Class 11 Mathematics | Written by Neeraj Anand
Published by ANAND TECHNICAL PUBLISHERS
Introduction
In Three-Dimensional Geometry, points are represented by three coordinates: x, y, and z. These coordinates help describe the exact position of a point in 3D space. One of the fundamental concepts in 3D geometry is finding the distance between two points in space using their coordinates.
Distance Between Two Points in Three Dimensional Space (3D)
For two points A(x1, y1, z1) and B(x2, y2, z2) located in three-dimensional space, the 3D distance formula is given as below in Figure.
The distance between two points A(x1, y1, z1) and B(x2, y2, z2) is given by :
AB = √[(x2 – x1)2 + (y2 – y1)2 + (z2 – z1)2]
Distance between a Point and Origin in 3D
In case of origin one of the points will be O(0, 0, 0) and hence we will use x1 = 0, y1 = 0 and z1 = 0 and other point A having coordinate x2 = x, y2 = y and z2 = z in the formula to calculate the distance OA.
OA = √[x2 + y2 + z2]
Proof : Derive The Formula For Distance Between 2 Points in 3D
Let the points O(0, 0, 0), P(x1, y1, z1) and Q (x2, y2, z2) be referred to a system of rectangular axes OX,OY and OZ as shown in the figure.
Through the points P and Q, we draw planes parallel to the rectangular coordinate plane such that we get a rectangular parallelepiped with PQ as the diagonal. ∠PAQ forms a right angle.
Using the Pythagoras theorem in triangle PAQ,
\(\begin{array}{l}PQ^2= PA^2+AQ^2 …(1)\end{array} \)
Also, in triangle ANQ, ∠ANQ is a right angle.
Similarly, applying the Pythagoras theorem in ΔANQ we get,
\(\begin{array}{l}AQ^2=AN^2+NQ^2 ….(2)\end{array} \)
From equations 1 and 2 we have,
\(\begin{array}{l}PQ^2=PA^2+NQ^2+AN^2\end{array} \)
As coordinates of the points, P and Q are known,
\(\begin{array}{l}PA=y_{2}-y_{1},\ AN=x_{2}-x_{1} \ and \ NQ=z_{2}-z_{1}\end{array} \)
Therefore,
\(\begin{array}{l}PQ^2=(x_2-x_1 )^2+(y_2-y_1 )^2+(z_2-z_1 )^2\end{array} \)
Thus the formula gives us the distance between two points P(x1, y1, z1) and Q (x2, y2, z2) in three dimensions is given by :
\(\begin{array}{l}PQ=\sqrt{(x_2-x_1 )^2+(y_2-y_1 )^2+(z_2-z_1 )^2}\end{array} \)
Distance of any point Q(x, y, z) in space from origin O(0, 0, 0), is given by,
\(\begin{array}{l}OQ=\sqrt{(x^2+y^2+z^2)}\end{array} \)
Solved Examples
Example.1 : Calculate the distance between P(-1, 2, 4) and origin O.
Solution:
We know that the coordinates for origin is O(0, 0, 0)
Hence, the Distance between origin O(0, 0, 0) and P(-1, 2, 4) is given as
OP = √(x2 + y2 + z2)
⇒ OP = √{(-1)2 + 22 + 42}
⇒ OP = √{1 + 4 + 16} = √21 units
Example.2 : Find the distance between two points P(2, 5, 6) and Q(3, 4, 7)
Answer:
For point P, x1 = 2, y1 = 5, z1 = 6
For point Q, x2 = 3, y2 = 4, z2 = 7
Distance between P and Q is given as
PQ = √[(x2 – x1)2 + (y2 – y1)2 + (z2 – z1)2]
Now filling the values of coordinates in the above formula we get
PQ = √[(3 – 2)2 + (4 – 5)2 + (7 – 6)2]
⇒ PQ = √(1 + 1 + 1) = √3 units
Example.3 : Find the distance between the two points given by P(6, 4, -3) and Q(2, -8, 3).
Solution:
Let the given points be:
P(6, 4, -3) = (x1, y1, z1)
Q(2, -8, 3) = (x2, y2, z2)
Using distance formula to find distance between the points P and Q,
\(\begin{array}{l}PQ=\sqrt{((x_2-x_1 )^2+(y_2-y_1 )^2+(z_2-z_1 )^2 )}\end{array} \)
\(\begin{array}{l}PQ=\sqrt{(6-2)^2+(4-(-8))^2+(-3-3)^2}\end{array} \)
\(\begin{array}{l}PQ=\sqrt{(16+144+36)}\end{array} \)
PQ = √196
PQ = 14
Example 4 : A, B, C are three points lying on the axes x,y and z respectively, and their distances from the origin are given as respectively; then find coordinates of the point which is equidistant from A, B, C and O.
Solution:
Let the required point be P(x, y, z).
Co-ordinates of the points A,B and C are given as (a, 0, 0), (0, b, 0), (0, 0, c) and (0, 0, 0).
As we know that the point P is equidistant from the given points.
Hence, PA = PB = PC = PO
Now, applying the distance formula for PO = PA, we get
\(\begin{array}{l}\sqrt{x^2+y^2+z^2}=\sqrt{(a-x)^2+y^2+z^2}\end{array} \)
\(\begin{array}{l}x^2+y^2+z^2=(a-x)^2+y^2+z^2\end{array} \)
\(\begin{array}{l}x^2=(a-x)^2\end{array} \)
\(\begin{array}{l}x= a/2\end{array} \)
Similarly applying the distance formula for PO = PB and PO=PC, we get
\(\begin{array}{l}y= \frac{b}{2}\end{array} \)
\(\begin{array}{l}z= \frac{c}{2}\end{array} \)
Therefore co-ordinates of the point which are equidistant from the points A,B,C and O is given by
\(\begin{array}{l}(\frac{a}{2},\;\frac{b}{2},\;\frac{c}{2}).\end{array} \)
NCERT Solutions
1. Find the distance between the following pairs of points:
(i) (2, 3, 5) and (4, 3, 1)
(ii) (–3, 7, 2) and (2, 4, –1)
(iii) (–1, 3, –4) and (1, –3, 4)
(iv) (2, –1, 3) and (–2, 1, 3)
Solution:
(i) (2, 3, 5) and (4, 3, 1)
Let P be (2, 3, 5) and Q be (4, 3, 1)
By using the formula,
Distance PQ = √[(x2 – x1)2 + (y2 – y1)2 + (z2 – z1)2]
So, here
x1 = 2, y1 = 3, z1 = 5
x2 = 4, y2 = 3, z2 = 1
Distance PQ = √[(4 – 2)2 + (3 – 3)2 + (1 – 5)2]
= √[(2)2 + 02 + (-4)2]
= √[4 + 0 + 16]
= √20
= 2√5
∴ The required distance is 2√5 units.
(ii) (–3, 7, 2) and (2, 4, –1)
Let P be (– 3, 7, 2) and Q be (2, 4, –1)
By using the formula,
Distance PQ = √[(x2 – x1)2 + (y2 – y1)2 + (z2 – z1)2]
So, here
x1 = – 3, y1 = 7, z1 = 2
x2 = 2, y2 = 4, z2 = – 1
Distance PQ = √[(2 – (-3))2 + (4 – 7)2 + (-1 – 2)2]
= √[(5)2 + (-3)2 + (-3)2]
= √[25 + 9 + 9]
= √43
∴ The required distance is √43 units.
(iii) (–1, 3, –4) and (1, –3, 4)
Let P be (– 1, 3, – 4) and Q be (1, – 3, 4)
By using the formula,
Distance PQ = √[(x2 – x1)2 + (y2 – y1)2 + (z2 – z1)2]
So here,
x1 = – 1, y1 = 3, z1 = – 4
x2 = 1, y2 = – 3, z2 = 4
Distance PQ = √[(1 – (-1))2 + (-3 – 3)2 + (4 – (-4))2]
= √[(2)2 + (-6)2 + (8)2]
= √[4 + 36 + 64]
= √104
= 2√26
∴ The required distance is 2√26 units.
(iv) (2, –1, 3) and (–2, 1, 3)
Let P be (2, – 1, 3) and Q be (– 2, 1, 3)
By using the formula,
Distance PQ = √[(x2 – x1)2 + (y2 – y1)2 + (z2 – z1)2]
So here,
x1 = 2, y1 = – 1, z1 = 3
x2 = – 2, y2 = 1, z2 = 3
Distance PQ = √[(-2 – 2)2 + (1 – (-1))2 + (3 – 3)2]
= √[(-4)2 + (2)2 + (0)2]
= √[16 + 4 + 0]
= √20
= 2√5
∴ The required distance is 2√5 units.
2. Show that the points (–2, 3, 5), (1, 2, 3) and (7, 0, –1) are collinear.
Solution:
If three points are collinear, then they lie on the same line.
First, let us calculate the distance between the 3 points
i.e., PQ, QR and PR
Calculating PQ
P ≡ (–2, 3, 5) and Q ≡ (1, 2, 3)
By using the formula,
Distance PQ = √[(x2 – x1)2 + (y2 – y1)2 + (z2 – z1)2]
So, here
x1 = – 2, y1 = 3, z1 = 5
x2 = 1, y2 = 2, z2 = 3
Distance PQ = √[(1 – (-2))2 + (2 – 3)2 + (3 – 5)2]
= √[(3)2 + (-1)2 + (-2)2]
= √[9 + 1 + 4]
= √14
Calculating QR
Q ≡ (1, 2, 3) and R ≡ (7, 0, – 1)
By using the formula,
Distance QR = √[(x2 – x1)2 + (y2 – y1)2 + (z2 – z1)2]
So, here
x1 = 1, y1 = 2, z1 = 3
x2 = 7, y2 = 0, z2 = – 1
Distance QR = √[(7 – 1)2 + (0 – 2)2 + (-1 – 3)2]
= √[(6)2 + (-2)2 + (-4)2]
= √[36 + 4 + 16]
= √56
= 2√14
Calculating PR
P ≡ (–2, 3, 5) and R ≡ (7, 0, – 1)
By using the formula,
Distance PR = √[(x2 – x1)2 + (y2 – y1)2 + (z2 – z1)2]
So here,
x1 = –2, y1 = 3, z1 = 5
x2 = 7, y2 = 0, z2 = –1
Distance PR = √[(7 – (-2))2 + (0 – 3)2 + (-1 – 5)2]
= √[(9)2 + (-3)2 + (-6)2]
= √[81 + 9 + 36]
= √126
= 3√14
Thus, PQ = √14, QR = 2√14 and PR = 3√14
So, PQ + QR = √14 + 2√14
= 3√14
= PR
∴ The points P, Q and R are collinear.
3. Verify the following:
(i) (0, 7, –10), (1, 6, –6), and (4, 9, –6) are the vertices of an isosceles triangle.
(ii) (0, 7, 10), (–1, 6, 6), and (–4, 9, 6) are the vertices of a right-angled triangle.
(iii) (–1, 2, 1), (1, –2, 5), (4, –7, 8), and (2, –3, 4) are the vertices of a parallelogram.
Solution:
(i) (0, 7, –10), (1, 6, –6), and (4, 9, – 6) are the vertices of an isosceles triangle.
Let us consider the points,
P(0, 7, –10), Q(1, 6, –6) and R(4, 9, –6)
If any 2 sides are equal, it will be an isosceles triangle
So, first, let us calculate the distance of PQ, QR
Calculating PQ
P ≡ (0, 7, – 10) and Q ≡ (1, 6, – 6)
By using the formula,
Distance PQ = √[(x2 – x1)2 + (y2 – y1)2 + (z2 – z1)2]
So, here,
x1 = 0, y1 = 7, z1 = – 10
x2 = 1, y2 = 6, z2 = – 6
Distance PQ = √[(1 – 0)2 + (6 – 7)2 + (-6 – (-10))2]
= √[(1)2 + (-1)2 + (4)2]
= √[1 + 1 + 16]
= √18
Calculating QR
Q ≡ (1, 6, – 6) and R ≡ (4, 9, – 6)
By using the formula,
Distance QR = √[(x2 – x1)2 + (y2 – y1)2 + (z2 – z1)2]
So, here
x1 = 1, y1 = 6, z1 = – 6
x2 = 4, y2 = 9, z2 = – 6
Distance QR = √[(4 – 1)2 + (9 – 6)2 + (-6 – (-6))2]
= √[(3)2 + (3)2 + (-6+6)2]
= √[9 + 9 + 0]
= √18
Hence, PQ = QR
18 = 18
2 sides are equal
∴ PQR is an isosceles triangle.
(ii) (0, 7, 10), (–1, 6, 6), and (–4, 9, 6) are the vertices of a right-angled triangle.
Let the points be
P(0, 7, 10), Q(– 1, 6, 6) & R(– 4, 9, 6)
First, let us calculate the distance of PQ, OR and PR
Calculating PQ
P ≡ (0, 7, 10) and Q ≡ (– 1, 6, 6)
By using the formula,
Distance PQ = √[(x2 – x1)2 + (y2 – y1)2 + (z2 – z1)2]
So here,
x1 = 0, y1 = 7, z1 = 10
x2 = –1, y2 = 6, z2 = 6
Distance PQ = √[(-1 – 0)2 + (6 – 7)2 + (6 – 10)2]
= √[(-1)2 + (-1)2 + (-4)2]
= √[1 + 1 + 16]
= √18
Calculating QR
Q ≡ (1, 6, –6) and R ≡ (4, 9, –6)
By using the formula,
Distance QR = √[(x2 – x1)2 + (y2 – y1)2 + (z2 – z1)2]
So, here
x1 = 1, y1 = 6, z1 = –6
x2 = 4, y2 = 9, z2 = –6
Distance QR = √[(4 – 1)2 + (9 – 6)2 + (-6 – (-6))2]
= √[(3)2 + (3)2 + (-6+6)2]
= √[9 + 9 + 0]
= √18
Calculating PR
P ≡ (0, 7, 10) and R ≡ (– 4, 9, 6)
By using the formula,
Distance PR = √[(x2 – x1)2 + (y2 – y1)2 + (z2 – z1)2]
So, here
x1 = 0, y1 = 7, z1 = 10
x2 = – 4, y2 = 9, z2 = 6
Distance PR = √[(-4 – 0)2 + (9 – 7)2 + (6 – 10)2]
= √[(-4)2 + (2)2 + (-4)2]
= √[16 + 4 + 16]
= √36
Now,
PQ2 + QR2 = 18 + 18
= 36
= PR2
By using the converse of Pythagoras theorem,
∴ The given vertices P, Q & R are the vertices of a right–angled triangle at Q.
(iii) (–1, 2, 1), (1, –2, 5), (4, –7, 8), and (2, –3, 4) are the vertices of a parallelogram.
Let the points: A(–1, 2, 1), B(1, –2, 5), C(4, –7, 8) & D(2, –3, 4)
ABCD can be vertices of parallelogram only if opposite sides are equal.
i.e., AB = CD and BC = AD
First, let us calculate the distance
Calculating AB
A ≡ (–1, 2, 1) and B ≡ (1, –2, 5)
By using the formula,
Distance AB = √[(x2 – x1)2 + (y2 – y1)2 + (z2 – z1)2]
So, here
x1 = – 1, y1 = 2, z1 = 1
x2 = 1, y2 = – 2, z2 = 5
Distance AB = √[(1 – (-1))2 + (-2 – 2)2 + (5 – 1)2]
= √[(2)2 + (-4)2 + (4)2]
= √[4 + 16 + 16]
= √36
= 6
Calculating BC
B ≡ (1, –2, 5) and C ≡ (4, –7, 8)
By using the formula,
Distance BC = √[(x2 – x1)2 + (y2 – y1)2 + (z2 – z1)2]
So, here
x1 = 1, y1 = – 2, z1 = 5
x2 = 4, y2 = – 7, z2 = 8
Distance BC = √[(4 – 1)2 + (-7 – (-2))2 + (8 – 5)2]
= √[(3)2 + (-5)2 + (3)2]
= √[9 + 25 + 9]
= √43
Calculating CD
C ≡ (4, –7, 8) and D ≡ (2, –3, 4)
By using the formula,
Distance CD = √[(x2 – x1)2 + (y2 – y1)2 + (z2 – z1)2]
So here,
x1 = 4, y1 = –7, z1 = 8
x2 = 2, y2 = –3, z2 = 4
Distance CD = √[(2 – 4)2 + (-3 – (-7))2 + (4 – 8)2]
= √[(-2)2 + (4)2 + (-4)2]
= √[4 + 16 + 16]
= √36
= 6
Calculating DA
D ≡ (2, –3, 4) and A ≡ (–1, 2, 1)
By using the formula,
Distance DA = √[(x2 – x1)2 + (y2 – y1)2 + (z2 – z1)2]
So, here
x1 = 2, y1 = – 3, z1 = 4
x2 = –1, y2 = 2, z2 = 1
Distance DA = √[(-1 – 2)2 + (2 – (-3))2 + (1 – 4)2]
= √[(-3)2 + (5)2 + (-3)2]
= √[9 + 25 + 9]
= √43
Since AB = CD and BC = DA (given),
In ABCD, both pairs of opposite sides are equal.
∴ ABCD is a parallelogram.
4. Find the equation of the set of points which are equidistant from the points (1, 2, 3) and (3, 2, –1).
Solution:
Let A (1, 2, 3) & B (3, 2, –1)
Let point P be (x, y, z)
Since it is given that point P(x, y, z) is equal distance from point A(1, 2, 3) & B(3, 2, –1)
i.e. PA = PB
First, let us calculate
Calculating PA
P ≡ (x, y, z) and A ≡ (1, 2, 3)
By using the formula,
Distance PA = √[(x2 – x1)2 + (y2 – y1)2 + (z2 – z1)2]
So, here
x1 = x, y1 = y, z1 = z
x2 = 1, y2 = 2, z2 = 3
Distance PA = √[(1 – x)2 + (2 – y)2 + (3 – z)2]
Calculating PB
P ≡ (x, y, z) and B ≡ (3, 2, –1)
By using the formula,
Distance PB = √[(x2 – x1)2 + (y2 – y1)2 + (z2 – z1)2]
So, here
x1 = x, y1 = y, z1 = z
x2 = 3, y2 = 2, z2 = –1
Distance PB = √[(3 – x)2 + (2 – y)2 + (-1 – z)2]
Since PA = PB
Square on both sides, we get
PA2 = PB2
(1 – x)2 + (2 – y)2 + (3 – z)2 = (3 – x)2 + (2 – y)2 + (– 1 – z)2
(1 + x2 – 2x) + (4 + y2 – 4y) + (9 + z2 – 6z)
(9 + x2 – 6x) + (4 + y2 – 4y) + (1 + z2 + 2z)
– 2x – 4y – 6z + 14 = – 6x – 4y + 2z + 14
4x – 8z = 0
x – 2z = 0
∴ The required equation is x – 2z = 0.
5. Find the equation of the set of points P, the sum of whose distances from A(4, 0, 0) and B(–4, 0, 0) is equal to 10.
Solution:
Let A(4, 0, 0) & B(– 4, 0, 0)
Let the coordinates of point P be (x, y, z)
Calculating PA
P ≡ (x, y, z) and A ≡ (4, 0, 0)
By using the formula,
Distance PA = √[(x2 – x1)2 + (y2 – y1)2 + (z2 – z1)2]
So, here
x1 = x, y1 = y, z1 = z
x2 = 4, y2 = 0, z2 = 0
Distance PA = √[(4– x)2 + (0 – y)2 + (0 – z)2]
Calculating PB,
P ≡ (x, y, z) and B ≡ (– 4, 0, 0)
By using the formula,
Distance PB = √[(x2 – x1)2 + (y2 – y1)2 + (z2 – z1)2]
So here,
x1 = x, y1 = y, z1 = z
x2 = – 4, y2 = 0, z2 = 0
Distance PB = √[(-4– x)2 + (0 – y)2 + (0 – z)2]
It is given that,
PA + PB = 10
PA = 10 – PB
Square on both sides, we get
PA2 = (10 – PB)2
PA2 = 100 + PB2 – 20 PB
(4 – x)2 + (0 – y)2 + (0 – z)2
100 + (– 4 – x)2 + (0 – y)2 + (0 – z)2 – 20 PB
(16 + x2 – 8x) + (y2) + (z2)
100 + (16 + x2 + 8x) + (y2) + (z2) – 20 PB
20 PB = 16x + 100
5 PB = (4x + 25)
Square on both sides again, we get
25 PB2 = 16x2 + 200x + 625
25 [(– 4 – x)2 + (0 – y)2 + (0 – z)2] = 16x2 + 200x + 625
25 [x2 + y2 + z2 + 8x + 16] = 16x2 + 200x + 625
25x2 + 25y2 + 25z2 + 200x + 400 = 16x2 + 200x + 625
9x2 + 25y2 + 25z2 – 225 = 0
∴ The required equation is 9x2 + 25y2 + 25z2 – 225 = 0.
Practice Problems
- Show that the points P (–2, 3, 5), Q (1, 2, 3) and R (7, 0, –1) are collinear using the distance formula.
- What is the distance between two points (3, 4, 5) and (–1, 3, –7)?
- Verify that (0, 7, –10), (1, 6, – 6) and (4, 9, – 6) are the vertices of an isosceles triangle.
- Calculate the distance between the points P(-1, 0, 1) and Q(2, -1, 3)
- Calculate the distance of the point (3, -5, 4) from the Origin.
3D Distance Formula FAQs
How is a Point represented in 3D?
A Point is represented in 3D by using three coordinates as P (x, y, z) where, x, y, z are the coordinates of point P along the x, y, and z axis
What is 3D Distance Formula?
3D Distance Formula is used to calculate the distance between two points, between a point and a line, and between a point and a plane in three-dimensional space.
What is Distance Formula between Two Points in 3D?
Distance formula between two points is 3D is given as PQ = √[(x2 – x1)2 + (y2 – y1)2 + (z2 – z1)2]
What is Distance Formula between a Point and the Origin in 3D?
Distance between a point and origin in 3D is given as PO = √(x2 + y2 +z2)
How to Derive Distance Formula for Two Points in 3D?
Distance formula for two points in 3D can be derived by drawing planes parallel to the coordinate axes such that the distance between two points appears as a part of a rectangular parallelepiped.
Applications of the Distance Formula
- Physics: Calculating the shortest path between two objects in space.
- Engineering: Used in designing 3D structures and models.
- Computer Graphics: Essential for rendering objects at accurate distances.
- Astronomy: Measuring distances between celestial bodies.
Summary
Understanding the Distance Between Two Points in Three-Dimensional Space is essential for mastering 3D geometry concepts. It is not only crucial for solving mathematical problems but also plays a significant role in real-world applications like architecture, physics, and computer graphics.
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