NCERT Class 11 Mathematics Chapter 11 Introduction of Three Dimension Geomenter Solutions, NCERT Solutions For Class 11 Maths, NCERT Class 11 Mathematics Chapter 11 Introduction of Three Dimension Geomenter Notes to each chapter is provided in the list so that you can easily browse throughout different chapter NCERT Class 11 Mathematics Chapter 11 Introduction of Three Dimension Geomenter Notes and select needs one.
NCERT Class 11 Mathematics Chapter 11 Introduction of Three Dimension Geomenter
Also, you can read the CBSE book online in these sections Solutions by Expert Teachers as per (CBSE) Book guidelines. NCERT Class 11 Mathematics Chapter 11 Introduction of Three Dimension Geomenter Textual Question Answer. These solutions are part of NCERT All Subject Solutions. Here we have given NCERT Class 11 Mathematics Chapter 11 Introduction of Three Dimension Geomenter Solutions for All Subject, You can practice these here.
Introduction of Three Dimension Geomenter
Chapter – 11
| Exercise 11.1 |
1. A point is on the x-axis. What are its y-coordinate and z-coordinates?
Ans: If a point is on the x-axis, then its y-coordinates and z-coordinates are zero.
2. A point is in the XZ-plane. What can you say about its y-coordinate?
Ans: If a point is in the XZ plane, then its y-coordinate is zero.
3. 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).
Ans: The x-coordinate, y-coordinate, and z-coordinate of points (1, 2, 3) are all positive. Therefore, this point lies in octant I.
The x-coordinate, y-coordinate, and z-coordinate of points (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 points (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 points (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 points (-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 points (-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 points (2, -4, -7) are positive, negative, and negative respectively. Therefore, this point lies in octant VIII.
4. Fill in the blanks:
(i) The x-axis and y-axis taken together determine a plane known as___________.
Ans: 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_____________.
Ans: The coordinates of points in the XY-plane are of the form (x,y,0)
(iii) Coordinate planes divide the space into__________ octants.
Ans: Coordinate planes divide the space into eight octants.
| Exercise 11.2 |
1. Find the distance between the following pairs of points:
(i) (2, 3, 5) and (4, 3, 1)
Ans: 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
= 2√5
(ii) (-3, 7, 2) and (2, 4, -1)
Ans: 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
(iii) (-1,3,-4) and (1, -3, 4)
Ans: 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 = 2√26
(iv) (2,-1, 3) and (-2, 1, 3)
Ans: 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
= 2√5
2. Show that the points (-2, 3, 5), (1, 2, 3) and (7, 0, -1) are collinear.
Ans: 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 = √(1 + 2)² + (2 – 3)² + (3 – 5)²
= √(3)² + (- 1)² + (- 2)²
= √9 + 1 + 4
= √14
QR = √(7 – 1)² + (0 – 2)² + (- 1 – 3)²
= √(6)² + (- 2)² + (- 4)²
= √36 + 4 + 16
= √56
= 2√14
PR = √(7 + 2)² + (0 – 3)² + (- 1 – 5)²
= √(9)² + (- 3)² + (- 6)²
= √(81 + 9 + 36)
= √126
= 3√14
Here, PQ + QR = √14 + 2√14 =3 √14 = PR,
Hence, points P(- 2, 3, 5), Q(1, 2, 3), and R(7, 0, – 1) are collinear.
3. Verify the following:
(i) (0,7,-10), (1, 6, -6) and (4, 9, -6) are the vertices of an isosceles triangle.
Ans: Let points (0, 7, – 10), (1, 6, – 6), and (4, 9, – 6) be denoted by A, B, and C respectively.
AB = √(1 + 0)² + (6 – 7)² + (- 6 + 10)²
= ᵃ√(1)² + (- 1)² + (4)²
= √1 + 1 + 16
= √18
= 3√2
BC = √(4 – 1)² + (9 – 6)² + (- 6 + 6)²
= √(3)² + (3)²
= √9 + 9 = √18 = 3√2
CA = √(0 – 4)² + (7 – 9)² + (- 10 + 6)²
= √(- 4)² + (- 2)² + (- 4)²
= √16 + 4 + 16 = √36 = 6
Here, AB =BC ≠ CA
Thus, the given points 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.
Ans: Let (0, 7, 10), ( – 1,6,6) and ( – 4,9,6) be denoted by A, B, and C respectively.
AB = √(- 1 – 0)² + (6 – 7)² + (6 – 10)²
= √(- 1)² + (- 1)² + (- 4)²
= √1 + 1 + 16 = √18
= 3√2
BC = √(- 4 + 1)² + (9 – 6)² + (6 – 6)²
= √(- 3)² + (3)² + (0)²
= √9 + 9 = √18
= 3√2
CA = √(0 + 4)² + (7 – 9)² + (10 – 6)²
= √(4)² + (- 2)² + (4)²
= √16 + 4 + 16
= √36
= 6
Now, AB² + BC² = (3√2)² + (3√2)² = 18 + 18 = 36 = AC²
Therefore, by Pythagoras theorem, ABC is a right triangle.
Hence, the given points 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.
Ans: 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.
4. Find the equation of the set of points which are equidistant from the points (1, 2, 3) and (3, 2, -1).
Ans: Let P(x, y, z) be the point that is equidistant from points A(1, 2, 3) and B( 3, 2 ,- 1)
Accordingly, PA = PB
⇒ PA² = PB²
⇒ (x – 1)² + (y – 2)² + (z – 3)² = (x – 3)² + (y – 2)² + (z + 1)²
⇒ x² – 2x + 1 + y² – 4y + 4 + z² – 6z + 9 = x² – 6x + 9 + y² – 4y + 4 + z² + 2z + 1
⇒ – 2x – 4y – 6z + 14 = – 6x – 4y + 2z + 14
⇒ – 2x – 6z + 6x – 2z = 0
⇒ 4x – 8z = 0
⇒ x – 2z = 0
Thus, the required equation is x – 2z = 0
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.
Ans: 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
⇒ √(x – 4)² + y² + z² + √(x + 4)² + y² + z² = 10
⇒ √(x – 4)² + y² + z² = 10 – √(x + 4)² + y² + z²
On squaring both sides, we obtain
⇒ (x – 4)² + y² + z² = 100 – 20√(x + 4)² + y² + z² + (x + 4)² + y² + z²
⇒ x² – 8x + 16 + y² + z² = 100 – 20√x² + 8x + 16 + y² + z² + x² + 8x + 16 + y² + z²
⇒ 20√x² + 8x + 16 + y² + z² = 100 + 16x
⇒ 5√x² + 8x + 16 + y² + z² = (25 + 4x)
On squaring both sides again, we obtain
25(x² + 8x + 16 + y² + z²) = 625 + 16x² + 200x
⇒ 25x² + 200x + 400 + 25y² + 25z² = 625 + 16x² + 200x
⇒ 9x² + 25y² + 25z² – 225 = 0
Thus, the required equation is 9x² + 25y² + 25z² – 225 = 0
| Miscellaneous Exercise on Chapter 11 |
1. 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.
Ans: 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).
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
⇒ x = 1 , y = – 2 and z = 8
Thus, the coordinates of the fourth vertex are ( 1 ,- 2,8)
2. Find the lengths of the medians of the triangle with vertices A (0, 0, 6), B (0,4, 0) and (6, 0, 0).
Ans: Let AD, BE, and CF be the medians of the given triangle ABC.
Since AD is the median, D is the midpoint of BC.
∴ Coordinates of point D
= (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 midpoint of AC.
∴ Coordinates of point E
= (3,0,3)
BE = √(3 + 0)² + (0 – 4)² + (3 + 0)² = √9 + 16 + 9) = √34
Since CF is the median, F is the midpoint of AB.
∴ Coordinates of point F =
= (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
3. If the origin is the centroid of the ∆PQR with vertices P (2a, 2, 6) ) (- 4, 3b, – 10) and R(8, 14, 2c) then find the values of a, b and c.
Ans:
It is known that the coordinates of the centroid of the triangle, whose vertices are (x₁, y₁, z₁) (x₂, y₂, z₂) and (x₃, y₃, z₃) are
Therefore, coordinates of the centroid of
∆PQR =
It is given that origin is the centroid of ∆PQR
Thus, the respective values of a, b, and c are
4. If A and B are the points (3, 4, 5) and (-1, 3, -7), respectively, find the equation of the set of points P such that PA² + PB² = k² where k is a constant.
Ans: 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
PA² = (x – 3)² + (y – 4)² + (z – 5)²
= x² + 9 – 6x + y² + 16 – 8y + z² + 25 – 10z
= x² – 6x + y² – 8y + z² – 10z + 50
PB² = (x + 1)² + (y – 3)² + (z + 7)²
= x² + 2x + y² – 6y + z² + 14z + 59
Now, if PA² + PB² = k², then
(x² – 6x + y² – 8y + z² – 10z + 50) + (x² + 2x + y² – 6y + z² + 14z + 59) = k²
⇒ 2x² + 2y² + 2z² – 4x – 14y + 4z + 109 = k²
⇒ 2(x² + y² + z² – 2x – 7y + 2z) = k² – 109
⇒ x² + y² + z² – 2x – 7y + 2z =
Thus, the required equation is x² + y² + z² – 2x – 7y + 2z =
