CBSE Class 11 Maths Chapter 12: Introduction to Three Dimensional Geometry NCERT Solutions

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CBSE Class 11 Maths Chapter 12 delves into the fascinating world of three-dimensional geometry. This chapter introduces the fundamental concepts needed to navigate and understand space. You'll learn how to pinpoint locations in 3D using coordinates, and how the signs of these coordinates determine which of the eight octants a point resides in. The solutions also guide you through finding the exact spot where a perpendicular line from a point meets the coordinate axes and planes. A crucial skill covered is calculating the distance between any two points in this 3D realm. These explanations are crafted to make grasping the principles of 3D geometry straightforward, building a solid base for spatial reasoning and coordinate systems. They offer clear, step-by-step methods to master the textbook's problems, ensuring you're well-prepared for your exams.

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Quick info

BoardCBSE
ClassClass 11
SubjectMaths Exemplar
Session2026
LanguageEnglish
TypeNCERT Solutions
ChapterChapter 12

Chapter summary

Chapter 12 of the NCERT Class 11 Maths textbook, 'Introduction to Three Dimensional Geometry,' lays the groundwork for understanding spatial relationships. The NCERT Solutions for this chapter focus on visualizing and representing points in three-dimensional space. Key exercises involve plotting points, determining octants, finding projections onto axes and planes, and applying the distance formula in 3D. These solutions offer a clear path to mastering these foundational concepts.

Learning outcomes

  • Understand the concept of coordinates in three-dimensional space.
  • Locate points in different octants of the Cartesian coordinate system.
  • Determine the octant of a point based on the signs of its coordinates.
  • Find the coordinates of the feet of perpendiculars from a point to the axes.
  • Find the coordinates of the feet of perpendiculars from a point to the coordinate planes.
  • Apply the distance formula to calculate the distance between two points in 3D space.

Topics covered

Paper topics

  • Introduction to 3D Coordinate System
  • Locating Points in 3D Space
  • Octants
  • Coordinates of a Point
  • Feet of Perpendiculars to Axes
  • Feet of Perpendiculars to Coordinate Planes
  • Distance Formula in 3D

Important topics

  • Locating Points and Octants
  • Feet of Perpendiculars to Axes
  • Feet of Perpendiculars to Coordinate Planes
  • Distance Formula in 3D

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Questions and Solutions

Question 1

Locate the following points in the three-dimensional coordinate system:
  1. (1, -1, 3)
  2. (-2, -4, -7)
  3. (-1, 2, 4)
  4. (-4, 2, -5)
Solution:

To locate these points, we consider their signs and magnitudes along the x, y, and z axes.

  1. (1, -1, 3): This point is located 1 unit along the positive x-axis, 1 unit along the negative y-axis, and 3 units along the positive z-axis.
  2. (-2, -4, -7): This point is located 2 units along the negative x-axis, 4 units along the negative y-axis, and 7 units along the negative z-axis.
  3. (-1, 2, 4): This point is located 1 unit along the negative x-axis, 2 units along the positive y-axis, and 4 units along the positive z-axis.
  4. (-4, 2, -5): This point is located 4 units along the negative x-axis, 2 units along the positive y-axis, and 5 units along the negative z-axis.

Visualizing these movements from the origin (0,0,0) helps in plotting the points in the 3D space.

Question 2

Name the octant in which each of the following points lies:
  1. (1, 2, 3)
  2. (4, -2, 3)
  3. (4, -2, -5)
  4. (4, 2, -5)
  5. (-4, 2, 5)
  6. (-3, -1, 6)
  7. (2, -4, -7)
  8. (-4, 2, -5)
Solution:

The octant in which a point lies is determined by the signs of its coordinates (x, y, z).

  1. (1, 2, 3): (+, +, +) - First Octant
  2. (4, -2, 3): (+, -, +) - Fourth Octant
  3. (4, -2, -5): (+, -, -) - Eighth Octant
  4. (4, 2, -5): (+, +, -) - Fifth Octant
  5. (-4, 2, 5): (-, +, +) - Second Octant
  6. (-3, -1, 6): (-, -, +) - Third Octant
  7. (2, -4, -7): (+, -, -) - Eighth Octant
  8. (-4, 2, -5): (-, +, -) - Sixth Octant

Question 3

If A, B, C are the feet of the perpendiculars from a point P to the X, Y and Z axes respectively, find the coordinates of A, B and C for the point P given by:
  1. P (3, 4, 2)
  2. P (-5, 3, 7)
  3. P (4, -3, -5)
Solution:

The feet of the perpendiculars from a point P(x, y, z) to the axes are found by setting the other two coordinates to zero.

  1. For P (3, 4, 2):
    • A (feet on X-axis): (3, 0, 0)
    • B (feet on Y-axis): (0, 4, 0)
    • C (feet on Z-axis): (0, 0, 2)
  2. For P (-5, 3, 7):
    • A (feet on X-axis): (-5, 0, 0)
    • B (feet on Y-axis): (0, 3, 0)
    • C (feet on Z-axis): (0, 0, 7)
  3. For P (4, -3, -5):
    • A (feet on X-axis): (4, 0, 0)
    • B (feet on Y-axis): (0, -3, 0)
    • C (feet on Z-axis): (0, 0, -5)

Question 4

If A, B, and C are the feet of the perpendiculars from a point P to the XY, YZ and ZX-planes respectively, find the coordinates of A, B and C for the point P given by:
  1. P (3, 4, 5)
  2. P (-5, 3, 7)
  3. P (4, -3, -5)
Solution:

The feet of the perpendiculars from a point P(x, y, z) to the coordinate planes are found by setting the coordinate corresponding to the perpendicular axis to zero.

  • On the XY-plane, z = 0.
  • On the YZ-plane, x = 0.
  • On the ZX-plane, y = 0.
  1. For P (3, 4, 5):
    • A (feet on XY-plane): (3, 4, 0)
    • B (feet on YZ-plane): (0, 4, 5)
    • C (feet on ZX-plane): (3, 0, 5)
  2. For P (-5, 3, 7):
    • A (feet on XY-plane): (-5, 3, 0)
    • B (feet on YZ-plane): (0, 3, 7)
    • C (feet on ZX-plane): (-5, 0, 7)
  3. For P (4, -3, -5):
    • A (feet on XY-plane): (4, -3, 0)
    • B (feet on YZ-plane): (0, -3, -5)
    • C (feet on ZX-plane): (4, 0, -5)

Question 5

How far apart are the points (2, 0, 0) and (–3, 0, 0)?
Solution:

We use the distance formula between two points (x_1, y_1, z_1) and (x_2, y_2, z_2), which is given by:

d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}

Let the two points be A (2, 0, 0) and B (–3, 0, 0).

Here, x_1 = 2, y_1 = 0, z_1 = 0 and x_2 = -3, y_2 = 0, z_2 = 0.

Substituting these values into the distance formula:

AB = \sqrt{(-3 - 2)^2 + (0 - 0)^2 + (0 - 0)^2}

AB = \sqrt{(-5)^2 + 0^2 + 0^2}

AB = \sqrt{25}

AB = 5

Thus, the distance between the points (2, 0, 0) and (–3, 0, 0) is 5 units.

Common mistakes

  • Incorrectly identifying the octant based on coordinate signs.
  • Errors in applying the distance formula, especially with negative signs.
  • Confusing feet of perpendiculars to axes with feet of perpendiculars to planes.
  • Mistakes in determining the coordinates of projections onto axes and planes.

Revision tips

  • Visualize the 3D coordinate system and octants to better understand point locations.
  • Practice identifying octants by systematically checking the signs of x, y, and z coordinates.
  • Memorize and correctly apply the distance formula for 3D points.
  • Work through the examples of finding feet of perpendiculars to axes and planes to solidify the concept.

Practice MCQs

Q1. In which octant does the point (-4, 2, -5) lie?

Q2. What are the coordinates of the feet of the perpendicular from point P(3, 4, 2) to the Y-axis?

Q3. What are the coordinates of the feet of the perpendicular from point P(3, 4, 5) to the XY-plane?

Q4. What is the distance between the points (2, 0, 0) and (-3, 0, 0)?

Q5. The point (4, -2, 3) lies in which octant?

Frequently asked questions

What is the main focus of Chapter 12, Introduction to Three Dimensional Geometry for Class 11 Maths?

This chapter introduces the basic concepts of 3D geometry, including how to represent and locate points in three-dimensional space using coordinates, understanding octants, and calculating distances between points.

How do I determine the octant of a point in 3D space?

The octant is determined by the signs of the x, y, and z coordinates. Each combination of three signs (+ or -) corresponds to a specific octant, similar to quadrants in 2D.

What does it mean to find the feet of the perpendicular from a point to an axis or a plane?

It means finding the coordinates of the point on the axis or plane that is closest to the given point. For an axis, two coordinates become zero; for a plane, one coordinate becomes zero.

Which formula is used to find the distance between two points in 3D?

The distance formula in 3D is used, which is an extension of the 2D distance formula: d = sqrt((x2-x1)^2 + (y2-y1)^2 + (z2-z1)^2).

How can these NCERT Solutions help with exam preparation?

These solutions provide clear, step-by-step explanations for each problem, helping students understand the methods and concepts. Practicing these solutions aids in building confidence and accuracy for exams.

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