# NCERT Solutions for Exercise 11.3 Class 12 Maths Chapter 11- Three Dimensional Geometry

NCERT solutions for exercise 11.3 class 12 maths chapter 11 move around the topic plane. The questions in NCERT solutions for class 12 maths chapter 11 exercise 11.3 are related to exercise 11.3 class 12 maths equation of a plane in different conditions, the concept of coplanarity of two lines, the angle between two planes and the exercise 11.3 class 12 maths also covers the distance between a point and a plane. One should grasp the concepts well before solving class 12 maths chapter 11 exercise 11.3. And to get more idea about steps involved in solving the problems under the topic plane, one can go through the solved example given in the NCERT and then crack the class 12th maths chapter 11 exercise 11.3. The below-mentioned exercises are also present in the NCERT book.

NCERT solutions for class 12 maths chapter 11 three dimensional geometry 11.1

NCERT solutions for class 12 maths chapter 11 three dimensional geometry 11.2

- NCERT solutions for class 12 maths chapter 11 three dimensional geometry miscellaneous exercise

## ** NCERT solutions for class 12 maths chapter 11 three dimensional geometry-Exercise: 11.3 **

** Question:1(a) ** In each of the following cases, determine the direction cosines of the normal to the plane and the distance from the origin.

** Answer: **

Equation of plane Z=2, i.e.

The direction ratio of normal is 0,0,1

Divide equation by 1 from both side

We get,

Hence, direction cosins are 0,0,1.

The distance of the plane from the origin is 2.

** Question:1(b) ** In each of the following cases, determine the direction cosines of the normal to the plane and the distance from the origin.

** Answer: **

Given the equation of the plane is or we can write

So, the direction ratios of normal from the above equation are, .

Therefore

Then dividing both sides of the plane equation by , we get

So, this is the form of the plane, where are the direction cosines of normal to the plane and d is the distance of the perpendicular drawn from the origin.

The direction cosines of the given line are and the distance of the plane from the origin is units.

** Question:1(c) ** In each of the following cases, determine the direction cosines of the normal to the plane and the distance from the origin.

** Answer: **

Given the equation of plane is

So, the direction ratios of normal from the above equation are, .

Therefore

Then dividing both sides of the plane equation by , we get

So, this is the form of the plane, where are the direction cosines of normal to the plane and d is the distance of the perpendicular drawn from the origin.

The direction cosines of the given line are and the distance of the plane from the origin is units.

** Question:1(d) ** In each of the following cases, determine the direction cosines of the normal to the plane and the distance from the origin.

** Answer: **

Given the equation of plane is or we can write

So, the direction ratios of normal from the above equation are, .

Therefore

Then dividing both sides of the plane equation by , we get

So, this is the form of the plane, where are the direction cosines of normal to the plane and d is the distance of the perpendicular drawn from the origin.

The direction cosines of the given line are and the distance of the plane from the origin is units.

** Answer: **

We have given the distance between the plane and origin equal to 7 units and normal to the vector .

So, it is known that the equation of the plane with position vector is given by, the relation,

, where d is the distance of the plane from the origin.

Calculating ;

** is the vector equation of the required plane. **

** Question:3(a) ** Find the Cartesian equation of the following planes:

** Answer: **

Given the equation of the plane

So we have to find the Cartesian equation,

Any point on this plane will satisfy the equation and its position vector given by,

Hence we have,

Or,

** Therefore this is the required Cartesian equation of the plane. **

** Question:3(b) ** Find the Cartesian equation of the following planes:

** Answer: **

Given the equation of plane

So we have to find the Cartesian equation,

Any point on this plane will satisfy the equation and its position vector given by,

Hence we have,

Or,

** Therefore this is the required Cartesian equation of the plane. **

** Question:3(c) ** Find the Cartesian equation of the following planes:

** Answer: **

Given the equation of plane

So we have to find the Cartesian equation,

Any point on this plane will satisfy the equation and its position vector given by,

Hence we have,

Or,

** Therefore this is the required Cartesian equation of the plane. **

** Question:4(a) ** In the following cases, find the coordinates of the foot of the perpendicular drawn from the origin.

** Answer: **

Let the coordinates of the foot of perpendicular P from the origin to the plane be

Given a plane equation ,

Or,

The direction ratios of the normal of the plane are 2, 3 and 4 .

Therefore

So, now dividing both sides of the equation by we will obtain,

This equation is similar to where, are the directions cosines of normal to the plane and d is the distance of normal from the origin.

Then finding the coordinates of the foot of the perpendicular are given by .

The coordinates of the foot of the perpendicular are;

or

** Question:4(b) ** In the following cases, find the coordinates of the foot of the perpendicular drawn from the origin.

** Answer: **

Let the coordinates of the foot of perpendicular P from the origin to the plane be

Given a plane equation ,

Or,

The direction ratios of the normal of the plane are 0,3 and 4 .

Therefore

So, now dividing both sides of the equation by we will obtain,

This equation is similar to where, are the directions cosines of normal to the plane and d is the distance of normal from the origin.

Then finding the coordinates of the foot of the perpendicular are given by .

The coordinates of the foot of the perpendicular are;

or

** Question:4(c) ** In the following cases, find the coordinates of the foot of the perpendicular drawn from the origin.

** Answer: **

Let the coordinates of the foot of perpendicular P from the origin to the plane be

Given plane equation .

The direction ratios of the normal of the plane are 1,1 and 1 .

Therefore

So, now dividing both sides of the equation by we will obtain,

This equation is similar to where, are the directions cosines of normal to the plane and d is the distance of normal from the origin.

Then finding the coordinates of the foot of the perpendicular are given by .

The coordinates of the foot of the perpendicular are;

or ..

** Question: ** ** 4(d) ** In the following cases, find the coordinates of the foot of the perpendicular drawn from the origin.

** Answer: **

Let the coordinates of the foot of perpendicular P from the origin to the plane be

Given plane equation .

or written as

The direction ratios of the normal of the plane are 0, -5 and 0 .

Therefore

So, now dividing both sides of the equation by we will obtain,

Then finding the coordinates of the foot of the perpendicular are given by .

The coordinates of the foot of the perpendicular are;

or .

** Answer: **

Given the point and the normal vector which is perpendicular to the plane is

The position vector of point A is

So, the vector equation of the plane would be given by,

Or

where is the position vector of any arbitrary point in the plane.

Therefore, the equation we get,

or

** So, this is the required Cartesian equation of the plane. **

** Question:5(b) ** Find the vector and cartesian equations of the planes

that passes through the point (1,4, 6) and the normal vector to the plane is .

** Answer: **

Given the point and the normal vector which is perpendicular to the plane is

The position vector of point A is

So, the vector equation of the plane would be given by,

Or

where is the position vector of any arbitrary point in the plane.

Therefore, the equation we get,

** So, this is the required Cartesian equation of the plane. **

** Question:6(a) ** Find the equations of the planes that passes through three points.

(1, 1, – 1), (6, 4, – 5), (– 4, – 2, 3)

** Answer: **

The equation of the plane which passes through the three points is given by;

Determinant method,

Or,

Here, these three points A, B, C are collinear points.

** Hence there will be an infinite number of planes possible which passing through the given points. **

** Question:6(b) ** Find the equations of the planes that passes through three points.

(1, 1, 0), (1, 2, 1), (– 2, 2, – 1)

** Answer: **

The equation of the plane which passes through the three points is given by;

Determinant method,

As determinant value is not equal to zero hence there must be a plane that passes through the points A, B, and C.

Finding the equation of the plane through the points,

After substituting the values in the determinant we get,

** So, this is the required Cartesian equation of the plane. **

** Question:7 ** Find the intercepts cut off by the plane 2x + y – z = 5.

** Answer: **

Given plane

We have to find the intercepts that this plane would make so,

Making it look like intercept form first:

By dividing both sides of the equation by 5 (as we have to make the R.H.S =1) , we get then,

So, as we know that from the equation of a plane in intercept form, where a,b,c are the intercepts cut off by the plane at x,y, and z-axes respectively.

Therefore after comparison, we get the values of a,b, and c.

.

** Hence the intercepts are . **

** Question:8 ** Find the equation of the plane with intercept 3 on the y-axis and parallel to ZOX plane.

** Answer: **

Given that the plane is parallel to the ZOX plane.

So, we have the equation of plane ZOX as .

And an intercept of 3 on the y-axis

Intercept form of a plane given by;

So, here the plane would be parallel to the x and z-axes both.

we have any plane parallel to it is of the form, .

Equation of the plane required is .

** Answer: **

The equation of any plane through the intersection of the planes,

Can be written in the form of; , where

So, the plane passes through the point , will satisfy the above equation.

That implies

Now, substituting the value of in the equation above we get the final equation of the plane;

** is the required equation of the plane. **

** Answer: **

Here and

and and

Hence, using the relation , we get

or ..............(1)

where, is some real number.

Taking , we get

or

or .............(2)

Given that the plane passes through the point , it must satisfy (2), i.e.,

or

Putting the values of in (1), we get

or

or

** which is the required vector equation of the plane. **

** Answer: **

The equation of the plane through the intersection of the given two planes, and is given in Cartesian form as;

or ..................(1)

So, the direction ratios of (1) plane are which are .

Then, the plane in equation (1) is perpendicular to whose direction ratios are .

As planes are perpendicular then,

we get,

or

or

Then we will substitute the values of in the equation (1), we get

or

** This is the required equation of the plane. **

** Question:12 ** Find the angle between the planes whose vector equations are and .

** Answer: **

Given two vector equations of plane

and .

Here, and

The formula for finding the angle between two planes,

.............................(1)

and

Now, we can substitute the values in the angle formula (1) to get,

or

or

7x + 5y + 6z + 30 = 0 and 3x – y – 10z + 4 = 0

** Answer: **

Two planes

whose direction ratios are and whose direction ratios are ,

are said to ** Parallel: **

If,

and ** Perpendicular: **

If,

And the angle between is given by the relation,

So, given two planes

Here,

and

So, applying each condition to check:

** Parallel check: **

Clearly, the given planes are ** NOT ** parallel.

** Perpendicular check: **

.

Clearly, the given planes are ** NOT ** perpendicular.

Then find the angle between them,

2x + y + 3z – 2 = 0 and x – 2y + 5 = 0

** Answer: **

Two planes

whose direction ratios are and whose direction ratios are ,

are said to ** Parallel: **

If,

and ** Perpendicular: **

If,

And the angle between is given by the relation,

So, given two planes

Here,

and

So, applying each condition to check:

** Perpendicular check: **

.

** Thus, the given planes are perpendicular to each other. **

2x – 2y + 4z + 5 = 0 and 3x – 3y + 6z – 1 = 0

** Answer: **

Two planes

whose direction ratios are and whose direction ratios are ,

are said to ** Parallel: **

If,

and ** Perpendicular: **

If,

And the angle between is given by the relation,

So, given two planes

Here,

and

So, applying each condition to check:

** Parallel check: **

** Thus, the given planes are parallel as **

2x – y + 3z – 1 = 0 and 2x – y + 3z + 3 = 0

** Answer: **

Two planes

whose direction ratios are and whose direction ratios are ,

are said to ** Parallel: **

If,

and ** Perpendicular: **

If,

And the angle between is given by the relation,

So, given two planes

Here,

and

So, applying each condition to check:

** Parallel check: **

Therefore

** Thus, the given planes are parallel to each other. **

4x + 8y + z – 8 = 0 and y + z – 4 = 0

** Answer: **

Two planes

whose direction ratios are and whose direction ratios are ,

are said to ** Parallel: **

If,

and ** Perpendicular: **

If,

And the angle between is given by the relation,

So, given two planes

Here,

and

So, applying each condition to check:

** Parallel check: **

Clearly, the given planes are ** NOT ** parallel as .

** Perpendicular check: **

## ** NCERT solutions for class 12 maths chapter 11 three dimensional geometry-Exercise: 11.3 **

** Question:1(a) ** In each of the following cases, determine the direction cosines of the normal to the plane and the distance from the origin.

** Answer: **

Equation of plane Z=2, i.e.

The direction ratio of normal is 0,0,1

Divide equation by 1 from both side

We get,

Hence, direction cosins are 0,0,1.

The distance of the plane from the origin is 2.

** Question:1(b) ** In each of the following cases, determine the direction cosines of the normal to the plane and the distance from the origin.

** Answer: **

Given the equation of the plane is or we can write

So, the direction ratios of normal from the above equation are, .

Therefore

Then dividing both sides of the plane equation by , we get

The direction cosines of the given line are and the distance of the plane from the origin is units.

** Question:1(c) ** In each of the following cases, determine the direction cosines of the normal to the plane and the distance from the origin.

** Answer: **

Given the equation of plane is

So, the direction ratios of normal from the above equation are, .

Therefore

Then dividing both sides of the plane equation by , we get

The direction cosines of the given line are and the distance of the plane from the origin is units.

** Question:1(d) ** In each of the following cases, determine the direction cosines of the normal to the plane and the distance from the origin.

** Answer: **

Given the equation of plane is or we can write

So, the direction ratios of normal from the above equation are, .

Therefore

Then dividing both sides of the plane equation by , we get

The direction cosines of the given line are and the distance of the plane from the origin is units.

** Answer: **

We have given the distance between the plane and origin equal to 7 units and normal to the vector .

So, it is known that the equation of the plane with position vector is given by, the relation,

, where d is the distance of the plane from the origin.

Calculating ;

** is the vector equation of the required plane. **

** Question:3(a) ** Find the Cartesian equation of the following planes:

** Answer: **

Given the equation of the plane

So we have to find the Cartesian equation,

Any point on this plane will satisfy the equation and its position vector given by,

Hence we have,

Or,

** Therefore this is the required Cartesian equation of the plane. **

** Question:3(b) ** Find the Cartesian equation of the following planes:

** Answer: **

Given the equation of plane

So we have to find the Cartesian equation,

Any point on this plane will satisfy the equation and its position vector given by,

Hence we have,

Or,

** Therefore this is the required Cartesian equation of the plane. **

** Question:3(c) ** Find the Cartesian equation of the following planes:

** Answer: **

Given the equation of plane

So we have to find the Cartesian equation,

Any point on this plane will satisfy the equation and its position vector given by,

Hence we have,

Or,

** Therefore this is the required Cartesian equation of the plane. **

** Question:4(a) ** In the following cases, find the coordinates of the foot of the perpendicular drawn from the origin.

** Answer: **

Let the coordinates of the foot of perpendicular P from the origin to the plane be

Given a plane equation ,

Or,

The direction ratios of the normal of the plane are 2, 3 and 4 .

Therefore

So, now dividing both sides of the equation by we will obtain,

Then finding the coordinates of the foot of the perpendicular are given by .

The coordinates of the foot of the perpendicular are;

or

** Question:4(b) ** In the following cases, find the coordinates of the foot of the perpendicular drawn from the origin.

** Answer: **

Let the coordinates of the foot of perpendicular P from the origin to the plane be

Given a plane equation ,

Or,

The direction ratios of the normal of the plane are 0,3 and 4 .

Therefore

So, now dividing both sides of the equation by we will obtain,

Then finding the coordinates of the foot of the perpendicular are given by .

The coordinates of the foot of the perpendicular are;

or

** Question:4(c) ** In the following cases, find the coordinates of the foot of the perpendicular drawn from the origin.

** Answer: **

Let the coordinates of the foot of perpendicular P from the origin to the plane be

Given plane equation .

The direction ratios of the normal of the plane are 1,1 and 1 .

Therefore

So, now dividing both sides of the equation by we will obtain,

Then finding the coordinates of the foot of the perpendicular are given by .

The coordinates of the foot of the perpendicular are;

or ..

** Question: ** ** 4(d) ** In the following cases, find the coordinates of the foot of the perpendicular drawn from the origin.

** Answer: **

Let the coordinates of the foot of perpendicular P from the origin to the plane be

Given plane equation .

or written as

The direction ratios of the normal of the plane are 0, -5 and 0 .

Therefore

So, now dividing both sides of the equation by we will obtain,

Then finding the coordinates of the foot of the perpendicular are given by .

The coordinates of the foot of the perpendicular are;

or .

** Answer: **

Given the point and the normal vector which is perpendicular to the plane is

The position vector of point A is

So, the vector equation of the plane would be given by,

Or

where is the position vector of any arbitrary point in the plane.

Therefore, the equation we get,

or

** So, this is the required Cartesian equation of the plane. **

** Question:5(b) ** Find the vector and cartesian equations of the planes

that passes through the point (1,4, 6) and the normal vector to the plane is .

** Answer: **

Given the point and the normal vector which is perpendicular to the plane is

The position vector of point A is

So, the vector equation of the plane would be given by,

Or

where is the position vector of any arbitrary point in the plane.

Therefore, the equation we get,

** So, this is the required Cartesian equation of the plane. **

** Question:6(a) ** Find the equations of the planes that passes through three points.

(1, 1, – 1), (6, 4, – 5), (– 4, – 2, 3)

** Answer: **

The equation of the plane which passes through the three points is given by;

Determinant method,

Or,

Here, these three points A, B, C are collinear points.

** Hence there will be an infinite number of planes possible which passing through the given points. **

** Question:6(b) ** Find the equations of the planes that passes through three points.

(1, 1, 0), (1, 2, 1), (– 2, 2, – 1)

** Answer: **

The equation of the plane which passes through the three points is given by;

Determinant method,

As determinant value is not equal to zero hence there must be a plane that passes through the points A, B, and C.

Finding the equation of the plane through the points,

After substituting the values in the determinant we get,

** So, this is the required Cartesian equation of the plane. **

** Question:7 ** Find the intercepts cut off by the plane 2x + y – z = 5.

** Answer: **

Given plane

We have to find the intercepts that this plane would make so,

Making it look like intercept form first:

By dividing both sides of the equation by 5 (as we have to make the R.H.S =1) , we get then,

So, as we know that from the equation of a plane in intercept form, where a,b,c are the intercepts cut off by the plane at x,y, and z-axes respectively.

Therefore after comparison, we get the values of a,b, and c.

.

** Hence the intercepts are . **

** Question:8 ** Find the equation of the plane with intercept 3 on the y-axis and parallel to ZOX plane.

** Answer: **

Given that the plane is parallel to the ZOX plane.

So, we have the equation of plane ZOX as .

And an intercept of 3 on the y-axis

Intercept form of a plane given by;

So, here the plane would be parallel to the x and z-axes both.

we have any plane parallel to it is of the form, .

Equation of the plane required is .

** Answer: **

The equation of any plane through the intersection of the planes,

Can be written in the form of; , where

So, the plane passes through the point , will satisfy the above equation.

That implies

Now, substituting the value of in the equation above we get the final equation of the plane;

** is the required equation of the plane. **

** Answer: **

Here and

and and

Hence, using the relation , we get

or ..............(1)

where, is some real number.

Taking , we get

or

or .............(2)

Given that the plane passes through the point , it must satisfy (2), i.e.,

or

Putting the values of in (1), we get

or

or

** which is the required vector equation of the plane. **

** Answer: **

The equation of the plane through the intersection of the given two planes, and is given in Cartesian form as;

or ..................(1)

So, the direction ratios of (1) plane are which are .

Then, the plane in equation (1) is perpendicular to whose direction ratios are .

As planes are perpendicular then,

we get,

or

or

Then we will substitute the values of in the equation (1), we get

or

** This is the required equation of the plane. **

** Question:12 ** Find the angle between the planes whose vector equations are and .

** Answer: **

Given two vector equations of plane

and .

Here, and

The formula for finding the angle between two planes,

.............................(1)

and

Now, we can substitute the values in the angle formula (1) to get,

or

or

7x + 5y + 6z + 30 = 0 and 3x – y – 10z + 4 = 0

** Answer: **

Two planes

whose direction ratios are and whose direction ratios are ,

are said to ** Parallel: **

If,

and ** Perpendicular: **

If,

And the angle between is given by the relation,

So, given two planes

Here,

and

So, applying each condition to check:

** Parallel check: **

Clearly, the given planes are ** NOT ** parallel.

** Perpendicular check: **

.

Clearly, the given planes are ** NOT ** perpendicular.

Then find the angle between them,

2x + y + 3z – 2 = 0 and x – 2y + 5 = 0

** Answer: **

Two planes

whose direction ratios are and whose direction ratios are ,

are said to ** Parallel: **

If,

and ** Perpendicular: **

If,

And the angle between is given by the relation,

So, given two planes

Here,

and

So, applying each condition to check:

** Perpendicular check: **

.

** Thus, the given planes are perpendicular to each other. **

2x – 2y + 4z + 5 = 0 and 3x – 3y + 6z – 1 = 0

** Answer: **

Two planes

whose direction ratios are and whose direction ratios are ,

are said to ** Parallel: **

If,

and ** Perpendicular: **

If,

And the angle between is given by the relation,

So, given two planes

Here,

and

So, applying each condition to check:

** Parallel check: **

** Thus, the given planes are parallel as **

2x – y + 3z – 1 = 0 and 2x – y + 3z + 3 = 0

** Answer: **

Two planes

whose direction ratios are and whose direction ratios are ,

are said to ** Parallel: **

If,

and ** Perpendicular: **

If,

And the angle between is given by the relation,

So, given two planes

Here,

and

So, applying each condition to check:

** Parallel check: **

Therefore

** Thus, the given planes are parallel to each other. **

4x + 8y + z – 8 = 0 and y + z – 4 = 0

** Answer: **

Two planes

whose direction ratios are and whose direction ratios are ,

are said to ** Parallel: **

If,

and ** Perpendicular: **

If,

And the angle between is given by the relation,

So, given two planes

Here,

and

So, applying each condition to check:

** Parallel check: **

Clearly, the given planes are ** NOT ** parallel as .

** Perpendicular check: **

## More about NCERT solutions for class 12 maths chapter 11 exercise 11.3

Fourteen questions in total are given in the exercise 11.3 class 12 maths.

There are sub-questions to certain question numbers.

All these 14 questions are detailed in the NCERT solutions for class 12 maths chapter 11 exercise 11.3

## Significance of NCERT solutions for class 12 maths chapter 11 exercise 11.3

The topic plane covers many concepts and the questions from this part are important for the CBSE board exam preparation for class 12.

Exercise 11.3 is a part of the topic plane and the NCERT solutions for class 12 maths chapter 11 exercise 11.3 will be useful to score well in the exam.

### Also see-

## NCERT Solutions Subject Wise

## Subject wise NCERT Exemplar solutions

## Frequently Asked Question (FAQs) - NCERT Solutions for Exercise 11.3 Class 12 Maths Chapter 11- Three Dimensional Geometry

**Question: **What is the main topic that is to be covered to solve exercise 11.3 class 12 maths?

**Answer: **

The topic 11.6 plane

**Question: **What is discussed after class 12 maths chapter 11 exercise 11.3

**Answer: **

Miscellaneous examples are given after class 12th maths chapter 11 exercise 11.3

**Question: **Who solved the NCERT solutions for class 12 maths chapter 11 exercise 11.3

**Answer: **

A team of mathematics experts solved exercise 11.3 discussed here

**Question: **Why students should solve class 12 maths chapter 11 exercise 11.3?

**Answer: **

To understand how much students have grasped the concepts of plane discussed in the NCERT mathematics book, it is good to solve exercise 11.3

**Question: **Is there any supporting NCERT material for more practice questions?

**Answer: **

Yes, NCERT exemplars have a good number of practice questions and will be useful in the preparation of the chapter.

**Question: **Are NCERT solutions helpful in the CBSE board examination?

**Answer: **

Yes, for the CBSE board there will be a good number of similar questions as discussed in the NCERT book.

**Question: **Is three-dimensional geometry important for JEE main examination?

**Answer: **

Yes. Questions are asked from three-dimensional geometry in the JEE main papers.

**Question: **Which NCERT class 12 chapter explains the concepts of vectors?

**Answer: **

Chapter 10 of class 12 NCERT book

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