x1, y1, z1

You can also write the position vector in the given form 

→           ^       ^

a= x1î + y1j + z1k

For finding the distance of point A from plane by using the formula given. You can also find the normal vector to the plane.  

→.    ^    ^     ^

N= Ai + Bj + Ck

When you use the formula, the perpendicular distance can be given as follows:

            →       →

d= | a. N -D| / N

Substituting the given equation you get

 |(x1iˆ+y1jˆ+z1kˆ)|

d= (Aiˆ+Bjˆ+Ckˆ)−D/ √A2+B2+C2

Solved Examples On Perpendicular Distance Of a Point From a Line in 3D Formula

 

  • Question:

Determine the distance between M(0, 2, 3) and line

x – 3

y – 1

    =

z + 1

2

1

2

Solution:

You can find the answer using the line equation, 

s = {2; 1; 2} – directing vector of line;

M1(3; 1; -1) – coordinates of point on line.

Then

M0M1 = {3 – 0; 1 – 2; -1 – 3} = {3; -1; -4}

M0M1×s = 

i

j

k

  3  

  -1  

  -4  

  2  

  1  

  2  

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

d = 

|M0M1×s|

√22 + (-14)2 + 52

√225

15

 

|s|

√22 + 12 + 22

√9

3

Distance from point to line is equal to 5.

 

  • Question:

State the distance between  3x + 4y = 9 and 6x + 8y = 15.

Solution:

Given equations of lines are:

3x + 4y = 9….(i)

6x + 8y = 15 Or 3x + 4y = 15/2 ….(ii)

Let us find whether the given lines are parallel or not.

From (i),

4y  = -3x + 9

y = (-¾)x + (9/4)

Here, slope = m1 = -¾

From (ii),

8y = -6x + 15

y = (-6/8)x + (15/8)

y = (-¾)x + (15/8)

Here, slope = m2 = -¾

Thus, the slope of the given lines is equal so they are parallel to each other.

Now, by comparing with the standard form of parallel lines equations, we get:

A = 3, B = 4, C1 = -9, C2 = -15/2

d = |C1 – C2|/√(A2 + B2)

= |-9 + (15/2)|/√(9 + 16)

= |-18 + 15|/2√25

= |-3|/(2 × 5)

= 3/10

Therefore, the distance between the given lines is 3/10 units.

 

  • Question

Determine the distance from the point K(−3,7) to the line PQ y=(6/5)​ x + 2 using the distance of the point from a line formula.

Solution:

Let us express the given line in the standard form first.

The line PQ can be simplified as:

y=(6/5)​ x + 2

5y = 6x +10

Thus, 6x – 5y + 10 = 0

As per the distance of the point from the line formula, d = |A

x1 + B.y1 + C| / √(A2 + B2

Here, P= (x1,y1)

= (-3, 7), and A = 6, B =-5 and C = 10

d = |(6)(-3) + (-5)(7) + 10|/ √((6)2+(-5)2)

= |-18 -35 + 10|/ √(36 + 25)

= |-43|/√(61)

= |-5.506|

Conclusion

When two distinct lines intersect at 90° or form a right angle, the lines are said to be perpendicular to each other and are referred to as “perpendicular lines.” In this article, we have read about the perpendicular distance of a point from a line in 3d and the distance of a line from a point in 3d. 

faq

Frequently asked questions

Get answers to the most common queries related to the JEE Examination Preparation.

How do you calculate the perpendicular?

Ans: Since perpendicular lines have opposite-reciprocal slopes, we can check whether lines are para...Read full

How to determine the distance between a line and a plane.

Ans: If the solid line and the plane are parallel, the distance between them is measured by deducti...Read full

Define perpendicular vectors.

Ans: A vector perpendicular to a given vector is one in which and forms a right angle (voiced ̶...Read full