Class 11 NCERT Solutions Chapter 12 Introduction to three dimensional Geometry Exercise 12.3 (original) (raw)
Last Updated : 23 Jul, 2025
Chapter 12 of the Class 11 NCERT Mathematics textbook, titled "Introduction to Three-Dimensional Geometry," provides an essential foundation for understanding the spatial relationships between points, lines, and planes in three dimensions. This chapter introduces the concepts of coordinates in three-dimensional space, the distance between points, and the section formula. Exercise 12.3 specifically focuses on problems related to these concepts.
Class 11 NCERT Solutions - Chapter 12 Introduction to Three-Dimensional Geometry - Exercise 12.3
This section offers comprehensive solutions to Exercise 12.3 from Chapter 12 of the Class 11 NCERT Mathematics textbook. The solutions are designed to help students navigate through the complexities of three-dimensional geometry, ensuring a solid grasp of the concepts and their applications.
**Explanation:
Chapter 12 "Introduction to Three Dimensional Geometry " in the Class 11 NCERT Mathematics textbook provides students with the foundational knowledge of three-dimensional space covering concepts like coordinate systems, direction cosines, and distance between points in space. Exercise 12.3 focuses on the problems related to finding the coordinates of a point in space, distances between the points, and understanding the 3D coordinate system. Mastering this exercise is crucial as it forms the basis for the more advanced topics in three-dimensional geometry which are essential for the students aspiring to excel in mathematics.
Introduction to Three-dimensional Geometry
Introduction to Three-Dimensional Geometry in Class 11 Mathematics (Chapter 12) provides the students with a fundamental understanding of representing points in three-dimensional space using coordinates. Exercise 12.3 focuses on calculating distances between the points and understanding the spatial relationships between them. This chapter is crucial as it lays the groundwork for the more advanced topics in geometry and vector algebra which are essential for higher studies in mathematics, physics, and engineering.
Class 11 NCERT Mathematics Solutions - Exercise 12.3
Problem 1: Find the coordinates of the point that divides the line segment joining the points (– 2, 3, 5) and (1, – 4, 6) in the ratio.
(i) 2: 3 internally
**Solution:
By using Section formula,
(x, y, z) = \mathbf{(\frac{(mx_2 + nx_1)}{(m + n)}, \frac{(my_2 + ny_1)}{(m + n)}, \frac{(mz_2 + nz_1)}{(m + n)})}
Upon comparing we get,
x1 = -2, y1 = 3, z1 = 5
x2 = 1, y2 = -4, z2 = 6 and
m = 2, n = 3
So, the coordinates of the point which divides the line segment joining the points P (– 2, 3, 5) and Q (1, – 4, 6) in the ratio 2 : 3 internally is given by:
((2 * 1 + 3 * -2)/(2 + 3), (2 * -4 + 3 * 3)/(2 + 3), (2 * 6 + 3 * 5)/(2 + 3))
****((-4/5), (1/5), (27/5))**
**Hence, the coordinates of the point which divides the line segment joining the points (-2, 3, 5) and (1, -4, 6) is (-4/5, 1/5, 27/5)
(ii) 2: 3 externally
**Solution:
By using Section formula,
(x, y, z) = \mathbf{(\frac{(mx_2 - nx_1)}{(m - n)}, \frac{(my_2 - ny_1)}{(m - n)}, \frac{(mz_2 - nz_1)}{(m - n)})}
Upon comparing we get,
x1 = -2, y1 = 3, z1 = 5;
x2 = 1, y2 = -4, z2 = 6 and
m = 2, n = 3
So, the coordinates of the point which divides the line segment joining the points P (– 2, 3, 5) and Q (1, – 4, 6) in the ratio 2: 3 externally is given by:
((2 * 1 - 3 * -2)/(2 + 3), (2 * -4 - 3 * 3)/(2 + 3), (2 * 6 - 3 * 5)/(2 + 3))
((2 + 6)/(-1), (-8 - 9)/(-1), (12 - 15)/(-1)
****(-8), (17), (3)**
**∴ The co-ordinates of the point which divides the line segment joining the points (-2, 3, 5) and (1, -4, 6) is (-8, 17, 3).
Problem 2: Given that P (3, 2, – 4), Q (5, 4, – 6) and R (9, 8, –10) are collinear. Find the ratio in which Q divides PR.
**Solution:
Let us consider Q divides PR in the ratio k: 1.
By using Section formula,
(x, y, z) = \mathbf{(\frac{(mx_2 + nx_1)}{(m + n)}, \frac{(my_2 + ny_1)}{(m + n)}, \frac{(mz_2 + nz_1)}{(m + n)})}
Upon comparing we get,
x1 = 3, y1 = 2, z1 = -4;
x2 = 9, y2 = 8, z2 = -10 and
m = k, n = 1
So, we have
((k * 9 + 1 * 3)/(k + 1), (k * 8 + 1 * 2)/(k + 1), (-10 * k + 1 * -4)/(k + 1) = (5, 4, -6)
(9k + 3)/(k + 1) = 5, (8k + 2)/(k + 1) = 4, (-10k + -4)/(k + 1) = -6
(8k + 2)/(k + 1) = 4
8k + 2 = 4k + 4
4k = 2
**k = 1/2
**Hence, the ratio in which Q divides PR is 1: 2
Problem 3: Find the ratio in which the YZ-plane divides the line segment formed by joining the points (–2, 4, 7) and (3, –5, 8).
**Solution:
Let the line segment formed by joining the points P (-2, 4, 7) and Q (3, -5, 8) be PQ.
We know that any point on the YZ-plane is of the form (0, y, z).
So now, let R (0, y, z) divides the line segment PQ in the ratio k: 1.
Then,
Upon comparing we get,
x1 = -2, y1 = 4, z1 = 7;
x2 = 3, y2 = -5, z2 = 8 and
m = k, n = 1
By using the Section formula,
(x, y, z) = \mathbf{(\frac{(mx_2 + nx_1)}{(m + n)}, \frac{(my_2 + ny_1)}{(m + n)}, \frac{(mz_2 + nz_1)}{(m + n)})}
So we have,
((3k - 2)/(k + 1), (-5k + 4)/(k + 1), (8k + 7)/(k + 1)) = (0, y, z)
3k – 2 = 0
3k = 2
**k = 2/3
**Hence, the ratio in which the YZ-plane divides the line segment formed by joining the points (–2, 4, 7) and (3, –5, 8) is 2:3.
Problem 4: Using section formula, show that the points A (2, –3, 4), B (–1, 2, 1) and C (0, 1/3, 2) are collinear.
**Solution:
Let the point P divides AB in the ratio k: 1.
Upon comparing we get,
x1 = 2, y1 = -3, z1 = 4;
x2 = -1, y2 = 2, z2 = 1 and
m = k, n = 1
By using Section formula,
(x, y, z) = \mathbf{(\frac{(mx_2 + nx_1)}{(m + n)}, \frac{(my_2 + ny_1)}{(m + n)}, \frac{(mz_2 + nz_1)}{(m + n)})}
So we have,
The coordinates of P are:
((-k + 2)/(k + 1), (2k - 3)/(k + 1), (k + 4)/(k + 1))
Now, we check if for some value of k, the point coincides with the point C.
Put (-k + 2)/(k + 1) = 0
-k + 2 = 0
k = 2
When k = 2, then (2k - 3)/(k + 1) = (2(2) - 3)/(2 + 1)
= (4 - 3)/3
= 1/3
And, (k + 4)/(k +1) = (2 + 4)/(2 +1)
= 6/3
= 2
**∴ C (0, 1/3, 2) is a point which divides AB in the ratio 2: 1 and is same as P.
**Hence, A, B, C are collinear.
Problem 5: Find the coordinates of the points which trisect the line segment joining the points P (4, 2, – 6) and Q (10, –16, 6).
**Solution:
Let A (x1, y1, z1) and B (x2, y2, z2) trisect the line segment joining the points P (4, 2, -6) and Q (10, -16, 6).
A divides the line segment PQ in the ratio 1: 2.
Upon comparing we get,
x1 = 4, y1 = 2, z1 = -6;
x2 = 10, y2 = -16, z2 = 6 and
m = 1, n = 2
By using the Section formula,
(x, y, z) = \mathbf{(\frac{(mx_2 + nx_1)}{(m + n)}, \frac{(my_2 + ny_1)}{(m + n)}, \frac{(mz_2 + nz_1)}{(m + n)})}
So we have,
The coordinates of A are:
((1*10 + 2*4)/(1 + 2), (1*(-16) + 2*2)/(1 + 2), (1*6 + 2*(-6))/(1 + 2)
((18/3), (-12/3), (-6/3))
**The coordinates of A are (6, -4, -2)
Similarly, we know that B divides the line segment PQ in the ratio 2: 1.
Upon comparing we get,
x1 = 4, y1 = 2, z1 = -6;
x2 = 10, y2 = -16, z2 = 6 and
m = 2, n = 1
By using the Section formula,
(x, y, z) = \mathbf{(\frac{(mx_2 + nx_1)}{(m + n)}, \frac{(my_2 + ny_1)}{(m + n)}, \frac{(mz_2 + nz_1)}{(m + n)})}
So we have,
The coordinates of B are:
((2 * 10 + 1 * 4)/(2 + 1), (2 * -16 + 1 * 2)/(2 + 1), (2 * 6 + 1 * -6)/(2 + 1)
**The coordinates of B are (8, -10, 2)
**∴ The coordinates of the points which trisect the line segment joining the points P (4, 2, – 6) and Q (10, –16, 6) are (6, -4, -2) and (8, -10, 2).