Introduction to Three Dimensional Geometry — Class 11 Maths Notes

Introduction to Three Dimensional Geometry · Class 11 Maths · 6 topics.

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Topics covered in Introduction to Three Dimensional Geometry

1.Introduction of Three Dimensional Geometry
Three-dimensional geometry, also known as 3D geometry, deals with objects that have three dimensions: length, width, and height. This branch of geometry extends the concepts of two-dimensional shapes to three-dimensional space, allowing us to study the properties and relationships of points, lines, planes, and figures in space.
Key Concepts
1. Coordinates in Space:
  - Point: Represented as (𝑥,𝑦,𝑧)(x,y,z) in three-dimensional space.
  - Coordinate Axes: The three perpendicular axes are the x-axis, y-axis, and z-axis.
  - Coordinate Planes: The xy-plane, yz-plane, and zx-plane divide the space into eight octants.
2. Distance Formula: To find the distance between two points (𝑥1,𝑦1,𝑧1)(x1,y1,z1) and (𝑥2,𝑦2,𝑧2)(x2,y2,z2):
  𝑑=(𝑥2−𝑥1)2+(𝑦2−𝑦1)2+(𝑧2−𝑧1)2d=(x2−x1)2+(y2−y1)2+(z2−z1)2
3. Section Formula: To find the coordinates of a point that divides the line segment joining (𝑥1,𝑦1,𝑧1)(x1,y1,z1) and (𝑥2,𝑦2,𝑧2)(x2,y2,z2) in the ratio 𝑚:𝑛m:n:
  (𝑚𝑥2+𝑛𝑥1𝑚+𝑛,𝑚𝑦2+𝑛𝑦1𝑚+𝑛,𝑚𝑧2+𝑛𝑧1𝑚+𝑛)(m+nmx2+nx1,m+nmy2+ny1,m+nmz2+nz1)
4. Direction Ratios and Cosines:
  - Direction Ratios (DRs): Proportional to the direction numbers of a line.
  - Direction Cosines (DCs): Cosines of the angles that a line makes with the coordinate axes.
5. Plane Equation: A plane in three-dimensional space can be represented as:
  𝑎𝑥+𝑏𝑦+𝑐𝑧+𝑑=0ax+by+cz+d=0
  where 𝑎,𝑏,𝑐a,b,c are the direction ratios of the normal to the plane.
Real-Life Example
Consider a drone flying in the air. To describe its position, we need three coordinates: its distance east (x), its distance north (y), and its altitude (z). This three-dimensional positioning allows us to accurately locate the drone in space, much like how 3D geometry allows us to analyze and locate points, lines, and planes in a three-dimensional context.
Step-by-Step Example
Let's find the distance between two points in space:
- Point A: (1,2,3)(1,2,3)
- Point B: (4,6,8)(4,6,8)
1. Identify the coordinates:
  - 𝑥1=1,𝑦1=2,𝑧1=3x1=1,y1=2,z1=3
  - 𝑥2=4,𝑦2=6,𝑧2=8x2=4,y2=6,z2=8
2. Apply the distance formula:
  𝑑=(4−1)2+(6−2)2+(8−3)2d=(4−1)2+(6−2)2+(8−3)2
3. Calculate the differences:
  𝑑=32+42+52d=32+42+52
4. Square the differences:
  𝑑=9+16+25d=9+16+25
5. Add and take the square root:
  𝑑=50=52d=50=52
Thus, the distance between points A and B is 5252 units.
Careers and Industries
- Engineering: Designing and analyzing structures, machines, and systems.
- Architecture: Creating 3D models of buildings and spaces.
- Computer Graphics: Developing 3D animations and visual effects.
- Geology: Studying the spatial distribution of geological formations.
- Robotics: Programming and navigating robots in three-dimensional space.
2.Coordinate Axes and Coordinate Planes in Three Dimensional Space
We have the three-dimensional coordinate system illustrated. Let's break down the key elements and their significance:
Coordinate Axes and Planes
1. Coordinate Axes:
  - X-axis: Represented by the line X'OX.
  - Y-axis: Represented by the line Y'OY.
  - Z-axis: Represented by the line Z'OZ.
2. Coordinate Planes:
  - XY-plane (XOY): The plane formed by the x-axis and y-axis.
  - YZ-plane (YOZ): The plane formed by the y-axis and z-axis.
  - ZX-plane (ZOX): The plane formed by the z-axis and x-axis.
Origin
- The point 𝑂O where the three axes intersect is called the origin of the coordinate system.
Octants
- The three coordinate planes divide the space into eight octants:
  1. I Octant (XOYZ): Positive x, positive y, positive z.
  2. II Octant (X'OYZ): Negative x, positive y, positive z.
  3. III Octant (X'OY'Z): Negative x, negative y, positive z.
  4. IV Octant (XOY'Z): Positive x, negative y, positive z.
  5. V Octant (XOYZ'): Positive x, positive y, negative z.
  6. VI Octant (X'OYZ'): Negative x, positive y, negative z.
  7. VII Octant (X'OY'Z'): Negative x, negative y, negative z.
  8. VIII Octant (XOY'Z'): Positive x, negative y, negative z.
Positivity and Negativity of Coordinates
- Distances measured upwards from the XY-plane along OZ are positive, and downwards along OZ' are negative.
- Distances measured to the right of the ZX-plane along OY are positive, and to the left along OY' are negative.
- Distances measured in front of the YZ-plane along OX are positive, and to the back along OX' are negative.
This three-dimensional coordinate system allows us to describe the location of points in space precisely. It's widely used in fields such as physics, engineering, computer graphics, and many more.
Real-Life Application Example
Consider a GPS system that determines your location using latitude, longitude, and altitude. These three parameters are analogous to the x, y, and z coordinates in the three-dimensional coordinate system. The GPS calculates your exact position in three-dimensional space, allowing for precise navigation and mapping.
3.Coordinates of a Point in Space
In three-dimensional space, the position of a point is described by three coordinates
(𝑥,𝑦,𝑧)(x,y,z). These coordinates are measured along three mutually perpendicular axes: X, Y, and Z.
Key Concepts
1. Origin: The point where all three axes intersect is called the origin, denoted by 𝑂O. Its coordinates are (0,0,0)(0,0,0).
2. Coordinate Axes:
  - X-axis: Runs horizontally from left to right (positive direction).
  - Y-axis: Runs horizontally from front to back (positive direction).
  - Z-axis: Runs vertically from bottom to top (positive direction).
3. Coordinate Planes:
  - XY-plane (XOY): Formed by the X and Y axes.
  - YZ-plane (YOZ): Formed by the Y and Z axes.
  - ZX-plane (ZOX): Formed by the Z and X axes.
Representation of a Point
A point 𝑃P in space can be represented by its coordinates (𝑥,𝑦,𝑧)(x,y,z), where:
- 𝑥x is the distance from the YZ-plane (measured along the X-axis).
- 𝑦y is the distance from the ZX-plane (measured along the Y-axis).
- 𝑧z is the distance from the XY-plane (measured along the Z-axis).
Diagram Explanation
Here is a diagram to help you visualize the coordinates of a point in space:
In this diagram:
- O is the origin.
- X, Y, Z are the coordinate axes.
- P is a point in space with coordinates (𝑥,𝑦,𝑧)(x,y,z).
- Lines are drawn from 𝑃P perpendicular to each of the axes, intersecting the axes at points 𝑀M, 𝑁N, and 𝐿L.
  - Point 𝑀M has coordinates (𝑥,𝑦,0)(x,y,0).
  - Point 𝑁N has coordinates (0,𝑦,𝑧)(0,y,z).
  - Point 𝐿L has coordinates (𝑥,0,𝑧)(x,0,z).
Steps to Locate a Point in Space
1. Start at the Origin: Begin at the origin 𝑂(0,0,0)O(0,0,0).
2. Move Along the X-axis: Move horizontally along the X-axis by 𝑥x units. If 𝑥x is positive, move to the right; if 𝑥x is negative, move to the left.
3. Move Along the Y-axis: From the new position, move horizontally along the Y-axis by 𝑦y units. If 𝑦y is positive, move forward; if 𝑦y is negative, move backward.
4. Move Along the Z-axis: From the new position, move vertically along the Z-axis by 𝑧z units. If 𝑧z is positive, move upwards; if 𝑧z is negative, move downwards.
The final position after these movements is the location of point 𝑃P with coordinates (𝑥,𝑦,𝑧)(x,y,z).
Real-Life Example
Consider a drone flying in a room. To specify its location, you need:
- How far it is from the left wall (X-coordinate).
- How far it is from the front wall (Y-coordinate).
- How high it is from the floor (Z-coordinate).
This three-dimensional coordinate system helps to accurately describe the drone's position in space.
4.Exercise Questions
1. A point is on the 𝑋X-axis. What are its 𝑦y-coordinate and 𝑧z-coordinates?
Explanation: A point on the 𝑋X-axis only has a non-zero 𝑥x-coordinate. Therefore, its 𝑦y-coordinate and 𝑧z-coordinates are both zero.
Answer:
- 𝑦y-coordinate: 0
- 𝑧z-coordinate: 0
2. A point is in the 𝑋𝑍XZ-plane. What can you say about its 𝑦y-coordinate?
Explanation: A point in the 𝑋𝑍XZ-plane has 𝑦y-coordinate equal to zero because it lies entirely in the plane formed by the 𝑋X and 𝑍Z axes.
Answer:
- 𝑦y-coordinate: 0
3. Name the octants in which the following points lie:
To determine the octant in which a point lies, consider the signs of its coordinates (𝑥,𝑦,𝑧)(x,y,z):
- I: 𝑥>0,𝑦>0,𝑧>0x>0,y>0,z>0
- II: 𝑥<0,𝑦>0,𝑧>0x<0,y>0,z>0
- III: 𝑥<0,𝑦<0,𝑧>0x<0,y<0,z>0
- IV: 𝑥>0,𝑦<0,𝑧>0x>0,y<0,z>0
- V: 𝑥>0,𝑦>0,𝑧<0x>0,y>0,z<0
- VI: 𝑥<0,𝑦>0,𝑧<0x<0,y>0,z<0
- VII: 𝑥<0,𝑦<0,𝑧<0x<0,y<0,z<0
- VIII: 𝑥>0,𝑦<0,𝑧<0x>0,y<0,z<0
Points:
1. (1,2,3)(1,2,3):
  - 𝑥>0,𝑦>0,𝑧>0x>0,y>0,z>0
  - Octant: I
2. (4,−2,3)(4,−2,3):
  - 𝑥>0,𝑦<0,𝑧>0x>0,y<0,z>0
  - Octant: IV
3. (4,−2,−5)(4,−2,−5):
  - 𝑥>0,𝑦<0,𝑧<0x>0,y<0,z<0
  - Octant: VIII
4. (−4,2,5)(−4,2,5):
  - 𝑥<0,𝑦>0,𝑧>0x<0,y>0,z>0
  - Octant: II
5. (−4,2,−5)(−4,2,−5):
  - 𝑥<0,𝑦>0,𝑧<0x<0,y>0,z<0
  - Octant: VI
6. (−4,−2,5)(−4,−2,5):
  - 𝑥<0,𝑦<0,𝑧>0x<0,y<0,z>0
  - Octant: III
7. (−3,−1,6)(−3,−1,6):
  - 𝑥<0,𝑦<0,𝑧>0x<0,y<0,z>0
  - Octant: III
8. (−2,−4,−7)(−2,−4,−7):
  - 𝑥<0,𝑦<0,𝑧<0x<0,y<0,z<0
  - Octant: VII
Summary of Octants:
1. (1,2,3)(1,2,3): I
2. (4,−2,3)(4,−2,3): IV
3. (4,−2,−5)(4,−2,−5): VIII
4. (−4,2,5)(−4,2,5): II
5. (−4,2,−5)(−4,2,−5): VI
6. (−4,−2,5)(−4,−2,5): III
7. (−3,−1,6)(−3,−1,6): III
8. (−2,−4,−7)(−2,−4,−7): VII
5.Distance Between Two Points
To understand the distance between two points in three-dimensional space, we can use the distance formula. Let's explain this concept using the provided figure.
Figure Explanation
In the figure, we have a rectangular box with the following points labeled:
- O: Origin (0,0,0)(0,0,0)
- P, Q, A, N: Points in space with coordinates
We'll derive the distance between points 𝑃P and 𝑄Q.
Distance Formula
The distance 𝑑d between two points (𝑥1,𝑦1,𝑧1)(x1,y1,z1) and (𝑥2,𝑦2,𝑧2)(x2,y2,z2) in three-dimensional space is given by: 𝑑=(𝑥2−𝑥1)2+(𝑦2−𝑦1)2+(𝑧2−𝑧1)2d=(x2−x1)2+(y2−y1)2+(z2−z1)2
Step-by-Step Calculation
1. Identify Coordinates:
  - Let’s assume the coordinates of points 𝑃P and 𝑄Q are:
    - 𝑃(𝑥1,𝑦1,𝑧1)P(x1,y1,z1)
    - 𝑄(𝑥2,𝑦2,𝑧2)Q(x2,y2,z2)
2. Coordinate Differences:
  - Calculate the differences in the coordinates:
    - Δ𝑥=𝑥2−𝑥1Δx=x2−x1
    - Δ𝑦=𝑦2−𝑦1Δy=y2−y1
    - Δ𝑧=𝑧2−𝑧1Δz=z2−z1
3. Square the Differences:
  - (Δ𝑥)2=(𝑥2−𝑥1)2(Δx)2=(x2−x1)2
  - (Δ𝑦)2=(𝑦2−𝑦1)2(Δy)2=(y2−y1)2
  - (Δ𝑧)2=(𝑧2−𝑧1)2(Δz)2=(z2−z1)2
4. Sum of Squares:
  - Sum these squares: (Δ𝑥)2+(Δ𝑦)2+(Δ𝑧)2=(𝑥2−𝑥1)2+(𝑦2−𝑦1)2+(𝑧2−𝑧1)2(Δx)2+(Δy)2+(Δz)2=(x2−x1)2+(y2−y1)2+(z2−z1)2
5. Square Root:
  - Take the square root of the sum to find the distance: 𝑑=(𝑥2−𝑥1)2+(𝑦2−𝑦1)2+(𝑧2−𝑧1)2d=(x2−x1)2+(y2−y1)2+(z2−z1)2
Application to the Figure
Assume the coordinates are:
- 𝑃(𝑥1,𝑦1,𝑧1)P(x1,y1,z1)
- 𝑄(𝑥2,𝑦2,𝑧2)Q(x2,y2,z2)
To find the distance between 𝑃P and 𝑄Q: 𝑑𝑃𝑄=(𝑥2−𝑥1)2+(𝑦2−𝑦1)2+(𝑧2−𝑧1)2dPQ=(x2−x1)2+(y2−y1)2+(z2−z1)2
Example Calculation
Let's assume:
- 𝑃(2,3,4)P(2,3,4)
- 𝑄(5,7,9)Q(5,7,9)
Then: Δ𝑥=5−2=3Δx=5−2=3 Δ𝑦=7−3=4Δy=7−3=4 Δ𝑧=9−4=5Δz=9−4=5
Calculate the squares: (Δ𝑥)2=32=9(Δx)2=32=9 (Δ𝑦)2=42=16(Δy)2=42=16 (Δ𝑧)2=52=25(Δz)2=52=25
Sum the squares: (Δ𝑥)2+(Δ𝑦)2+(Δ𝑧)2=9+16+25=50(Δx)2+(Δy)2+(Δz)2=9+16+25=50
Take the square root: 𝑑𝑃𝑄=50=52dPQ=50=52
Thus, the distance between points 𝑃P and 𝑄Q is 5252 units.
6.Exercise Questions
1. Find the distance between the following pairs of points:
The distance formula between two points (𝑥1,𝑦1,𝑧1)(x1,y1,z1) and (𝑥2,𝑦2,𝑧2)(x2,y2,z2) is given by: 𝑑=(𝑥2−𝑥1)2+(𝑦2−𝑦1)2+(𝑧2−𝑧1)2d=(x2−x1)2+(y2−y1)2+(z2−z1)2
(i) (2,3,5)(2,3,5) and (4,3,1)(4,3,1)
𝑑=(4−2)2+(3−3)2+(1−5)2=22+0+(−4)2=4+16=20=25d=(4−2)2+(3−3)2+(1−5)2=22+0+(−4)2=4+16=20=25
(ii) (−3,7,2)(−3,7,2) and (2,4,−1)(2,4,−1)
𝑑=(2−(−3))2+(4−7)2+(−1−2)2=(2+3)2+(−3)2+(−3)2=52+32+32=25+9+9=43d=(2−(−3))2+(4−7)2+(−1−2)2=(2+3)2+(−3)2+(−3)2=52+32+32=25+9+9=43
(iii) (−1,3,−4)(−1,3,−4) and (−1,−3,4)(−1,−3,4)
𝑑=(−1−(−1))2+(−3−3)2+(4−(−4))2=0+(−6)2+82=36+64=100=10d=(−1−(−1))2+(−3−3)2+(4−(−4))2=0+(−6)2+82=36+64=100=10
(iv) (2,−1,3)(2,−1,3) and (−2,1,3)(−2,1,3)
𝑑=(−2−2)2+(1−(−1))2+(3−3)2=(−4)2+22+0=16+4=20=25d=(−2−2)2+(1−(−1))2+(3−3)2=(−4)2+22+0=16+4=20=25
2. Show that the points (−2,3,5)(−2,3,5), (1,2,3)(1,2,3), and (7,0,−1)(7,0,−1) are collinear.
To show that points are collinear, we need to check if the direction ratios of the lines formed by these points are proportional.
Let the points be 𝐴(−2,3,5)A(−2,3,5), 𝐵(1,2,3)B(1,2,3), and 𝐶(7,0,−1)C(7,0,−1).
Vector 𝐴𝐵→AB
𝐴𝐵→=(1+2,2−3,3−5)=(3,−1,−2)AB=(1+2,2−3,3−5)=(3,−1,−2)
Vector 𝐵𝐶→BC
𝐵𝐶→=(7−1,0−2,−1−3)=(6,−2,−4)BC=(7−1,0−2,−1−3)=(6,−2,−4)
Check if 𝐴𝐵→AB and 𝐵𝐶→BC are proportional: 63=−2−1=−4−2=236=−1−2=−2−4=2
Since the direction ratios are proportional, the points are collinear.
3. Verify the following:
(i) (0,7,−10)(0,7,−10), (1,6,−6)(1,6,−6), and (4,9,−6)(4,9,−6) are the vertices of an isosceles triangle.
Calculate the distances between the points:
- 𝐴(0,7,−10)A(0,7,−10), 𝐵(1,6,−6)B(1,6,−6), and 𝐶(4,9,−6)C(4,9,−6).
Distance 𝐴𝐵AB:
𝑑𝐴𝐵=(1−0)2+(6−7)2+(−6−(−10))2=12+(−1)2+42=1+1+16=18=32dAB=(1−0)2+(6−7)2+(−6−(−10))2=12+(−1)2+42=1+1+16=18=32
Distance 𝐵𝐶BC:
𝑑𝐵𝐶=(4−1)2+(9−6)2+(−6−(−6))2=32+32+0=9+9=18=32dBC=(4−1)2+(9−6)2+(−6−(−6))2=32+32+0=9+9=18=32
Distance 𝐶𝐴CA:
𝑑𝐶𝐴=(4−0)2+(9−7)2+(−6−(−10))2=42+22+42=16+4+16=36=6dCA=(4−0)2+(9−7)2+(−6−(−10))2=42+22+42=16+4+16=36=6
Since 𝐴𝐵=𝐵𝐶=32AB=BC=32 and 𝐶𝐴=6CA=6, △𝐴𝐵𝐶△ABC is isosceles.
(ii) (0,7,10)(0,7,10), (−1,6,6)(−1,6,6), and (−4,9,6)(−4,9,6) are the vertices of a right-angled triangle.
Calculate the distances between the points:
- 𝐴(0,7,10)A(0,7,10), 𝐵(−1,6,6)B(−1,6,6), and 𝐶(−4,9,6)C(−4,9,6).
Distance 𝐴𝐵AB:
𝑑𝐴𝐵=(−1−0)2+(6−7)2+(6−10)2=1+1+16=18=32dAB=(−1−0)2+(6−7)2+(6−10)2=1+1+16=18=32
Distance 𝐵𝐶BC:
𝑑𝐵𝐶=(−4−(−1))2+(9−6)2+(6−6)2=9+9=18=32dBC=(−4−(−1))2+(9−6)2+(6−6)2=9+9=18=32
Distance 𝐶𝐴CA:
𝑑𝐶𝐴=(−4−0)2+(9−7)2+(6−10)2=16+4+16=36=6dCA=(−4−0)2+(9−7)2+(6−10)2=16+4+16=36=6
If 𝑎2+𝑏2=𝑐2a2+b2=c2 holds for the sides, it confirms a right triangle.
(32)2+(32)2=62(32)2+(32)2=62 18+18=3618+18=36
Since the condition holds, △𝐴𝐵𝐶△ABC is a right-angled triangle.
4. Find the equation of the set of points which are equidistant from the points (1,2,3)(1,2,3) and (3,2,−1)(3,2,−1).
Let 𝑃(𝑥,𝑦,𝑧)P(x,y,z) be the point equidistant from 𝐴(1,2,3)A(1,2,3) and 𝐵(3,2,−1)B(3,2,−1).
The distance 𝑃𝐴=𝑃𝐵PA=PB gives: (𝑥−1)2+(𝑦−2)2+(𝑧−3)2=(𝑥−3)2+(𝑦−2)2+(𝑧+1)2(x−1)2+(y−2)2+(z−3)2=(x−3)2+(y−2)2+(z+1)2
Squaring both sides: (𝑥−1)2+(𝑦−2)2+(𝑧−3)2=(𝑥−3)2+(𝑦−2)2+(𝑧+1)2(x−1)2+(y−2)2+(z−3)2=(x−3)2+(y−2)2+(z+1)2
Expand and simplify: 𝑥2−2𝑥+1+𝑦2−4𝑦+4+𝑧2−6𝑧+9=𝑥2−6𝑥+9+𝑦2−4𝑦+4+𝑧2+2𝑧+1x2−2x+1+y2−4y+4+z2−6z+9=x2−6x+9+y2−4y+4+z2+2z+1
−2𝑥+1−6𝑧+9=−6𝑥+9+2𝑧+1−2x+1−6z+9=−6x+9+2z+1
Combine like terms: −2𝑥−6𝑧+10=−6𝑥+2𝑧+10−2x−6z+10=−6x+2z+10
4𝑥+8𝑧=04x+8z=0
𝑥+2𝑧=0x+2z=0
So, the equation of the set of points is: 𝑥+2𝑧=0x+2z=0
5. Find the equation of the set of points 𝑃P, the sum of whose distances from 𝐴(4,0,0)A(4,0,0) and 𝐵(−4,0,0)B(−4,0,0) is equal to 10.
Let 𝑃(𝑥,𝑦,𝑧)P(x,y,z) be the point such that the sum of its distances from 𝐴(4,0,0)A(4,0,0) and 𝐵(−4,0,0)B(−4,0,0) is 10.
𝑃𝐴+𝑃𝐵=10PA+PB=10
(𝑥−4)2+𝑦2+𝑧2+(𝑥+4)2+𝑦2+𝑧2=10(x−4)2+y2+z2+(x+4)2+y2+z2=10
Let 𝑑1=(𝑥−4)2+𝑦2+𝑧2d1=(x−4)2+y2+z2 and 𝑑2=(𝑥+4)2+𝑦2+𝑧2d2=(x+4)2+y2+z2
We have: 𝑑1+𝑑2=10d1+d2=10
This is the equation of an ellipse with foci at 𝐴(4,0,0)A(4,0,0) and 𝐵(−4,0,0)B(−4,0,0), and the major axis length equal to 10.

Introduction to Three Dimensional Geometry — Class 11 Maths Notes

Topics covered in Introduction to Three Dimensional Geometry

1.Introduction of Three Dimensional Geometry

Key Concepts

Real-Life Example

Step-by-Step Example

Careers and Industries

2.Coordinate Axes and Coordinate Planes in Three Dimensional Space

Coordinate Axes and Planes

Origin

Octants

Positivity and Negativity of Coordinates

Real-Life Application Example

3.Coordinates of a Point in Space

In three-dimensional space, the position of a point is described by three coordinates

Key Concepts

Representation of a Point

Diagram Explanation

Steps to Locate a Point in Space

Real-Life Example

4.Exercise Questions

1. A point is on the 𝑋X-axis. What are its 𝑦y-coordinate and 𝑧z-coordinates?

2. A point is in the 𝑋𝑍XZ-plane. What can you say about its 𝑦y-coordinate?

3. Name the octants in which the following points lie:

Summary of Octants:

5.Distance Between Two Points

To understand the distance between two points in three-dimensional space, we can use the distance formula. Let's explain this concept using the provided figure.

Figure Explanation

Distance Formula

Step-by-Step Calculation

Application to the Figure

Example Calculation

6.Exercise Questions

1. Find the distance between the following pairs of points:

(i) (2,3,5)(2,3,5) and (4,3,1)(4,3,1)

(ii) (−3,7,2)(−3,7,2) and (2,4,−1)(2,4,−1)

(iii) (−1,3,−4)(−1,3,−4) and (−1,−3,4)(−1,−3,4)

(iv) (2,−1,3)(2,−1,3) and (−2,1,3)(−2,1,3)

2. Show that the points (−2,3,5)(−2,3,5), (1,2,3)(1,2,3), and (7,0,−1)(7,0,−1) are collinear.

Vector 𝐴𝐵→AB

Vector 𝐵𝐶→BC

3. Verify the following:

(i) (0,7,−10)(0,7,−10), (1,6,−6)(1,6,−6), and (4,9,−6)(4,9,−6) are the vertices of an isosceles triangle.

Distance 𝐴𝐵AB:

Distance 𝐵𝐶BC:

Distance 𝐶𝐴CA:

(ii) (0,7,10)(0,7,10), (−1,6,6)(−1,6,6), and (−4,9,6)(−4,9,6) are the vertices of a right-angled triangle.

Distance 𝐴𝐵AB:

Distance 𝐵𝐶BC:

Distance 𝐶𝐴CA:

4. Find the equation of the set of points which are equidistant from the points (1,2,3)(1,2,3) and (3,2,−1)(3,2,−1).

5. Find the equation of the set of points 𝑃P, the sum of whose distances from 𝐴(4,0,0)A(4,0,0) and 𝐵(−4,0,0)B(−4,0,0) is equal to 10.

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