Chapter 11 Introduction to Three-dimensional Geometry

NCERT Solutions Class 11 Maths Chapter 11 – Introduction to Three-dimensional Geometry

NCERT Solutions for Class 11 Maths Chapter 11 (Introduction to Three-dimensional Geometry) take students beyond the flat 2D Cartesian plane and into the world of spatial geometry. While 2D geometry uses two axes (x and y), 3D geometry introduces the z-axis, allowing us to describe the position of any object in physical space. This is the foundational chapter for Vector Algebra and 3D Geometry in Class 12, as well as for multi-variable calculus and engineering mechanics.

The ALLEN NCERT Solutions for Class 11 Maths Chapter 11 is prepared by expert faculty and aims to help students visualise the eight regions of space, called Octants. The solutions concern the extension of the classic formulas for distance, and for a section, into three dimensions, thus providing a natural extension from plane geometry.

It is essential to have a basic understanding of 3D coordinates for competitive exams. These solutions provide a natural logical understanding of points moving in space and the computing of the ratio in which a point divides a line segment, which are basic topics in high-level spatial reasoning.

1.0Class 11 Maths Chapter 11: Key Concepts

This chapter focuses on defining the position of points in space and the mathematical relationships between them. Key lessons include:

Coordinate Axes and Planes in 3D: Understanding the three mutually perpendicular lines (x, y, and z axes) and the three coordinate planes (XY, YZ, and ZX planes).
Coordinates of a Point in Space: Representing a point as an ordered triplet (x, y, z).
- Any point on the x-axis is (x, 0, 0).
- Any point in the XY-plane is (x, y, 0).
Octants: The three coordinate planes divide the entire space into eight parts called octants.
Distance Formula in 3D: Extending the 2D formula to find the distance between $P (x_{1}, y_{1}, z_{1}) an d Q (x_{2}, y_{2}, z_{2}) PQ = (x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2} + (z_{2} - z_{1})^{2}$
Section Formula in 3D: Finding the coordinates of a point R that divides the line segment joining P and Q in the ratio m:n.
- Internal Division: $R = (\frac{m x _{2} + n x _{1}}{m + n}, \frac{m y _{2} + n y _{1}}{m + n}, \frac{m z _{2} + n z _{1}}{m + n})$
- External Division: $R = (\frac{m x _{2} - n x _{1}}{m - n}, \frac{m y _{2} - n y _{1}}{m - n}, \frac{m z _{2} - n z _{1}}{m - n})$

2.0Detailed NCERT Textbook Solutions : Class 11 Maths Chapter 11

EXERCISE - 11.1

A point is on the x-axis. What are its y-coordinates and z-coordinates? Sol. If a point is on the x-axis (i.e. x, 0, 0), then its y-coordinates and z-coordinates are zero.
A point is in the XZ-plane. What can you say about its y-coordinate? Sol. If a point is in the XZ-plane (i.e. x, 0, 2), then its y-coordinate is zero.
Name the octants in which the following points lie: (1, 2, 3), (4, -2, 3), (4, -2, -5), (4, 2, -5), (-4, 2, -5), (-4, 2, 5), (-3, -1, 6), (2, -4, -7) Sol. Point (1, 2, 3) lies in octant I. Point (4, -2, 3) lies in octant IV. Point (4, -2, -5) lies in octant VIII. Point (4, 2, -5) lies in octant V. Point (-4, 2, -5) lies in octant VI. Point (-4, 2, 5) lies in octant II. Point (-3, -1, 6) lies in octant III. Point (2, -4, -7) lies in octant VIII.
Fill in the blanks: (i) The x-axis and y-axis taken together determine a plane known as $____$ . (ii) The coordinates of points in the XY-plane are of the form $____$ . (iii) Coordinate planes divide the space into $____$ octants.

Sol.

(i) The x -axis and y -axis taken together determine a plane known as xy-plane. (ii) The coordinates of points in the XY-plane are of the form $(x, y, 0)$ . (iii) Coordinate planes divide the space into eight octants.

EXERCISE - 11.2

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) Sol. The distance between points $P (x_{1}, y_{1}, z_{1})$ and $P (x_{2}, y_{2}, z_{2})$ is given by

$PQ = (x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2} + (z_{2} - z_{1})^{2}$

(i) Distance between points (2, 3, 5) and (4, 3, 1)

$= (4 - 2)^{2} + (3 - 3)^{2} + (1 - 5)^{2} = (2)^{2} + (0)^{2} + (- 4)^{2} = 4 + 16 = 20 = 25 units$

(ii) Distance between points (-3, 7, 2) and (2, 4, -1)

$= (2 + 3)^{2} + (4 - 7)^{2} + (- 1 - 2)^{2} = (5)^{2} + (- 3)^{2} + (- 3)^{2} = 25 + 9 + 9 = 43 units$

(iii) Distance between points (-1, 3, -4) and (1, -3, 4)

$= (1 + 1)^{2} + (- 3 - 3)^{2} + (4 + 4)^{2} = (2)^{2} + (- 6)^{3} + (8)^{2} = 4 + 36 + 64 = 104 = 226 units$

(iv) Distance between points (2, -1, 3) and (-2, 1, 3)

$= (- 2 - 2)^{2} + (1 + 1)^{2} + (3 - 3)^{2} = (- 4)^{2} + (2)^{2} + (0)^{2} = 16 + 4 = 20 = 25 units$

Show that the points (-2, 3, 5), (1, 2, 3) and (7, 0,-1) are collinear. Sol. Let points (-2, 3, 5), (1, 2, 3), and (7, 0, -1) be denoted by P, Q, and R respectively. Points P, Q, and R are collinear if they lie on a line.

$PQ QR = (1 + 2)^{2} + (2 - 3)^{2} + (3 - 5)^{2} = (3)^{2} + (- 1)^{2} + (- 2)^{2} = 9 + 1 + 4 = 14 = (7 - 1)^{2} + (0 - 2)^{2} + (- 1 - 3)^{2} = (6)^{2} + (- 2)^{2} + (- 4)^{2} = 36 + 4 + 16 = 56 = 214$

$PR = (7 + 2)^{2} + (0 - 3)^{2} + (- 1 - 5)^{2} = (9)^{2} + (- 3)^{2} + (- 6)^{2} = 81 + 9 + 36 = 126 = 314$

Here, $PQ + QR = 14 + 214 = 314 = PR$ Hence, points $P (- 2, 3, 5), Q (1, 2, 3)$ , and R(7, 0, -1) are collinear.

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. Sol. (i) Let points (0, 7, -10), (1, 6, -6), and (4, 9, -6) be denoted by A, B, and C respectively.

$AB BC CA = (1 - 0)^{2} + (6 - 7)^{2} + (- 6 + 10)^{2} = (1)^{2} + (- 1)^{2} + (4)^{2} = 1 + 1 + 16 = 18 = 32 = (4 - 1)^{2} + (9 - 6)^{2} + (- 6 + 6)^{2} = (3)^{2} + (3)^{2} = 9 + 9 = 18 = 32 = (0 - 4)^{2} + (7 - 9)^{2} + (- 10 + 6)^{2} = (- 4)^{2} + (- 2)^{2} + (- 4)^{2} = 16 + 4 + 16 = 36 = 6$

Here, $AB = BC \neq = CA$ Thus, the given points are the vertices of an isosceles triangle. (ii) Let (0, 7, 10), (-1, 6, 6), and (- 4, 9, 6) be denoted by A, B, and C respectively. Therefore, by Pythagoras theorem, ABC is a right triangle.

$AB = (- 1 - 0)^{2} + (6 - 7)^{2} + (6 + 10)^{2} = (- 1)^{2} + (- 1)^{2} + (- 4)^{2} = 1 + 1 + 16 = 18 = 32$

$BC = (- 4 - 1)^{2} + (9 - 6)^{2} + (6 + 6)^{2} = (- 3)^{2} + (3)^{2} + (0)^{2} = 9 + 9 = 18 = 32$

$CA = (0 + 4)^{2} + (7 - 9)^{2} + (10 + 6)^{2} = (4)^{2} + (- 2)^{2} + (4)^{2} = 16 + 4 + 16 = 36 = 6$

Now, $AB^{2} + BC^{2} = (32)^{2} + (32)^{2}$

$= 18 + 18 = 36 = AC^{2}$

Therefore, by Pythagoras theorem, ABC is a right triangle. Hence, the given points are the vertices of a right-angled triangle. (iii) Let (-1, 2, 1), (1, -2, 5), (4, -7, 8), and (2, -3, 4) be denoted by $A, B, C$ , and D respectively.

$AB = (1 + 1)^{2} + (- 2 - 2)^{2} + (5 - 1)^{2} = 4 + 16 + 16 = 36 = 6$

$BC = (4 - 1)^{2} + (- 7 + 2)^{2} + (8 - 5)^{2} = 9 + 25 + 9 = 43$

$CD = (2 - 4)^{2} + (- 3 + 7)^{2} + (4 - 8)^{2} = 4 + 16 + 16 = 36 = 6$

$DA = (- 1 - 2)^{2} + (2 + 3)^{2} + (1 - 4)^{2} = 9 + 25 + 9 = 43$

Here, $AB = CD = 6, BC = AD = 43$

Hence, the opposite sides of quadrilateral ABCD, whose vertices are taken in order, are equal. Therefore, ABCD is a parallelogram. Hence, the given points are the vertices of a parallelogram.

Find the equation of the set of points which are equidistant from the points (1, 2, 3) and (3, 2, -1). Sol. Let P (x, y, z) be the point that is equidistant from points $A (1, 2, 3)$ and $B (3, 2, - 1)$ . Accordingly to question, $PA = PB$ $\Rightarrow PA^{2} = PB^{2}$ $\Rightarrow (x - 1)^{2} + (y - 2)^{2} + (z - 3)^{2}$ $= (x - 3)^{2} + (y - 2)^{2} + (z + 1)^{2}$ $\Rightarrow x^{2} - 2 x + 1 + y^{2} - 4 y + 4 + z^{2} - 6 z + 9$ $= x^{2} - 6 x + 9 + y^{2} - 4 y + 4 + z^{2} + 2 z + 1$ $\Rightarrow - 2 x - 4 y - 6 z + 14 = - 6 x - 4 y + 2 z + 14$ $\Rightarrow - 2 x - 6 z + 6 x + 2 z = 0$ $\Rightarrow 4 x - 8 z = 0$ $\Rightarrow x - 2 z = 0$

Thus, the required equation is $x - 2 z = 0$ .

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. Sol. Let the coordinates of P be (x, y, z). The coordinates of points A and B are (4, 0, 0) and (-4, 0, 0) respectively. It is given that $PA + PB = 10$ . $\Rightarrow (x - 4)^{2} + (y - 0)^{2} + (z - 0)^{2}$ $+ (x + 4)^{2} + (y - 0)^{2} + (z - 0)^{2} = 10$ $\Rightarrow (x - 4)^{2} + y^{2} + z^{2} = 10 - (x + 4)^{2} + y^{2} + z^{2}$

On squaring both sides, we obtain

$\Rightarrow (x - 4)^{2} + y^{2} + z^{2}$ $= 100 - 20 (x + 4)^{2} + y^{2} + z^{2}$ $+ (x + 4)^{2} + y^{2} + z^{2}$ $\Rightarrow x^{2} - 8 x + 16 + y^{2} + z^{2}$ $= 100 - 20 x^{2} + 8 x + 16 + y^{2} + z^{2}$ $+ x^{2} + 8 x + 16 + y^{2} + z^{2}$ $\Rightarrow 20 x^{2} + 8 x + 16 + y^{2} + z^{2} = 100 + 16 x$ $\Rightarrow 5 x^{2} + 8 x + 16 + y^{2} + z^{2} = (25 + 4 x)$

On squaring both sides again, we obtain $25 (x^{2} + 8 x + 16 + y^{2} + z^{2}) = 625 + 16 x^{2} + 200 x$

$\Rightarrow 25 x^{2} + 200 x + 400 + 25 y^{2} + 25 z^{2}$ $= 625 + 16 x^{2} + 200 x$ $\Rightarrow 9 x^{2} + 25 y^{2} + 25 z^{2} - 225 = 0$

Thus, the required equation is $9 x^{2} + 25 y^{2} + 25 z^{2} - 225 = 0$ .

MISCELLANEOUS EXERCISE

Three vertices of a parallelogram ABCD are A (3, -1, 2), B (1, 2, -4) and C (-1, 1, 2). Find the coordinates of the fourth vertex. Sol. The three vertices of a parallelogram ABCD are given as $A (3, - 1, 2), B (1, 2, - 4)$ , and C (-1, 1, 2). Let the coordinates of the fourth vertex be D (x, y, z).

ques-1-mis-exer-class-11-chapter-11-maths

We know that the diagonals of a parallelogram bisect each other. Therefore, in parallelogram ABCD, AC and BD bisect each other. ∴ Mid-point of AC = Mid-point of BD $\Rightarrow (\frac{3 - 1}{2}, \frac{- 1 + 1}{2}, \frac{2 + 2}{2}) = (\frac{x + 1}{2}, \frac{y + 2}{2}, \frac{z - 4}{2})$ $\Rightarrow (1, 0, 2) = (\frac{x + 1}{2}, \frac{y + 2}{2}, \frac{z - 4}{2})$ $\Rightarrow \frac{x + 1}{2} = 1, \frac{y + 2}{2} = 0$ and $\frac{z - 4}{2} = 2$ $\Rightarrow x = 1, y = - 2$ , and $z = 8$ Thus, the coordinates of the fourth vertex are (1, -2, 8).

Find the lengths of the medians of the triangle with vertices A (0, 0, 6), B (0, 4, 0) and (6, 0, 0). Sol. Let AD, BE, and CF be the medians of the given triangle ABC .

class-11-maths-chapter-11-ques-2-mis-exercise

Since AD is the median, D is the mid-point of BC . ∴ Coordinates of point

$D = (\frac{0 + 6}{2}, \frac{4 + 0}{2}, \frac{0 + 0}{2}) = (3, 2, 0) AD = (0 - 3)^{2} + (0 - 2)^{2} + (6 - 0)^{2} = 9 + 4 + 36 = 49 = 7$

Since BE is the median, E is the mid-point of AC . ∴ Coordinates of point

$E BE = (\frac{0 + 6}{2}, \frac{0 + 0}{2}, \frac{6 + 0}{2}) = (3, 0, 3) = (3 - 0)^{2} + (0 - 4)^{2} + (3 - 0)^{2} = 9 + 16 + 9 = 34$

Since CF is the median, F is the mid-point of AB. ∴ Coordinates of point

$F = (\frac{0 + 0}{2}, \frac{0 + 4}{2}, \frac{6 + 0}{2}) = (0, 2, 3)$

Length of

$CF = (6 - 0)^{2} + (0 - 2)^{2} + (0 - 3)^{2} = 36 + 4 + 9 = 49 = 7$

Thus, the lengths of the medians of $△ ABC$ are 7, $34$ and 7.

If the origin is the centroid of the triangle PQR with vertices P (2a, 2, 6), Q (- 4, 3b, -10) and R $(8, 14, 2 c)$ , then find the values of $a, b$ and c . Sol. It is known that the coordinates of the centroid of the triangle, whose vertices are $(x_{1}, y_{1}, z_{1})$ , $(x_{2}, y_{2}, z_{2})$ and $(x_{3}, y_{3}, z_{3})$ , are $(\frac{x _{1} + x _{2} + x _{3}}{3}, \frac{y _{1} + y _{2} + y _{3}}{3}, \frac{z _{1} + z _{2} + z _{3}}{3})$ .

ques-3-maths-class-11-chapter-11-mis-exer

Therefore, coordinates of the centroid of $∴ (0, 0, 0) = (\frac{2 a + 4}{3}, \frac{3 b + 16}{3}, \frac{2 c - 4}{3})$ $\Rightarrow \frac{2 a + 4}{3} = 0, \frac{3 b + 16}{3} = 0$ and $\frac{2 c - 4}{3} = 0$ $\Rightarrow a = - 2, b = - \frac{16}{3}$ and $c = 2$ Thus, the respective values of $a, b$ and c are $- 2, - \frac{16}{3}$ and 2.

If A and B be the points (3, 4, 5) and (-1, 3, -7), respectively, find the equation of the set of points P such that $PA^{2} + PB^{2} = k^{2}$ , where k is a constant. Sol. The coordinates of points A and B are given as (3, 4, 5) and (-1, 3, -7) respectively. Let the coordinates of point P be (x, y, z). On using distance formula, we obtain

$PA^{2} = (x - 3)^{2} + (y - 4)^{2} + (z - 5)^{2} = x^{2} + 9 - 6 x + y^{2} + 16 - 8 y + z^{2} + 25 - 10 z = x^{2} - 6 x + y^{2} - 8 y + z^{2} - 10 z + 50$

Now, if $PA^{2} + PB^{2} = k^{2}$ , then

$\Rightarrow \Rightarrow \Rightarrow (x^{2} - 6 x + y^{2} - 8 y + z^{2} - 10 z + 50) + (x^{2} + 2 x + y^{2} - 6 y + z^{2} + 14 z + 59) = k^{2} 2 x^{2} + 2 y^{2} + 2 z^{2} - 4 x - 14 y + 4 z + 109 = k^{2} 2 (x^{2} + y^{2} + z^{2} - 2 x - 7 y + 2 z) = k^{2} - 109 x^{2} + y^{2} + z^{2} - 2 x - 7 y + 2 z = \frac{k ^{2} - 109}{2}$

Thus, the required equation is

$x^{2} + y^{2} + z^{2} - 2 x - 7 y + 2 z = \frac{k ^{2} - 109}{2}$

3.0Class 11 Maths NCERT Solutions – Chapter-wise Links

Find chapter-wise NCERT Class 11 Maths solutions with accurate answers, important formulas, and simple explanations. These solutions help students understand each concept clearly, solve questions step by step, strengthen their problem-solving skills, and prepare effectively for exams.

Chapter Number	Chapterwise NCERT Solutions
Chapter 1	Sets
Chapter 2	Relations and Functions
Chapter 3	Trigonometric Functions
Chapter 4	Complex Numbers and Quadratic Equations
Chapter 5	Linear Inequalities
Chapter 6	Permutations and Combinations
Chapter 7	Binomial Theorem
Chapter 8	Sequence and Series
Chapter 9	Straight Lines
Chapter 10	Conic Sections
Chapter 12	Limits and Derivatives
Chapter 13	Statistics
Chapter 14	Probability

4.0Exercise-wise Questions and Key Topics in Class 11 Maths Chapter 11

Exercise	Questions with Solutions	Important Concepts Covered
Exercise 11.1	4 Questions	3D coordinate system, coordinates of points, distance formula, section formula
Exercise 11.2	5 Questions	Direction ratios, angle between two lines, equation of a plane, distance between a point and a plane
Miscellaneous Exercise	4 Questions	Mixed questions on 3D coordinates, direction ratios, equations of planes, angles, and distances

5.0Key Features of NCERT Solutions for Class 11 Maths Chapter 11 (Introduction to Three Dimensional Geometry)

Clear Understanding of 3D Coordinates: Learn how points are represented in three-dimensional space using the (x), (y), and (z) coordinates and how their signs determine the position of a point.
Octants and Coordinate Planes: Understand the eight octants in three-dimensional space and identify the position of a point using the signs of its coordinates.
Distance Between Two Points: Learn how to apply the three-dimensional distance formula to find the distance between two points with clear, step-by-step calculations.
Section Formula and Ratio: Understand how to find the coordinates of a point that divides a line segment in a given ratio, including cases involving the coordinate planes.
Midpoint and Collinearity: Learn how to find the midpoint of a line segment and use distance and coordinate methods to check whether points are collinear.
Geometrical Applications: Solve questions involving triangles, including finding side lengths, checking for special triangles, and determining important points using coordinates in three-dimensional space.
Prepared by ALLEN Subject Experts: These NCERT Solutions are prepared by ALLEN subject experts with accurate mathematical methods and alignment with the latest NCERT syllabus.
Complete NCERT Exercise Coverage: All questions from the NCERT exercises, including the miscellaneous exercise, are covered with clear, step-by-step solutions to help students understand and apply the concepts of three-dimensional geometry.

FAQ

Why are NCERT Solutions for Class 11 Maths Chapter 11 Introduction to Three-dimensional Geometry important for students?

How do Class 11 Maths NCERT Solutions for Chapter 11 help in board and JEE preparation?

What concepts are covered in NCERT Solutions for Class 11 Maths Chapter 11 Introduction to Three-dimensional Geometry?

Are NCERT Solutions for Class 11 Maths Chapter 11 by ALLEN useful for competitive exams?

Why is three-dimensional geometry important in mathematics?

What is the difference between internal and external division in 3D geometry?

Join ALLEN!

FAQ