Here we provide RD Sharma Class 12 Ex 29.7 Solutions Chapter 29 The Plane for English medium students, Which will very helpful for every student in their exams. Students can download the latest RD Sharma Class 12 Ex 29.7 Solutions Chapter 29 The Plane book pdf download. Now you will get step-by-step solutions to each question.

Textbook | NCERT |

Class | Class 12th |

Subject | Maths |

Chapter | 29 |

Exercise | 29.7 |

Category | RD Sharma Solutions |

Table of Contents

**RD Sharma Class 12 Ex 29.7 Solutions Chapter 29 The Plane**

**Question 1: Ten cards numbered 1 through 10 are placed in a box, mixed **

**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,

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