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NCERT Solutions Class 11 - Chapter 11 Introduction to three dimensional Geometry - Exercise 11.2

Last Updated : 22 Apr, 2024
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Problem 1: Find the distance between the following pairs of points:

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

Solution:

Let P be (2, 3, 5) and Q be (4, 3, 1)

Now, by using the distance formula,

Length of distance PQ = √[(x2 – x1)2 + (y2 – y1)2 + (z2 – z1)2]

So here,

x1 = 2, y1 = 3, z1 = 5

x2 = 4, y2 = 3, z2 = 1

Length of distance PQ = √[(4 – 2)2 + (3 – 3)2 + (1 – 5)2]

= √[(2)2 + (0)2 + (-4)2]

= √[4 + 0 + 16]

= √20

= 2√5

∴ The length of distance PQ is 2√5 units.

(ii) (–3, 7, 2) and (2, 4, –1)

Solution:

Let P be (– 3, 7, 2) and Q be (2, 4, – 1)

Now, by using the distance formula,

Length of distance PQ = √[(x2 – x1)2 + (y2 – y1)2 + (z2 – z1)2]

So here,

x1 = – 3, y1 = 7, z1 = 2

x2 = 2, y2 = 4, z2 = – 1

Length of distance PQ = √[(2 – (-3))2 + (4 – 7)2 + (-1 – 2)2]

= √[(5)2 + (-3)2 + (-3)2]

= √[25 + 9 + 9]

= √43

∴ The length of distance PQ is √43 units.

(iii) (–1, 3, – 4) and (1, –3, 4)

Solution:

Let P be (– 1, 3, – 4) and Q be (1, – 3, 4)

Now, by using the distance formula,

Length of distance PQ = √[(x2 – x1)2 + (y2 – y1)2 + (z2 – z1)2]

So here,

x1 = – 1, y1 = 3, z1 = – 4

x2 = 1, y2 = – 3, z2 = 4

Length of distance PQ = √[(1 – (-1))2 + (-3 – 3)2 + (4 – (-4))2]

= √[(2)2 + (-6)2 + (8)2]

= √[4 + 36 + 64]

= √104

= 2√26

∴ The length of distance PQ is 2√26 units.

(iv) (2, –1, 3) and (–2, 1, 3)

Solution:

Let P be (2, – 1, 3) and Q be (– 2, 1, 3)

Now, by using the distance formula,

Length of distance PQ = √[(x2 – x1)2 + (y2 – y1)2 + (z2 – z1)2]

So here,

x1 = 2, y1 = – 1, z1 = 3

x2 = – 2, y2 = 1, z2 = 3

Length of distance PQ = √[(-2 – 2)2 + (1 – (-1))2 + (3 – 3)2]

= √[(-4)2 + (2)2 + (0)2]

= √[16 + 4 + 0]

= √20

= 2√5

∴ The required distance is 2√5 units.

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

Solution:

If three points are collinear, then they lie on a line.

Firstly let us calculate distance between the 3 points

i.e. PQ, QR and PR

P ≡ (– 2, 3, 5) and Q ≡ (1, 2, 3)

Now, by using the distance formula,

Length of distance PQ = √[(x2 – x1)2 + (y2 – y1)2 + (z2 – z1)2]

So here,

x1 = – 2, y1 = 3, z1 = 5

x2 = 1, y2 = 2, z2 = 3

Length of distance PQ = √[(1 – (-2))2 + (2 – 3)2 + (3 – 5)2]

= √[(3)2 + (-1)2 + (-2)2]

= √[9 + 1 + 4]

= √14

Length of distance PQ is √14

Q ≡ (1, 2, 3) and R ≡ (7, 0, – 1)

Now, by using the distance formula,

Length of distance QR = √[(x2 – x1)2 + (y2 – y1)2 + (z2 – z1)2]

So here,

x1 = 1, y1 = 2, z1 = 3

x2 = 7, y2 = 0, z2 = – 1

Length of distance QR = √[(7 – 1)2 + (0 – 2)2 + (-1 – 3)2]

= √[(6)2 + (-2)2 + (-4)2]

= √[36 + 4 + 16]

= √56

= 2√14

Length of distance QR is 2√14

P ≡ (– 2, 3, 5) and R ≡ (7, 0, – 1)

Now, by using the distance formula,

Length of distance PR= √[(x2 – x1)2 + (y2 – y1)2 + (z2 – z1)2]

So here,

x1 = – 2, y1 = 3, z1 = 5

x2 = 7, y2 = 0, z2 = – 1

Length of distance PR = √[(7 – (-2))2 + (0 – 3)2 + (-1 – 5)2]

= √[(9)2 + (-3)2 + (-6)2]

= √[81 + 9 + 36]

= √126

= 3√14

Length of distance PR is 3√14

Thus, PQ = √14, QR = 2√14 and PR = 3√14

So, PQ + QR = √14 + 2√14

= 3√14

= PR

∴ The points P, Q and R are collinear.

Problem 3: Verify the following:

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

Solution:

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

Let us consider the points be

P(0, 7, –10), Q(1, 6, – 6) and R(4, 9, – 6)

If any 2 sides are equal, hence it will be an isosceles triangle

So firstly let us calculate the distance of PQ, QR

P ≡ (0, 7, – 10) and Q ≡ (1, 6, – 6)

Now, by using the distance formula,

Length of distance PQ = √[(x2 – x1)2 + (y2 – y1)2 + (z2 – z1)2]

So here,

x1 = 0, y1 = 7, z1 = – 10

x2 = 1, y2 = 6, z2 = – 6

Length of distance PQ = √[(1 – 0)2 + (6 – 7)2 + (-6 – (-10))2]

= √[(1)2 + (-1)2 + (4)2]

= √[1 + 1 + 16]

= √18

Calculating QR

Q ≡ (1, 6, – 6) and R ≡ (4, 9, – 6)

Now, by using the distance formula,

Length of distance QR = √[(x2 – x1)2 + (y2 – y1)2 + (z2 – z1)2]

So here,

x1 = 1, y1 = 6, z1 = – 6

x2 = 4, y2 = 9, z2 = – 6

Length of distance QR = √[(4 – 1)2 + (9 – 6)2 + (-6 – (-6))2]

= √[(3)2 + (3)2 + (-6+6)2]

= √[9 + 9 + 0]

= √18

Hence, 

Length of distance PQ = Length of distance QR i.e 

√18 = √18

∴ Length of 2 sides are equal

∴ PQR is an isosceles triangle.

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

Solution:

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

Let the points be

P(0, 7, 10), Q(– 1, 6, 6) & R(– 4, 9, 6)

Firstly let us calculate the distance of PQ, OR and PR

Calculating PQ

P ≡ (0, 7, 10) and Q ≡ (– 1, 6, 6)

Now, by using the distance formula,

Length of distance PQ = √[(x2 – x1)2 + (y2 – y1)2 + (z2 – z1)2]

So here,

x1 = 0, y1 = 7, z1 = 10

x2 = – 1, y2 = 6, z2 = 6

Length of distance PQ = √[(-1 – 0)2 + (6 – 7)2 + (6 – 10)2]

= √[(-1)2 + (-1)2 + (-4)2]

= √[1 + 1 + 16]

= √18

Length of distance PQ is √18cm

Q ≡ (1, 6, – 6) and R ≡ (4, 9, – 6)

Now, by using the distance formula,

Length of distance QR = √[(x2 – x1)2 + (y2 – y1)2 + (z2 – z1)2]

So here,

x1 = 1, y1 = 6, z1 = – 6

x2 = 4, y2 = 9, z2 = – 6

Length of distance QR = √[(4 – 1)2 + (9 – 6)2 + (-6 – (-6))2]

= √[(3)2 + (3)2 + (-6+6)2]

= √[9 + 9 + 0]

= √18

Length of distance QR is √18cm

P ≡ (0, 7, 10) and R ≡ (– 4, 9, 6)

Now, by using the distance formula,

Length of distance PR = √[(x2 – x1)2 + (y2 – y1)2 + (z2 – z1)2]

So here,

x1 = 0, y1 = 7, z1 = 10

x2 = – 4, y2 = 9, z2 = 6

Length of distance PR = √[(-4 – 0)2 + (9 – 7)2 + (6 – 10)2]

= √[(-4)2 + (2)2 + (-4)2]

= √[16 + 4 + 16]

= √36

Length of distance PR is √36cm

Now,

PQ2 + QR2 = 18 + 18

= 36

= PR2

By using converse of Pythagoras theorem,

∴ The given vertices P, Q & R are the vertices of a right-angled triangle at Q

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

Solution :

(–1, 2, 1), (1, –2, 5), (4, –7, 8) and (2, –3, 4) are the vertices of a parallelogram.

Let the points be: A(–1, 2, 1), B(1, –2, 5), C(4, –7, 8) & D(2, –3, 4)

ABCD can be vertices of parallelogram only if opposite sides are equal.

i.e. AB = CD and BC = AD

Firstly let us calculate the distance

A ≡ (– 1, 2, 1) and B ≡ (1, – 2, 5)

Now, by using the distance formula,

Length of distance AB = √[(x2 – x1)2 + (y2 – y1)2 + (z2 – z1)2]

So here,

x1 = – 1, y1 = 2, z1 = 1

x2 = 1, y2 = – 2, z2 = 5

Length of distance AB = √[(1 – (-1))2 + (-2 – 2)2 + (5 – 1)2]

= √[(2)2 + (-4)2 + (4)2]

= √[4 + 16 + 16]

= √36

= 6

Length of distance AB is 6cm

B ≡ (1, – 2, 5) and C ≡ (4, – 7, 8)

Now, by using the distance formula,

Length of distance BC = √[(x2 – x1)2 + (y2 – y1)2 + (z2 – z1)2]

So here,

x1 = 1, y1 = – 2, z1 = 5

x2 = 4, y2 = – 7, z2 = 8

Length of distance BC = √[(4 – 1)2 + (-7 – (-2))2 + (8 – 5)2]

= √[(3)2

 + (-5)2 + (3)2]

= √[9 + 25 + 9]

= √43

Length of distance BC is √43cm

C ≡ (4, – 7, 8) and D ≡ (2, – 3, 4)

Now, by using the distance formula,

Length of distance CD = √[(x2 – x1)2 + (y2 – y1)2 + (z2 – z1)2]

So here,

x1 = 4, y1 = – 7, z1 = 8

x2 = 2, y2 = – 3, z2 = 4

Length of distance CD = √[(2 – 4)2 + (-3 – (-7))2 + (4 – 8)2]

= √[(-2)2 + (4)2 + (-4)2]

= √[4 + 16 + 16]

= √36

= 6

Length of distance CD is 6cm

D ≡ (2, – 3, 4) and A ≡ (– 1, 2, 1)

By using the formula,

Length of distance DA = √[(x2 – x1)2 + (y2 – y1)2 + (z2 – z1)2]

So here,

x1 = 2, y1 = – 3, z1 = 4

x2 = – 1, y2 = 2, z2 = 1

Length of distance DA = √[(-1 – 2)2 + (2 – (-3))2 + (1 – 4)2]

= √[(-3)2 + (5)2 + (-3)2]

= √[9 + 25 + 9]

= √43

Length of distance DA is √43cm

Since AB = CD and BC = DA (given)

So, In ABCD both pairs of opposite sides are equal

∴ ABCD is a parallelogram

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

Solution:

Let A (1, 2, 3) & B (3, 2, – 1)

Let point P be (x, y, z)

Since it is given that point P(x, y, z) is equal distance from point A(1, 2, 3) & B(3, 2, – 1) i.e. PA = PB

P ≡ (x, y, z) and A ≡ (1, 2, 3)

Now, by using the distance formula,

Now, by using the distance formula, PA = √[(x2 – x1)2 + (y2 – y1)2 + (z2 – z1)2]

So here,

x1 = x, y1 = y, z1 = z

x2 = 1, y2 = 2, z2 = 3

Length of distance PA = √[(1 – x)2 + (2 – y)2 + (3 – z)2]

P ≡ (x, y, z) and B ≡ (3, 2, – 1)

Now, by using the distance formula,

Length of distance PB = √[(x2 – x1)2 + (y2 – y1)2 + (z2 – z1)2]

So here,

x1 = x, y1 = y, z1 = z

x2 = 3, y2 = 2, z2 = – 1

Length of distance PB = √[(3 – x)2 + (2 – y)2 + (-1 – z)2]

Since PA = PB

Square on both the sides, we get

PA2 = PB2

(1 – x)2 + (2 – y)2 + (3 – z)2 = (3 – x)2 + (2 – y)2 + (– 1 – z)2

(1 + x2 – 2x) + (4 + y2 – 4y) + (9 + z2 – 6z)

(9 + x2 – 6x) + (4 + y2 – 4y) + (1 + z2 + 2z)

– 2x – 4y – 6z + 14 = – 6x – 4y + 2z + 14

4x – 8z = 0

x – 2z = 0

∴ The required equation is x – 2z = 0

Problem 5: 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.

Solution:

Let A (4, 0, 0) & B (– 4, 0, 0)

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

Calculating PA

P ≡ (x, y, z) and A ≡ (4, 0, 0)

Now, by using the distance formula,

Length of distance PA = √[(x2 – x1)2 + (y2 – y1)2 + (z2 – z1)2]

So here,

x1 = x, y1 = y, z1 = z

x2 = 4, y2 = 0, z2 = 0

Length of distance PA = √[(4– x)2 + (0 – y)2 + (0 – z)2]

Calculating PB

P ≡ (x, y, z) and B ≡ (– 4, 0, 0)

Now, by using the distance formula,

Length of distance PB = √[(x2 – x1)2 + (y2 – y1)2 + (z2 – z1)2]

So here,

x1 = x, y1 = y, z1 = z

x2 = – 4, y2 = 0, z2 = 0

Length of distance PB = √[(-4– x)2 + (0 – y)2 + (0 – z)2]

Now it is given that:

PA + PB = 10

PA = 10 – PB

Square on both the sides, we get

PA2 = (10 – PB)2

PA2 = 100 + PB2 – 20 PB

(4 – x)2 + (0 – y)2 + (0 – z)2

100 + (– 4 – x)2 + (0 – y)2 + (0 – z)2 – 20 PB

(16 + x2 – 8x) + (y2) + (z2)

100 + (16 + x2 + 8x) + (y2) + (z2) – 20 PB

20 PB = 16x + 100

5 PB = (4x + 25)

Square on both the sides again, we get

25 PB2 = 16x2 + 200x + 625

25 [(– 4 – x)2 + (0 – y)2 + (0 – z)2] = 16x2 + 200x + 625

25 [x2 + y2 + z2 + 8x + 16] = 16x2 + 200x + 625

25x2 + 25y2 + 25z2 + 200x + 400 = 16x2 + 200x + 625

9x2 + 25y2 + 25z2 – 225 = 0

∴ The required equation is 9x2 + 25y2 + 25z2 – 225 = 0


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NCERT Solutions for Class 9 Maths Chapter 5 Introduction to Euclid’s Geometry
NCERT Solutions for Class 9 Maths Chapter 5 Introduction to Euclid’s Geometry is an article curated by the professional team of subject matter experts at GfG, to help students learn how to solve the problems given in the NCERT textbook. NCERT Solutions for Class 9 Maths Chapter 5- Introduction to Euclid’s Geometry on this page are regularly revised
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Class 9 RD Sharma Solutions - Chapter 7 Introduction to Euclid’s Geometry- Exercise 7.1
Question 1: Define the following terms.(i) Line segment(ii) Collinear points(iii) Parallel lines(iv) Intersecting lines(v) Concurrent lines(vi) Ray(vii) Half-line Solution: (i) A line segment is a one-dimensional line connecting two points. It is the shortest distance that is incident on both the points. (ii) Two or more points incident on the same
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Class 11 RD Sharma Solutions - Chapter 28 Introduction to 3D Coordinate Geometry - Exercise 28.2 | Set 1
Question 1. Find the distance between the following pairs of points:(i) P(1, −1, 0) and Q (2, 1, 2) Solution: Given: The points P(1, −1, 0) and Q(2, 1, 2) By using distance formula, the distance between any two points (a, b, c) and (d, e, f) is given as follows: [Tex]\sqrt{(a − d)^2 + (b − e)^2 + (c − f)^2}[/Tex] Hence, distance between P(1, −1, 0)
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Class 11 RD Sharma Solutions - Chapter 28 Introduction to 3D Coordinate Geometry - Exercise 28.3
Question 1. The vertices of the triangle are A(5, 4, 6), B(1, -1, 3), and C(4, 3, 2). The internal bisector of angle A meets BC at D. Find the coordinates of D and the Length AD. Solution: We know that the angle bisector divides opposite side in ratio of other two sides ⇒D divides BC in ratio of AB:AC A(5, 4, 6), B(1, -1, 3) and C(4, 3, 2) [Tex]AB=
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Class 11 RD Sharma Solutions - Chapter 28 Introduction to 3D Coordinate Geometry - Exercise 28.2 | Set 2
Question 13. Prove that the tetrahedron with vertices at points O(0,0,0), A(0,1,1), B(1,0,1) and C(1,1,0) is a regular one.Solution: Given: The points O(0,0,0), A(0,1,1), B(1,0,1) and C(1,1,0). A regular tetrahedron has all equal sides and diagonals. We know that the distance between two points (a, b, c) and (d, e, f) is given as follows: [Tex]\sqr
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Class 11 RD Sharma Solutions - Chapter 28 Introduction to 3D Coordinate Geometry - Exercise 28.1
Chapter 28 of RD Sharma's Class 11 mathematics textbook introduces students to the fascinating world of 3D Coordinate Geometry, with Exercise 28.1 serving as the foundational entry point into this three-dimensional mathematical landscape. This exercise extends the familiar concepts of 2D coordinate geometry into the third dimension, introducing the
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Real-life Applications of Three Dimensional Geometry
The word "geometry" itself comes from two Greek words meaning "earth" and "measurement." In this article, we're going to talk about how 3-D geometry is used in our daily lives. If you look around, you'll see geometry everywhere! Table of Content What is 3D Geometry?Applications of Three-Dimensional GeometryReal life Applications of Three-Dimensiona
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