RD Sharma Solutions Class 11 Maths Chapter 28 – Introduction To 3D Coordinate Geometry (Updated For 2024)

RD Sharma Solutions Class 11 Maths Chapter 28: So, to make it easier for you to understand the Chapter 28, you have RD Sharma Solutions for chapter 28 – Introduction to Coordinate Geometry. The RD Sharma Class 11 Solutions comprehensively present each topic. So, you get to know them better and be able to apply them when you solve the problems.

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RD Sharma Solutions Class 11 Maths Chapter 28

Exercise-wise RD Sharma Solutions Class 11 Maths Chapter 28

RD Sharma Class 11 Chapter 28A

RD Sharma Class 11 Chapter 28B

RD Sharma Class 11 Chapter 28C

Details of Exercises from RD Sharma Solutions Class 11 Maths Chapter 28

Each exercise is related to a concept that you learn in the chapter, such as the coordinates of a point and its signs and the distance and section formulae. By going through the following sections, you will know the types of problems you have in each exercise. So, make sure to go through and solve them so that you score well in the exams.

RD Sharma Class 11 Chapter 28 Exercise 28.1

There are seven questions in Exercise 28.1, and in each question, you will be given a set of coordinate points and asked to determine the distances between them. You may also have various three-dimensional figures for which you must estimate metrics such as area or the length of the sides or angles.

RD Sharma Class 11 Chapter 28 Exercise 28.2

Exercise 28.2 has 12 questions. Here, it would help if you solved each of the 12 problems using the distance and section formulae.

RD Sharma Class 11 Chapter 28 Exercise 28.3

In Exercise 28.3, you have a set of coordinate points using which you trace a three-dimensional figure. Once you do that, you determine the different metrics related to that three-dimensional figure using the distance and section formulae.

Access RD Sharma Solutions Class 11 Maths Chapter 28 

1. Name the octants in which the following points lie:
(i) (5, 2, 3)
(ii) (-5, 4, 3)
(iii) (4, -3, 5)
(iv) (7, 4, -3)
(v) (-5, -4, 7)
(vi) (-5, -3, -2)
(vii) (2, -5, -7)
(viii) (-7, 2, -5)

Solution:

(i) (5, 2, 3)

In this case, since x, y and z all three are positive then octant will be XOYZ

(ii) (-5, 4, 3)

In this case, since x is negative and y and z are positive then the octant will be X′OYZ

(iii) (4, -3, 5)

In this case, since y is negative and x and z are positive then the octant will be XOY′Z

(iv) (7, 4, -3)

In this case, since z is negative and x and y are positive then the octant will be XOYZ′

(v) (-5, -4, 7)

In this case, since x and y are negative and z is positive then the octant will be X′OY′Z

(vi) (-5, -3, -2)

In this case, since x, y and z all three are negative then octant will be X′OY′Z′

(vii) (2, -5, -7)

In this case, since z and y are negative and x is positive then the octant will be XOY′Z′

(viii) (-7, 2, -5)

In this case, since x and z are negative and x is positive then the octant will be X′OYZ′

2. Find the image of:
(i) (-2, 3, 4) in the yz-plane
(ii) (-5, 4, -3) in the xz-plane

(iii) (5, 2, -7) in the xy-plane
(iv) (-5, 0, 3) in the xz-plane
(v) (-4, 0, 0) in the xy-plane

Solution:

(i) (-2, 3, 4)

Since we need to find its image in yz-plane, a sign of its x-coordinate will change

So, Image of point (-2, 3, 4) is (2, 3, 4)

(ii)(-5, 4, -3)

Since we need to find its image in xz-plane, sign of its y-coordinate will change

So, Image of point (-5, 4, -3) is (-5, -4, -3)

(iii) (5, 2, -7)

Since we need to find its image in xy-plane, a sign of its z-coordinate will change

So, Image of point (5, 2, -7) is (5, 2, 7)

(iv) (-5, 0, 3)

Since we need to find its image in xz-plane, sign of its y-coordinate will change

So, Image of point (-5, 0, 3) is (-5, 0, 3)

(v) (-4, 0, 0)

Since we need to find its image in xy-plane, sign of its z-coordinate will change

So, Image of point (-4, 0, 0) is (-4, 0, 0)

3. A cube of side 5 has one vertex at the point (1, 0, 1), and the three edges from this vertex are, respectively, parallel to the negative x and y-axes and positive z-axis. Find the coordinates of the other vertices of the cube.

Solution:

Given: A cube has side 4 having one vertex at (1, 0, 1)

Side of cube = 5

We need to find the coordinates of the other vertices of the cube.

So let the Point A(1, 0, 1) and AB, AD and AE is parallel to –ve x-axis, -ve y-axis and +ve z-axis respectively.

Since side of cube = 5

Point B is (-4, 0, 1)

Point D is (1, -5, 1)

Point E is (1, 0, 6)

Now, EH is parallel to –ve y-axis

Point H is (1, -5, 6)

HG is parallel to –ve x-axis

Point G is (-4, -5, 6)

Now, again GC and GF is parallel to –ve z-axis and +ve y-axis respectively

Point C is (-4, -5, 1)

Point F is (-4, 0, 6)

4. Planes are drawn parallel to the coordinates planes through the points (3, 0, -1) and (-2, 5, 4). Find the lengths of the edges of the parallelepiped so formed.

Solution:

Given:

Points are (3, 0, -1) and (-2, 5, 4)

We need to find the lengths of the edges of the parallelepiped formed.

For point (3, 0, -1)

x₁ = 3, y₁ = 0 and z₁ = -1

For point (-2, 5, 4)

x₂ = -2, y₂ = 5 and z₂ = 4

Plane parallel to coordinate planes of x₁ and x₂ is yz-plane

Plane parallel to coordinate planes of y₁ and y₂ is xz-plane

Plane parallel to coordinate planes of z₁ and z₂ is xy-plane

Distance between planes x₁ = 3 and x₂ = -2 is 3 – (-2) = 3 + 2 = 5

Distance between planes x₁ = 0 and y₂ = 5 is 5 – 0 = 5

Distance between planes z₁ = -1 and z₂ = 4 is 4 – (-1) = 4 + 1 = 5

∴Theedges of parallelepiped is 5, 5, 5

5. Planes are drawn through the points (5, 0, 2) and (3, -2, 5) parallel to the coordinate planes. Find the lengths of the edges of the rectangular parallelepiped so formed.

Solution:

Given:

Points are (5, 0, 2) and (3, -2, 5)

We need to find the lengths of the edges of the parallelepiped formed

For point (5, 0, 2)

x₁ = 5, y₁ = 0 and z₁ = 2

For point (3, -2, 5)

x₂ = 3, y₂ = -2 and z₂ = 5

Plane parallel to coordinate planes of x₁ and x₂ is yz-plane

Plane parallel to coordinate planes of y₁ and y₂ is xz-plane

Plane parallel to coordinate planes of z₁ and z₂ is xy-plane

Distance between planes x₁ = 5 and x₂ = 3 is 5 – 3 = 2

Distance between planes x₁ = 0 and y₂ = -2 is 0 – (-2) = 0 + 2 = 2

Distance between planes z₁ = 2 and z₂ = 5 is 5 – 2 = 3

∴Theedges of parallelepiped is 2, 2, 3

6. Find the distances of the point P (-4, 3, 5) from the coordinate axes.

Solution:

Given:

The point P (-4, 3, 5)

The distance of the point from x-axis is given as:

The distance of the point from y-axis is given as:

The distance of the point from z-axis is given as:

7. The coordinates of a point are (3, -2, 5). Write down the coordinates of seven points such that the absolute values of their coordinates are the same as those of the coordinates of the given point.

Solution:

Given:

Point (3, -2, 5)

The Absolute value of any point(x, y, z) is given by,

√(x² + y² + z²)

We need to make sure that the absolute value is the same for all points.

So let the point A(3, -2, 5)

The remaining 7 points are:

Point B(3, 2, 5) (By changing the sign of y coordinate)

Point C(-3, -2, 5) (By changing the sign of x coordinate)

Point D(3, -2, -5) (By changing the sign of z coordinate)

Point E(-3, 2, 5) (By changing the sign of x and y coordinate)

Point F(3, 2, -5) (By changing the sign of y and z coordinate)

Point G(-3, -2, -5) (By changing the sign of x and z coordinate)

Point H(-3, 2, -5) (By changing the sign of x, y and z coordinate)

List of Important Topics from RD Sharma Solutions Class 11 Maths Chapter 28

Chapter 28 in Maths for CBSE class 11 is essential, not only for the annual exams and the class 12 board exams but also for various other competitive exams, such as the Olympiads. The chapter also serves as a foundation for many higher-order topics that you will be studying later when you pursue courses like engineering or architecture.

So, you must pay special attention to this chapter. In the following list, you will find those topics from chapter 28 that you need to focus on as they have appeared several times in the annual exams.

1. Coordinates of a Point in Space
2. Distance and Section Formulae

The topics mentioned above form the core of the chapter. So, don’t forget to go through and become thorough with them for you to write your exams.

RD Sharma Solutions Class 11 Maths Chapter 28

Chapter 28 in Maths for CBSE class 11, ‘Introduction to 3D Coordinate Geometry,’ has various essential concepts for the exams. Many of these topics are also form the basis for other, higher-order concepts that you will be learning in courses like engineering. 

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