NCERT Solutions for Class 12 Maths Chapter 11: Three Dimensional Geometry - Exercise 11.2
Introduction
Three Dimensional Geometry is an important chapter in Class 12 Mathematics that extends the concepts of coordinate geometry into three-dimensional space. Exercise 11.2 focuses on lines in three-dimensional space and various approaches to describe their position and relationship. This comprehensive solution guide will help students understand and solve all the problems in this exercise effectively.
Overview of Exercise 11.2
Exercise 11.2 deals with lines in 3D space and includes problems on:
- Equations of a line in space
- Direction ratios and direction cosines
- Relationship between a line and a plane
- Angle between two lines
- Skew lines and their shortest distance
Key Concepts
Before diving into the solutions, let's review some fundamental concepts:
1. Equation of a Line in 3D Space
Where (x, y, z) is a point on the line and a, b, c are the direction ratios.
2. Relation between Direction Ratios and Direction Cosines
If a, b, c are direction ratios of a line, then its direction cosines (l, m, n) are given by:
3. Angle Between Two Lines
If is the angle between two lines with direction ratios (a, b, c) and (a, b, c), then
4. Shortest Distance Between Two Lines
For lines with vector equations r = a + b and r = a + b, the shortest distance d is:
Detailed Solutions to Exercise 11.2
Question 1
Find the vector and Cartesian equations of the line that passes through the point with position vector 2 - + 4k and is in the direction + 2 - k.
Solution:
Given point: A(2, -1, 4) and direction vector: b = + 2 - k
Vector equation: r = a + b = (2 - + 4k) + ( + 2 - k)
Cartesian equation: (x-2)/1 = (y+1)/2 = (z-4)/(-1)
Question 2
Find the vector and the Cartesian equations of the line that passes through the point (3, -2, 1) and is parallel to the line joining the points A(1, 4, 6) and B(-1, 2, 4).
Solution:
Direction ratios of the given line AB are (-1-1, 2-4, 4-6) = (-2, -2, -2)
Since the required line is parallel to AB, its direction ratios are also (-2, -2, -2)
Using point (3, -2, 1), the Cartesian equation is:
(x-3)/(-2) = (y+2)/(-2) = (z-1)/(-2)
Or simply: (x-3)/1 = (y+2)/1 = (z-1)/1
The vector equation is: r = (3 - 2 + k) + ( + + k)
Question 3
Find the coordinates of the point where the line through (5, 1, 6) and (3, 4, 1) crosses the YZ-plane.
Solution:
First, we find the equation of the line passing through (5, 1, 6) and (3, 4, 1)
Direction ratios: (3-5, 4-1, 1-6) = (-2, 3, -5)
Using point (5, 1, 6), the equation is:
(x-5)/(-2) = (y-1)/3 = (z-6)/(-5) = (say)
For YZ-plane, x = 0
So, (0-5)/(-2) = = 5/2
Substituting = 5/2 in y and z:
y = 1 + 3 (5/2) = 17/2
z = 6 + (-5) (5/2) = -13/2
Therefore, the required point is (0, 17/2, -13/2)
Question 4
Find the vector and Cartesian equations of the line that passes through the point (1, 2, 3) and is parallel to the line (x-3)/2 = (y+2)/3 = (z-4)/5.
Solution:
The direction ratios of the given line are (2, 3, 5)
Since the required line is parallel to the given line, it will have the same direction ratios
Using point (1, 2, 3), the Cartesian equation is:
(x-1)/2 = (y-2)/3 = (z-3)/5
The vector equation is: r = ( + 2 + 3k) + (2 + 3 + 5k)
Question 5
Find the angle between the following pairs of lines:
- (x+4)/3 = (y-1)/2 = (z-3)/2 and (x+2)/4 = (y-3)/3 = (z+4)/5
- r = (2 - 3 + 4k) + (3 + + 2k) and r = ( - 2 + k) + (2 - + 3k)
Solution:
(i) Direction ratios of the first line: (3, 2, 2)
Direction ratios of the second line: (4, 3, 5)
Using the formula for angle between two lines:
cos = (34 + 23 + 25)/(9+4+4) (16+9+25)
cos = (12 + 6 + 10)/17 50 = 28/850
= cos(28/850)
(ii) Direction ratios of the first line: (3, 1, 2)
Direction ratios of the second line: (2, -1, 3)
cos = (32 + 1(-1) + 23)/(9+1+4) (4+1+9)
cos = (6 - 1 + 6)/14 14 = 11/14
= cos(11/14)
Question 6
Find the angle between the following pairs of lines:
- (x-2)/2 = (y-1)/5 = (z+3)/(-3) and (x+2)/(-1) = (y-4)/8 = (z-5)/4
- r = (5 - 2 + 3k) + (2 + + 2k) and r = ( - 2 + 3k) + (3 - - 2k)
Solution:
(i) Direction ratios of the first line: (2, 5, -3)
Direction ratios of the second line: (-1, 8, 4)
cos = (2(-1) + 58 + (-3)4)/(4+25+9) (1+64+16)
cos = (-2 + 40 - 12)/38 81 = 26/38 9
= cos(26/938)
(ii) Direction ratios of the first line: (2, 1, 2)
Direction ratios of the second line: (3, -1, -2)
cos = (23 + 1(-1) + 2(-2))/(4+1+4) (9+1+4)
cos = (6 - 1 - 4)/3 14 = 1/314
= cos(1/314)
Question 7
Find the angle between the lines whose direction cosines are given by the equations:
- l + m + n = 0, l + m - n = 0
- 2l - m + 2n = 0, lm + mn + nl = 0
Solution:
(i) From l + m + n = 0, we get n = -(l + m)
Substituting in l + m - n = 0:
l + m - (-(l+m)) = 0
l + m - (l + m + 2lm) = 0
-2lm = 0 lm = 0
Case 1: l = 0, then m + n = 0 and m - n = 0 m - (-m) = 0 0 = 0
This gives direction cosines (0, 1/2, -1/2) after normalization
Case 2: m = 0, then l + n = 0 and l - n = 0 l - (-l) = 0 0 = 0
This gives direction cosines (1/2, 0, -1/2) after normalization
The angle between these two lines is given by:
cos = 0 1/2 + 1/2 0 + (-1/2) (-1/2) = 1/2
= cos(1/2) = 60
(ii) From 2l - m + 2n = 0, we get m = 2l + 2n
Substituting in lm + mn + nl = 0:
l(2l + 2n) + (2l + 2n)n + nl = 0
2l + 2ln + 2ln + 2n + nl = 0
2l + 5ln + 2n = 0
Dividing by n (assuming n 0): 2(l/n) + 5(l/n) + 2 = 0
Let x = l/n, then 2x + 5x + 2 = 0
x = (-5 (25-16))/4 = (-5 3)/4
Case 1: x = (-5+3)/4 = -1/2 l/n = -1/2 l = -n/2
Then m = 2(-n/2) + 2n = -n + 2n = n
So direction cosines are (-n/2, n, n), which after normalization gives (-1/3, 2/3, 2/3)
Case 2: x = (-5-3)/4 = -2 l/n = -2 l = -2n
Then m = 2(-2n) + 2n = -4n + 2n = -2n
So direction cosines are (-2n, -2n, n), which after normalization gives (-2/3, -2/3, 1/3)
The angle between these two lines is given by:
cos = (-1/3)(-2/3) + (2/3)(-2/3) + (2/3)(1/3)
cos = (2/9) - (4/9) + (2/9) = 0
= cos(0) = 90
Question 8
Find the equation of the line passing through the point (-1, 3, -2) and perpendicular to the lines x/1 = y/2 = z/3 and (x+2)/(-3) = (y-1)/2 = (z+1)/5.
Solution:
For the required line to be perpendicular to both given lines, its direction ratios should be perpendicular to the direction ratios of both lines.
Direction ratios of first line: (1, 2, 3)
Direction ratios of second line: (-3, 2, 5)
Let (a, b, c) be the direction ratios of the required line.
For perpendicular lines, the dot product should be zero:
a + 2b + 3c = 0 and -3a + 2b + 5c = 0
Using cross product, we can find (a, b, c) as:
(a, b, c) = (25 - 32, -(-15 - 3(-3)), (-1)2 - 2(-3))
(a, b, c) = (10 - 6, -(-5 + 9), -2 + 6)
(a, b, c) = (4, -4, 4)
So the direction ratios are (4, -4, 4) or simply (1, -1, 1)
Using point (-1, 3, -2), the equation of the required line is:
(x+1)/1 = (y-3)/(-1) = (z+2)/1
Important Notes
When solving problems in three-dimensional geometry, always ensure that you:
- Correctly identify the direction ratios of lines
- Normalize direction cosines when required
- Be careful with signs when performing cross products
- Verify your solutions by checking if they satisfy the given conditions
- Pay special attention to cases where division by zero might occur
Conclusion
Exercise 11.2 of NCERT Class 12 Maths Chapter 11 provides students with fundamental practice in understanding lines in three-dimensional space. The problems covered in this exercise help develop skills in determining equations of lines, finding relationships between lines, and calculating angles between them. Mastering these concepts is essential for further study in 3D geometry and has applications in various fields including engineering, computer graphics, and physics.
Regular practice with these problems and understanding the underlying concepts will help students approach more complex problems in three-dimensional geometry with confidence. The solutions provided above serve as a guide, but students are encouraged to solve these problems independently for a deeper understanding.
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