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# Class 12 RD Sharma Solutions- Chapter 29 The Plane – Exercise 29.7

• Last Updated : 25 Jan, 2021

### Question 1. Find the vector equation of the following planes in scalar product form Solution:

(i) Here, We know that, represent a plane passing through a point having position vector and parallel to vectors and . Here, The given plane is perpendicular to a vector We know that vector equation of plane in scalar product form is, —(Equation-1)

Put and in (Equation-1), The equation is required form is, (ii) Here, We know that, represent a plane passing through a point having position vector and parallel to vectors and Here, The given plane is perpendicular to a vector We know that, vector equation of a plane is scalar product is, —(Equation-1)

Put value of and in (Equation-1) Multiplying both the sides by (-1), The equation in the required form, (iii) Given, equation of plane, We know that, is the equation of a plane passing through point and parallel to and .

Here, The given plane is perpendicular to a vector We know that, equation of plane in scalar product form is given by, Dividing by 3, we get Equation in required form is, (iv)  Plane is passing through and parallel to b and  ### Question 2. Find the cartesian form of the equation of the following planes:

Solution:

(i) Here, given equation of plane is, We know that, represents the equation of a plane passing through a vector and parallel to vector and .

Here, Given plane is perpendicular to vector We know that, equation of plane in the scalar product form, —Equation-1

Put the value of and in Equation-1, Put (x)(-3) + (y)(3) + (z)(-3) = -6

-3x + 3y – 3z = -6

Dividing by (-3), we get

x – y + z = 2

Equation in required form is,

x – y + z = 2

(ii) Given, equation of plane, We know that, represents the equation of a plane passing through the vector and parallel to vector and Here, The given plane is perpendicular to vector We know that, equation of plane in scalar product form is given by, —Equation-1

Put, the value of and in equation-1 Put  (x)(0) + (y)(-4) + (z)(2) = -2

-4y + 2z = -2

The equation in required form is,

2y – z = 1

### Question 3. Find the vector equation of the following planes in non-parametric form:

Solution:

(i) Given, equation of plane is, We know that, represents the equation of a plane passing through a point and parallel to vector and .

Given, The given plane is perpendicular to Vector equation of plane in non-parametric form is. = (0)(2) + (3)(-5) + (0)(-1)

= 0 – 15 + 0 The required form of equation is, (ii) Given, equation of plane is, We know that, represents the equation of a plane passing through a vector and parallel to vector and .

Here, The given plane is perpendicular to vector We know that, equation of a plane in non-parametric form is given by,  = (2)(20) + (2)(8) – (-1)(-12)

=40 + 16 + 12 Dividing by 4, Equation of plane in required form is, My Personal Notes arrow_drop_up
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