RD Sharma Solutions for Class 12 Maths Chapter 29 The Plane Exercise 29.3 is provided here. Our solution module utilizes various shortcut tips and practical examples to explain all the exercise questions in simple and easily understandable language. It helps students grasp the concepts faster and makes learning fun. The RD Sharma 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 3
Access RD Sharma Solutions for Class 12 Maths Chapter 29 Exercise 3
EXERCISE 29.3
Q1.
Solution:
Q2. i
Solution:
Given:
The vector equation of the plane is
12x – 3y + 4z + 5 = 0
Hence, the Cartesian form of the equation of the plane is 12x – 3y + 4z + 5 = 0.
ii.
Solution:
Given:
The equation of the plane is:
-x + y + 2z = 9
Hence, the Cartesian form of the equation of the plane is -x + y + 2z = 9.
Q3.
Solution:
Given:
For xy-plane,
This plane passes through the origin and is perpendicular to the z-axis.
So,
For xz-plane,
It passes through the origin and is perpendicular to the y-axis,
So,
For yz-plane,
It passes through the origin and is perpendicular to the x-axis,
So,
Hence, the equation of xy, yz and xz-plane is given by
Q4.i.
Solution:
Given:
The equation of the plane is:
2x – y + 2z = 8
By putting the values, we get
ii.
Solution:
Given:
The Cartesian equation of the plane is:
x + y – z = 5
By putting the values, we get
iii.
Solution:
Given:
The Cartesian equation of the plane is:
x + y = 3
By putting the values, we get
Q5.
Solution:
The given plane passes through the point (1, -1, 1) and normal to the line joining A(1, 2, 5) and B(-1, 3, 1).
So,
Now, multiply by (-1) on both sides
So,
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