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Edited By Ramraj Saini | Updated on Sep 25, 2023 07:33 PM IST

**NCERT solutions for class 11 maths chapter 12 Introduction to Three Dimensional Geometry **discussed here. In our previous classes, you have studied the basic concepts of geometry in two-dimensional space. In this article, you will get introduction to three dimensional geometry class 11 NCERT solutions. These NCERT solutions are prepared by subject matter expert at Careers360 keeping in my the latest CBSE syllabus 2023.

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This Story also Contains

- Introduction to Three Dimensional Geometry Class 11 Questions And Answers
- Introduction to Three Dimensional Geometry Class 11 Questions And Answers PDF Free Download
- Introduction to Three Dimensional Geometry Class 11 Solutions - Important Formulae
- Introduction to Three Dimensional Geometry Class 11 NCERT Solutions (Intext Questions and Exercise)
- NCERT Solutions for Class 11 Mathematics
- Key Features Of Introduction to Three Dimensional Geometry Class 11 Solutions
- NCERT Solutions for Class 11 - Subject wise
- NCERT Books and NCERT Syllabus

In introduction to three Dimensional geometry class 11 chapter, you will learn what is 3 D coordinate geometry, how a coordinate space can be determined, about coordinates of a point in space, the distance between two points and section formula. There are 14 questions in 3 exercises in this chapter. First, try to solve solve the problem on your own. If you are not able to do, you can take the help of NCERT solutions for class 11 maths chapter 12 which is prepared in a detailed manner. Students can find NCERT solutions for class 11 here.

Also Read

- NCERT Exemplar Solutions For Class 11 Maths Chapter 12 Introduction To Three Dimensional Geometry
- Introduction To Three Dimensional Geometry Class 11 NCERT Chapter Note

**Coordinate Axes in Three Dimensions:**

In three dimensions, a rectangular cartesian coordinate system has three mutually perpendicular axes: X, Y, and Z.

**Coordinate Planes:**

The pair of axes determines three coordinate planes: XY, YZ, and ZX planes. These planes divide space into eight regions called octants.

**Coordinates of a Point in Space:**

The coordinates of a point P in three-dimensional space are represented as a triplet (x, y, z), which corresponds to the perpendicular distances from P on three mutually perpendicular coordinate planes: YZ-plane, ZX-plane, and XY-plane.

Coordinates of points on coordinate axes are:

X-axis: (x, 0, 0)

Y-axis: (0, y, 0)

Z-axis: (0, 0, z)

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Download EBook**Distance Formula in 3D Space:**

The distance between two points A(x_{1}, y_{1}, z_{1}) and B(x_{2}, y_{2}, z_{2}) is given by:

AB = √((x

_{2}-x_{1})^{2}+ (y_{2}-y_{1})^{2}+ (z_{2}-z_{1})^{2})

**Distance from Origin in 3D Space:**

The distance of a point A(x, y, z) from the origin O(0, 0, 0) is given by:

OA = √(x

^{2}+ y^{2}+ z^{2})

**Section Formula for Internal and External Division in 3D Space:**

The coordinates of a point R that divides the line segment joining two points P(x_{1}, y_{1}, z_{1}) and Q(x_{2}, y_{2}, z_{2}) internally or externally in the ratio m : n are given by:

Internal Division: ((mx

_{2}+ nx_{1})/(m + n), (my_{2}+ ny_{1})/(m + n), (mz_{2}+ nz_{1})/(m + n))External Division: ((mx

_{2}- nx_{1})/(m - n), (my_{2}- ny_{1})/(m - n), (mz_{2}- nz_{1})/(m - n))

**Midpoint Formula in 3D Space:**

The midpoint of the line segment joining two points (x_{1}, y_{1}, z_{1}) and (x_{2}, y_{2}, z_{2}) is given by:

((x

_{1}+ x_{2})/2, (y_{1}+ y_{2})/2, (z_{1}+ z_{2})/2)

**Coordinates of Centroid of a Triangle in 3D Space:**

For a triangle with vertices (x_{1}, y_{1}, z_{1}), (x_{2}, y_{2}, z_{2}), and (x_{3}, y_{3}, z_{3}), the coordinates of the centroid are given by:

((x_{1} + x_{2} + x_{3})/3, (y_{1} + y_{2} + y_{3})/3, (z_{1} + z_{2} + z_{3})/3)

Free download **NCERT Solutions for Class 11 Maths Chapter 12 Introduction to Three Dimensional Geometry **for CBSE Exam.

** C lass 11 maths chapter 12 question answer - Exercise: 12.1 **

** Question:1 ** A point is on the x-axis. What are its y-coordinate and z-coordinates?

** Answer: **

Any point on x-axis have zero y coordinate and zero z coordinate.

** Question:2 ** A point is in the XZ-plane. What can you say about its y-coordinate?

** Answer: **

When a point is in XZ plane, the y coordinate of this point will always be zero.

** Answer: **

The x-coordinate, y-coordinate, and z-coordinate of point (1, 2, 3) are all positive. Therefore, this point lies in octant I.

The x-coordinate, y-coordinate, and z-coordinate of point (4, -2, 3) are positive, negative, and positive respectively. Therefore, this point lies in octant IV.

The x-coordinate, y-coordinate, and z-coordinate of point (4, -2, -5) are positive, negative, and negative respectively. Therefore, this point lies in octant VIII.

The x-coordinate, y-coordinate, and z-coordinate of point (4, 2, -5) are positive, positive, and negative respectively. Therefore, this point lies in octant V.

The x-coordinate, y-coordinate, and z-coordinate of point (-4, 2, -5) are negative, positive, and negative respectively. Therefore, this point lies in octant VI.

The x-coordinate, y-coordinate, and z-coordinate of point (-4, 2, 5) are negative, positive, and positive respectively. Therefore, this point lies in octant II.

The x-coordinate, y-coordinate, and z-coordinate of point (-3, -1, 6) are negative, negative, and positive respectively. Therefore, this point lies in octant III.

The x-coordinate, y-coordinate, and z-coordinate of point (2, -4, -7) are positive, negative, and negative respectively. Therefore, this point lies in octant VIII.

** Question:4 (i) ** Fill in the blanks: The x-axis and y-axis taken together determine a plane known as_______.

** Answer: **

The x-axis and y-axis taken together determine a plane known as ** XY ** ** Plane ** .

** Question:4 (ii) ** Fill in the blanks: The coordinates of points in the XY-plane are of the form _______.

** Answer: **

The coordinates of points in the XY-plane are of the form ** (x, y, 0 ) ** .

** Question:4 (iii) ** Fill in the blanks: Coordinate planes divide the space into ______ octants.

** Answer: **

Coordinate planes divide the space into ** Eight ** octants.

** Class 11 maths chapter 12 question answer - Exercise: 12.2 **

** Question:1 (i) ** Find the distance between the following pairs of points: (2, 3, 5) and (4, 3, 1)

** Answer: **

The distance between two points and is given by

So, distance between (2, 3, 5) and (4, 3, 1) is given by

** Question:1 (ii) ** Find the distance between the following pairs of points: (–3, 7, 2) and (2, 4, –1)

** Answer: **

The distance between two points and is given by

So, distance between (–3, 7, 2) and (2, 4, –1) is given by

** Question:1 (iii) ** Find the distance between the following pairs of points:(–1, 3, – 4) and (1, –3, 4)

** Answer: **

The distance between two points and is given by

So, distance between (–1, 3, – 4) and (1, –3, 4) is given by

** Question:1 (iv) ** Find the distance between the following pairs of points: (2, –1, 3) and (–2, 1, 3).

** Answer: **

The distance between two points and is given by

So, distance between (2, –1, 3) and (–2, 1, 3).) is given by

** Question:2 ** Show that the points (–2, 3, 5), (1, 2, 3) and (7, 0, –1) are collinear.

** Answer: **

Given,theree points A=(–2, 3, 5), B=(1, 2, 3) and C=(7, 0, –1)

The distance between two points and is given by

The distance AB :

The distance BC:

The distance CA

As we can see here,

Hence we can say that point A,B and C are colinear.

** Question:3 (i) ** Verify the following: (0, 7, –10), (1, 6, – 6) and (4, 9, – 6) are the vertices of an isosceles triangle.

** Answer: **

Given Three points A=(0, 7, –10), B=(1, 6, – 6) and C=(4, 9, – 6)

The distance between two points and is given by

The distance AB

The distance BC

The distance CA

As we can see

Hence we can say that ABC is an isosceles triangle.

** Question:3(ii) ** Verify the following

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

** Answer: **

Given Three points A=(0, 7, 10), B=(-1, 6, 6) and C=(-4, 9, 6)

The distance between two points and is given by

The distance AB

The distance BC

The distance CA

As we can see

Since this follows pythagorus theorem, we can say that ABC is a right angle triangle.

** Question:3(iii) ** Verify the following: (–1, 2, 1), (1, –2, 5), (4, –7, 8) and (2, –3, 4) are the vertices of a parallelogram.

** Answer: **

Given A=(–1, 2, 1), B=(1, –2, 5), C=(4, –7, 8) and D=(2, –3, 4)

Given Three points A=(0, 7, –10), B=(1, 6, – 6) and C=(4, 9, – 6)

The distance between two points and is given by

The distance AB

The distance BC

The distance CD

The distance DA

Here As we can see

And

As the opposite sides of quadrilateral are equal, we can say that ABCD is a parallelogram.

** Answer: **

Given, two points A=(1, 2, 3) and B=(3, 2, –1).

Let the point P= (x,y,z) be a point which is equidistance from the points A and B.

so,

The distance PA= The distance PB

Now lets apply the simplification property,

Hence locus of the point which is equidistant from A and B is .

** Answer: **

Given,

Two points A (4, 0, 0) and B (– 4, 0, 0)

let the point P(x,y,z) be a point sum of whose distance from A and B is 10.

So,

The distance PA+The distance PB=10

Squaring on both side :

Now again squaring both sides,

Hence the equation of the set of points P, the sum of whose distances from A and B is equal to 10 is .

** Class 11 maths chapter 12 NCERT solutions - Exercise: 12.3 **

** Answer: **

The line segment joining the points A (– 2, 3, 5) and B(1, – 4, 6)

Let point P(x,y,z) be the point that divides the line segment AB internally in the ratio 2:3.

Now, As we know by section formula, The coordinate of the point P which divides line segment And in ratio m:n is

Now the point that divides A (– 2, 3, 5) and B(1, – 4, 6) in ratio 2:3 is

Hence required point is

.

** Question:1 (ii) ** Find the coordinates of the point which divides the line segment joining the points (– 2, 3, 5) and (1, – 4, 6) in the ratio 2 : 3 externally.

** Answer: **

he line segment joining the points A (– 2, 3, 5) and B(1, – 4, 6)

Let point P(x,y,z) be the point that divides the line segment AB externally in the ratio 2:3.

Now, As we know by section formula , The coordinate of the point P which divides line segment And externally in ratio m:n is

Now the point that divides A (– 2, 3, 5) and B(1, – 4, 6) externally in ratio 2:3 is

Hence required point is

.

** Answer: **

Given Three points,

P (3, 2, – 4), Q (5, 4, – 6) and R (9, 8, –10)

Let point Q divides PR internally in the ratio

Now,

According to the section formula , The point Q in terms of P,Q and is:

Hence, point Q divides PR in ratio 1:2.

** Answer: **

Given,

two points A(–2, 4, 7) and B(3, –5, 8)

Let Y-Z plane divides AB in

So, According to the section formula, the point which divides AB in is

Since this point is in YZ plane, x coordinate of this point will be zero.

So,

Hence YZ plane divides Line segment AB in a ratio 2:3.

** Question:4 ** Using section formula, show that the points A (2, –3, 4), B (–1, 2, 1) and are collinear.

** Answer: **

Given,

three points, A (2, –3, 4), B (–1, 2, 1) and C (0, 1/3, 2)

Let a point P divides Line segment AB in the ratio

SO, according to the section formula, the point P will be

Now, let's compare this point P with point C.

From here, we see that for , point C divides the line segment AB in ratio 2:1. Since point C divides the line segment AB, it lies in the line joining A and B and Hence they are colinear.

** Answer: **

Given,

two points P (4, 2, – 6) and Q (10, –16, 6).

The point which trisects the line segment are the points which divide PQ in either 1:2 or 2:1

Let R (x,y,z) be the point which divides Line segment PR in ratio 1:2

Now, according to the section formula

Let S be the point which divides the Line segment PQ in ratio 2:1

So, The point S according to section formula is

Hence the points which trisect the line segment AB are (6,-4,-2) and (8,-10,2).

** Introduction to three dimensional geometry NCERT solutions - Miscellaneous Exercise **

** Answer: **

Given

Three vertices of the parallelogram, ABCD are:

A(3, – 1, 2), B (1, 2, – 4) and C (– 1, 1, 2).

Let the fourth vertice be (a, b, c )

Now, As we know the concept that diagonal of the parallelogram bisect each other,

Hence here in parallelogram ABCD, the midpoint of the line segment AC will be equal to the midpoint of the line segment BD.

So,

On comparing both points, we get

Hence the Fourth vertice of the is (1,-2,8)

** Question:2 ** Find the lengths of the medians of the triangle with vertices A (0, 0, 6), B (0,4, 0) and (6, 0, 0).

** Answer: **

Given,

Three vertices of the triangle,A (0, 0, 6), B (0,4, 0)and C(6, 0, 0).

Now,

Let D be the midpoint of the AB, E be the midpoint of the BC and F be the midpoint of the AC.

Vertice of the D =

Vertices of E =

Vertices of the F =

Now, Medians of the triangle are CD, AE, and BF

So the lengths of the medians are:

Hence lengths of the median are .

** Answer: **

Given,

Triangle PQR with vertices P (2a, 2, 6), Q (– 4, 3b, –10) and R(8, 14, 2c),

Now, As we know,

The centroid of a triangle is given by

Where coordinates of vertices of the triangle are

Since Centroid of the triangle, PQR is origin =(0,0,0)

On equating both coordinates, we get

** Question:4 ** Find the coordinates of a point on y-axis which are at a distance of from the point P (3, –2, 5).

** Answer: **

Let the point Q be (0,y,0)

Now Given

The distance of point Q From point P =

So,

The coordinates of the required point is (0,2,0) or (0,-6,0).

** Answer: **

Let R divides PQ in the ratio k: 1.

The coordinates of the point R are given by

now, as given the x coordinate of the R is 4.

So,

Hence the coordinates of R are:

** Answer: **

Given Points,

A (3, 4, 5) and B (–1, 3, –7),

Let the coordinates of point P be (x,y,z)

Now,

Given condition :

And

Now, Given Condition

Hence Equation of the set of the point P is

.

12.1 Introduction

12.2 Coordinate Axes and Coordinate Planes in Three Dimensional Space

12.3 Coordinates of a Point in Space

12.4 Distance between Two Points

12.5 Section Formula

Interested students can practice class 11 maths ch 12 question answer using the given exercises.

- NCERT Solutions Class 11 Maths Chapter 12 Exercise 12.1
- NCERT Solutions Class 11 Maths Chapter 12 Exercise 12.2
- NCERT Solutions Class 11 Maths Chapter 12 Exercise 12.3
- NCERT Solutions Class 11 Maths Chapter 12 Miscellaneous Exercise

Unlike 2D shapes, 3D shapes have length, width and depth or thickness. In three dimensional geometry, a coordinate space is determined by a vector perpendicular to that particular plane. Thus, a point in a three-dimensional plane will include the coordinates of three planes named as X- plane, Y- plane, and Z-plane.

There are some important formulas from NCERT solutions for class 11 maths chapter 12 introduction to three dimensional geometry which you should remember after studying this chapter.

- Distance between two points is

- The coordinates of the point R which divides the line segment joining two points internally in the ratio m : n

- The coordinates of the point R which divides the line segment joining two points externally in the ratio m : n

- Distance between origin is

- The coordinates of the mid-point of the line segment joining two points is

- The coordinates of the centroid of the triangle, whose vertices are

chapter-1 | Sets |

chapter-2 | Relations and Functions |

chapter-3 | Trigonometric Functions |

chapter-4 | Principle of Mathematical Induction |

chapter-5 | Complex Numbers and Quadratic equations |

chapter-6 | Linear Inequalities |

chapter-7 | Permutation and Combinations |

chapter-8 | Binomial Theorem |

chapter-9 | Sequences and Series |

chapter-10 | Straight Lines |

chapter-11 | Conic Section |

chapter-12 | Introduction to Three Dimensional Geometry |

chapter-13 | Limits and Derivatives |

chapter-14 | Mathematical Reasoning |

chapter-15 | Statistics |

chapter-16 | Probability |

**Step-by-step solution:** To understand the concepts of 3d geometry class 11, it is important to solve problems step-by-step, which helps students to grasp the concepts and develop problem-solving skills.

**Comprehensive coverage:** The chapter 12 maths class 11 is covered in a comprehensive manner in Class 11 Mathematics, with a focus on understanding the fundamental concepts and their applications in solving problems.

**Concepts:** The topic of introduction to coordinate geometry class 11 involves the study of various concepts such as co-ordinate geometry, lines and planes in space, distance formula, and direction ratios.

- NCERT solutions for class 11 biology
- NCERT solutions for class 11 maths
- NCERT solutions for class 11 chemistry
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** Happy Reading !!! **

1. Explain the concept of 3D Geometry covered in class 11 maths chapter 12.

Three Dimensional Geometry ch 12 maths class 11 is a field of Mathematics that focuses on the analysis of points, lines, and solid shapes in three-dimensional coordinate systems. Introduction to 3d geometry class 11 introduces students to the concept of the z-coordinate in addition to the x and y coordinates, which is used to determine the precise location of a point in the 3D coordinate plane. It is a fundamental theory with broad applications in other scientific disciplines and advanced Mathematics. Additionally, the trigonometric ratios concept has important applications in 3D Geometry.

2. Discuss the topics covered in chapter 12 class 11 maths.

maths chapter 12 class 11 includes the following topics:

- Introduction
- Coordinate Axes and Coordinate Planes in Three Dimensional Space
- Point Coordinates in 3D Space
- Distance Formula between Two Points
- Section Formula

3. Where can I find the complete solutions of NCERT for class 11 maths ?

Students can find detailed NCERT solutions for class 11 maths by clicking on the link. they can practice problems given in introduction to three dimensional geometry class 11 to get good hold on the concepts. that ultimately benefit during the exam and lead to good score.

4. How many chapters are there in CBSE class 11 maths ?

There are 16 chapters starting from sets to probability in the CBSE class 11 maths.

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