RD Sharma Solutions for Class 12 Maths Chapter 29 The Plane Exercise 29.1 is provided here for students to grasp the concepts easily and score good marks in their board exams. For a better understanding of the concepts, the team of expert faculty at BYJU’S has solved the problems in a step-by-step manner to help students understand the concepts easily. Students can practice the RD Sharma Class 12 Solutions on a daily basis to get good marks in the exam. The solutions to this exercise are provided in PDF format, which can be downloaded easily from the links provided below.
RD Sharma Solutions for Class 12 Maths Chapter 29 Exercise 1
Access RD Sharma Solutions For Class 12 Maths Chapter 29 Exercise 1
EXERCISE 29.1
Q1. i.
Solution:
Given:
The three points are:
(2, 1, 0), (3, -2, -2) and (3, 1, 7)
By using the formula, the equation of the plane passing through the three points is given as:
= -21x + 42 – 9y + 9 + 3z = 0
= -21x – 9y + 3z + 51 = 0
Let us divide by -3, and we get
Hence, the equation of the plane is 7x + 3y – z – 17 = 0.
ii.
Solution:
Given:
The three points are:
(-5, 0, -6), (-3, 10, -9) and (-2, 6, -6)
By using the formula, the equation of the plane passing through the three points is given as:
Divide by 9, and we get
Hence, the equation of the plane is 2x – y – 2z – 2 = 0.
iii.
Solution:
Given:
The three points are:
(1, 1, 1), (1, -1, 2) and (-2, -2, -2)
By using the formula, the equation of the plane passing through the three points is given as:
x – 3y – 6z + 8 = 0
Hence, the equation of the plane is x – 3y – 6z + 8 = 0.
iv.
Solution:
Given:
The three points are:
(2, 3, 4), (-3, 5, 1) and (4, -1, 2)
By using the formula, the equation of a plane passing through the three points is given as:
-16x – 16y + 16z + 16 = 0
Divide by -16, we get
Hence, the equation of the plane is x + y – z – 1 = 0.
v.
Solution:
Given:
The three points are:
(0, -1, 0), (3, 3, 0) and (1, 1, 1)
By using the formula, the equation of a plane passing through the three points is given as:
Hence, the equation of the plane is 4x – 3y + 2z – 3 = 0.
Q2.
Solution:
We have to prove that points (0, -1, -1), (4, 5, 1), (3, 9, 4) and (-4, 4, 4) are coplanar.
Now let us find the equation of the plane passing through the three points, i.e.:
(0, -1, -1), (4, 5, 1), (3, 9, 4)
By using the formula, the equation of the plane passing through the three points is given as:
10x – 14y + 22z + 8 = 0
Divide by 2, and we get
5x – 7y + 11z + 4 = 0 ……. (1)
By using the fourth point (-4, 4, 4),
Substitute the values as x = -4, y =4, z = 4 in equation (1), we get
0 = 0
LHS = RHS
Since the fourth point satisfies the equation of the plane passing through the three points. So, all the points are coplanar.
Hence, the equation of the common plane is 5x – 7y + 11z + 4 = 0.
Q3.i
Solution:
Given:
Four points are:
(0, -1, 0), (2, 1, -1), (1, 1, 1) and (3, 3, 0)
Now let us find the equation of the plane passing through the three points, i.e.:
(0, -1, 0), (2, 1, -1), (1, 1, 1)
By using the formula, the equation of the plane passing through the three points is given as:
4x – 3y – 3 + 2z = 0
4x – 3y + 2z – 3 = 0 ……. (1)
By using the fourth point (3, 3, 0),
Substitute the values as x = 3, y =3, z = 0 in equation (1), we get
0 = 0
LHS = RHS
Since the fourth point satisfies the equation of plane passing through three points. So, all the points are coplanar.
Hence, the equation of the common plane is 4x – 3y + 2z – 3 = 0.
ii.
Solution:
Given:
Four points are:
(0, 4, 3), (-1, -5, -3), (-2, -2, 1) and (1, 1, -1)
Now let us find the equation of the plane passing through the three points, i.e.:
(0, 4, 3), (-1, -5, -3), (-2, -2, 1)
By using the formula, the equation of the plane passing through the three points is given as:
-18x + 10y – 12z – 4 = 0 ………. (1)
By using the fourth point (1, 1, -1),
Substitute the values as x = 1, y =1, z = -1 in equation (1), we get
0 = 0
LHS = RHS
Since the fourth point satisfies the equation of the plane passing through the three points. So, all the points are coplanar.
Hence, the equation of the common plane is -18x + 10y – 12z – 4 = 0.
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