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

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.

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

### 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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