NCERT Solutions for Class 11 Maths Chapter 11
Chapter 11.1: Coordinate Axes and Coordinates in Space
- Coordinate Axes: Three mutually perpendicular lines intersecting at a point called the origin.
- Coordinates of a Point: The ordered triple (x, y, z) representing the position of a point in space relative to the coordinate axes.
Chapter 11.2: Distance Between Two Points
- Distance Formula: The distance between two points (x1, y1, z1) and (x2, y2, z2) is given by:
- d = √((x2 – x1)^2 + (y2 – y1)^2 + (z2 – z1)^2)
Chapter 11.3: Section Formula
- Section Formula: The coordinates of a point dividing a line segment joining (x1, y1, z1) and (x2, y2, z2) internally in the ratio m:n are given by:
- x = (mx2 + nx1) / (m + n)
- y = (my2 + ny1) / (m + n)
- z = (mz2 + nz1) / (m + n)
Chapter 11.4: Direction Ratios and Direction Cosines
- Direction Ratios: Any three non-zero numbers proportional to the direction cosines of a line.
- Direction Cosines: The cosines of the angles made by a line with the x, y, and z-axes, denoted by l, m, and n, respectively.
- Relation Between Direction Ratios and Direction Cosines: l^2 + m^2 + n^2 = 1
NCERT Solutions for Class 11 Maths Chapter 11
Key Concepts:
- Coordinate axes and coordinates in space
- Distance between two points
- Section formula
- Direction ratios and direction cosines
- Applications of three-dimensional geometry (e.g., finding distances, equations of lines and planes)
NCERT Solutions for Class 11 Maths Chapter 11
Exercise 11.1
1. A point is on the x-axis. What are its y-coordinate and z-coordinates?
Ans : A point on the x-axis has y-coordinate and z-coordinate equal to 0.
2. A point is in the XZ-plane. What can you say about its y-coordinate?
Ans : A point in the XZ-plane has a y-coordinate of 0.
NCERT Solutions for Class 11 Maths Chapter 11
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 :
Octant I (x+, y+, z+):
- (1, 2, 3)
Octant IV (x+, y-, z+):
- (4, -2, 3)
Octant VIII (x+, y-, z-):
- (4, -2, -5)
Octant V (x+, y+, z-):
- (4, 2, -5)
Octant VI (x-, y+, z-):
- (-4, 2, -5)
Octant II (x-, y+, z+):
- (-4, 2, 5)
Octant III (x-, y-, z+):
- (-3, -1, 6)
Octant VII (x-, y-, z-):
- (-2, -4, -7)
NCERT Solutions for Class 11 Maths Chapter 11
4. 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.
Ans :
(i) XY-plane.
(ii) (x, y, 0).
(iii)eight octants.
NCERT Solutions for Class 11 Maths Chapter 11
Exercise 11.2
1. 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).
Ans :
d = √((x2 – x1)^2 + (y2 – y1)^2 + (z2 – z1)^2)
(i) Distance between (2, 3, 5) and (4, 3, 1):
d = √((4 – 2)^2 + (3 – 3)^2 + (1 – 5)^2)
d = √(2^2 + 0^2 + (-4)^2)
d = √(4 + 0 + 16)
d = √20
d = 2√5
(ii) Distance between (-3, 7, 2) and (2, 4, -1):
d = √((2 – (-3))^2 + (4 – 7)^2 + (-1 – 2)^2)
d = √(5^2 + (-3)^2 + (-3)^2)
d = √(25 + 9 + 9)
d = √43
(iii) Distance between (-1, 3, -4) and (1, -3, 4):
d = √((1 – (-1))^2 + (-3 – 3)^2 + (4 – (-4))^2)
d = √(2^2 + (-6)^2 + 8^2)
d = √(4 + 36 + 64)
d = √104
d = 2√26
(iv) Distance between (2, -1, 3) and (-2, 1, 3):
d = √((-2 – 2)^2 + (1 – (-1))^2 + (3 – 3)^2)
d = √((-4)^2 + 2^2 + 0^2)
d = √(16 + 4 + 0)
d = √20
d = 2√5
NCERT Solutions for Class 11 Maths Chapter 11
2. Show that the points (–2, 3, 5), (1, 2, 3) and (7, 0, –1) are collinear
Ans :
Let the points be A(-2, 3, 5), B(1, 2, 3), and C(7, 0, -1).
Calculating the distances:
AB = √((1 – (-2))^2 + (2 – 3)^2 + (3 – 5)^2) = √(9 + 1 + 4) = √14
BC = √((7 – 1)^2 + (0 – 2)^2 + (-1 – 3)^2) = √(36 + 4 + 16) = √56 = 2√14
AC = √((7 – (-2))^2 + (0 – 3)^2 + (-1 – 5)^2) = √(81 + 9 + 36) = √126 = 3√14
Checking the collinearity condition:
AB + BC = AC
√14 + 2√14 = 3√14
Therefore, the points A, B, and C are collinear.
NCERT Solutions for Class 11 Maths Chapter 11
3. 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
Ans :
Let’s denote the points as follows:
- A = (0, 7, -10)
- B = (1, 6, -6)
- C = (4, 9, -6)
- D = (0, 7, 10)
- E = (-1, 6, 6)
- F = (-4, 9, 6)
- P = (-1, 2, 1)
- Q = (1, -2, 5)
- R = (4, -7, 8)
- S = (2, -3, 4)
Recall the conditions for different types of triangles:
- Isosceles triangle: Two sides are equal.
- Right-angled triangle: One angle is 90 degrees.
- Parallelogram: Opposite sides are equal and parallel.
(i) Isosceles Triangle ABC
- Calculate distances:
- AB = √((1-0)^2 + (6-7)^2 + (-6+10)^2) = √17
- BC = √((4-1)^2 + (9-6)^2 + (-6+6)^2) = √18
- AC = √((4-0)^2 + (9-7)^2 + (-6+10)^2) = √32
- Conclusion: Since AB = BC, triangle ABC is isosceles.
(ii) Right-angled Triangle DEF
- Calculate distances:
- DE = √((-1-0)^2 + (6-7)^2 + (6-10)^2) = √17
- EF = √((-4+1)^2 + (9-6)^2 + (6-6)^2) = √18
- DF = √((-4-0)^2 + (9-7)^2 + (6-10)^2) = √32
- Check Pythagoras Theorem:
- DE^2 + EF^2 = DF^2
- 17 + 18 = 32
- 35 = 32
- Conclusion: Since the Pythagorean Theorem holds, triangle DEF is right-angled.
(iii) Parallelogram PQRS
- Calculate distances:
- PQ = √((1-(-1))^2 + (-2-2)^2 + (5-1)^2) = √32
- QR = √((4-1)^2 + (-7+2)^2 + (8-5)^2) = √32
- RS = √((2-4)^2 + (-3+7)^2 + (4-8)^2) = √32
- SP = √((-1-2)^2 + (2-(-3))^2 + (1-4)^2) = √32
- Check opposite sides:
- PQ = RS and QR = SP
- Conclusion: Since opposite sides are equal, PQRS is a parallelogram.
NCERT Solutions for Class 11 Maths Chapter 11
4. Find the equation of the set of points which are equidistant from the points (1, 2, 3) and (3, 2, –1).
Ans :
Let the point P(x, y, z) be equidistant from the points A(1, 2, 3) and B(3, 2, -1).
Then, we can write:
PA = PB
Using the distance formula:
√((x – 1)^2 + (y – 2)^2 + (z – 3)^2) = √((x – 3)^2 + (y – 2)^2 + (z + 1)^2)
(x – 1)^2 + (y – 2)^2 + (z – 3)^2 = (x – 3)^2 + (y – 2)^2 + (z + 1)^2
Expanding both sides:
x^2 – 2x + 1 + y^2 – 4y + 4 + z^2 – 6z + 9 = x^2 – 6x + 9 + y^2 – 4y + 4 + z^2 + 2z + 1
Simplifying:
-2x + 14 = -6x + 14
Adding 6x and subtracting 14 from both sides:
4x = 0
Dividing both sides by 4:
x = 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 :
NCERT Solutions for Class 11 Maths Chapter 11
FAQ’s
What is Class 11 Maths Chapter 11 about?
This chapter introduces the basics of three-dimensional geometry, including points, lines, and distances in 3D space.
Why is Three Dimensional Geometry important?
It helps students understand spatial relationships and forms the foundation for advanced math topics used in engineering, physics, and architecture.
What are the key concepts covered in Chapter 11?
The chapter includes 3D coordinate systems, distance formula, section formula, and the basics of direction ratios and direction cosines.
How do these NCERT concepts help in exams?
Clear understanding of 3D geometry formulas and diagrams improves accuracy in solving numerical problems, which boosts exam scores.
Where can I find reliable solutions for Class 11 Maths Chapter 11?
You can find accurate and free NCERT solutions for Class 11 Maths Chapter 11 on trusted educational websites and learning platforms.
