# NCERT Solutions for Class 11 Maths Chapter 12 Introduction to Three Dimensional Geometry

NCERT solutions for class 11 maths chapter 12 Introduction to Three Dimensional Geometry: In our previous classes, you have studied the basic concepts of geometry in two-dimensional space. In this article, you will get solutions of NCERT for class 11 maths chapter 12 introduction to three dimensional geometry. A 3D shape can be defined as an object or a solid figure or shape that has 3 dimensions in length, width, and height. This chapter is new for you but you can easily understand the concepts with the help of CBSE NCERT solutions for class 11 maths chapter 12 introduction to three dimensional geometry. It is important to pay careful attention and try to relate the 2D geometry concepts with this chapter while studying this chapter. This chapter is very important for CBSE class 11 final examination and in various competitive exams like BITSAT, JEE Mains, etc. In this chapter, you will learn what is 3D coordinate geometry, how a coordinate space can be determined, about coordinates of a point in space, the distance between two points and section formula. There are 14 questions in 3 exercises in this chapter. First, try to solve it on your own. If you are not able to do, you can take the help of NCERT solutions for class 11 maths chapter 12 introduction to three dimensional geometry which is prepared in a detailed manner. Check all NCERT solutions from class 6 to class 12 to learn CBSE maths.

## What is the 3D Coordinate Geometry?

Unlike 2D shapes, 3D shapes have length, width and depth or thickness. In three dimensional geometry, a coordinate space is determined by a vector perpendicular to that particular plane. Thus, a point in a three-dimensional plane will include the coordinates of three planes named as X- plane, Y- plane, and Z-plane.

Topics of NCERT grade 11 maths chapter 12 Introduction to Three Dimensional Geometry

12.1 Introduction

12.2 Coordinate Axes and Coordinate Planes in Three Dimensional Space

12.3 Coordinates of a Point in Space

12.4 Distance between Two Points

12.5 Section Formula

The complete solutions of NCERT class 11 mathematics chapter 12 is provided below:

## NCERT solutions for class 11 maths chapter 12 introduction to three dimensional geometry-Exercise: 12.1

Any point on x-axis have zero y coordinate and zero z coordinate.

When a point is in XZ plane, the y coordinate of this point will always be zero.

The x-coordinate, y-coordinate, and z-coordinate of point (1, 2, 3) are all positive. Therefore, this point lies in octant I.

The x-coordinate, y-coordinate, and z-coordinate of point (4, -2, 3) are positive, negative, and positive respectively. Therefore, this point lies in octant IV.

The x-coordinate, y-coordinate, and z-coordinate of point (4, -2, -5) are positive, negative, and negative respectively. Therefore, this point lies in octant VIII.

The x-coordinate, y-coordinate, and z-coordinate of point (4, 2, -5) are positive, positive, and negative respectively. Therefore, this point lies in octant V.

The x-coordinate, y-coordinate, and z-coordinate of point (-4, 2, -5) are negative, positive, and negative respectively. Therefore, this point lies in octant VI.

The x-coordinate, y-coordinate, and z-coordinate of point (-4, 2, 5) are negative, positive, and positive respectively. Therefore, this point lies in octant II.

The x-coordinate, y-coordinate, and z-coordinate of point (-3, -1, 6) are negative, negative, and positive respectively. Therefore, this point lies in octant III.

The x-coordinate, y-coordinate, and z-coordinate of point (2, -4, -7) are positive, negative, and negative respectively. Therefore, this point lies in octant VIII.

The x-axis and y-axis taken together determine a plane known as XY Plane.

The coordinates of points in the XY-plane are of the form (x, y, 0 ).

Coordinate planes divide the space into  Eight octants.

Solutions of NCERT for class 11 maths chapter 12 introduction to three dimensional geometry-Exercise: 12.2

The distance between two points  and  is given by

So, distance between  (2, 3, 5) and (4, 3, 1) is given by

The distance between two points  and  is given by

So, distance between  (–3, 7, 2) and (2, 4, –1) is given by

The distance between two points  and  is given by

So, distance between  (–1, 3, – 4) and (1, –3, 4) is given by

The distance between two points  and  is given by

So, distance between  (2, –1, 3) and (–2, 1, 3).) is given by

Given,theree points A=(–2, 3, 5), B=(1, 2, 3) and C=(7, 0, –1)

The distance between two points  and  is given by

The distance AB :

The distance BC:

The distance CA

As we can see here,

Hence we can say that point A,B and C are colinear.

Given Three points  A=(0, 7, –10), B=(1, 6, – 6) and C=(4, 9, – 6)

The distance between two points  and  is given by

The distance AB

The distance BC

The distance CA

As we can see

Hence we can say that ABC is an isosceles triangle.

Question:3(ii) Verify the following

(0, 7, 10), (- 1, 6,  6) and (- 4, 9,  6) are the vertices of right angled  triangle.

Given Three points  A=(0, 7, 10), B=(-1, 6, 6) and C=(-4, 9, 6)

The distance between two points  and  is given by

The distance AB

The distance BC

The distance CA

As we can see

Since this follows pythagorus theorem,  we can say that ABC is a right angle triangle.

Given A=(–1, 2, 1), B=(1, –2, 5), C=(4, –7, 8) and D=(2, –3, 4)

Given Three points  A=(0, 7, –10), B=(1, 6, – 6) and C=(4, 9, – 6)

The distance between two points  and  is given by

The distance AB

The distance BC

The distance CD

The distance DA

Here As we can see

And

As the opposite sides of quadrilateral are equal, we can say that ABCD is a parallelogram.

Given, two points A=(1, 2, 3) and B=(3, 2, –1).

Let the point P= (x,y,z) be a point which is equidistance from the points A and B.

so,

The distance PA= The distance PB

Now lets apply the simplification property,

Hence locus of the point which is equidistant from A and B is .

Given,

Two points   A (4, 0, 0) and B (– 4, 0, 0)

let the point P(x,y,z) be a point sum of whose distance from A and B is 10.

So,

The distance PA+The distance PB=10

Squaring on both side :

Now again squaring both sides,

Hence the equation of the set of points P, the sum of whose distances from A and B is equal to 10 is .

CBSE NCERT solutions for class 11 maths chapter 12 introduction to three dimensional geometry-Exercise: 12.3

The line segment joining the points A (– 2, 3, 5) and B(1, – 4, 6)

Let point P(x,y,z) be the point that divides the line segment AB internally in the ratio 2:3.

Now, As we know by section formula, The coordinate of the point P which divides line segment  And  in ratio m:n is

Now the point that divides A (– 2, 3, 5) and B(1, – 4, 6) in ratio 2:3 is

Hence required point is

he line segment joining the points A (– 2, 3, 5) and B(1, – 4, 6)

Let point P(x,y,z) be the point that  divides the line segment AB externally in the ratio 2:3.

Now, As we know by section formula , The coordinate of the point P which divides line segment  And   externally in ratio m:n is

Now the point that divides A (– 2, 3, 5) and B(1, – 4, 6) externally in ratio 2:3 is

Hence required point is

Given Three points,

P (3, 2, – 4), Q (5, 4, – 6) and R (9, 8, –10)

Let point Q divides PR internally in the ratio

Now,

According to the section formula , The point Q in terms of P,Q and  is:

Hence, point Q divides PR in ratio 1:2.

Given,

two points A(–2, 4, 7) and B(3, –5, 8)

Let Y-Z plane divides AB in

So, According to the section formula, the point which divides AB in  is

Since this point is in YZ plane, x coordinate of this point will be zero.

So,

Hence YZ plane divides Line segment AB in a ratio 2:3.

Given,

three points, A (2, –3, 4), B (–1, 2, 1) and C (0, 1/3, 2)

Let a point P divides Line segment AB in the ratio

SO, according to the section formula, the point P will be

Now, let's compare this point P with point C.

From here, we see that for , point C divides the line segment AB in ratio 2:1. Since point C divides the line segment AB, it lies in the line joining A and B and Hence they are colinear.

Given,

two points P (4, 2, – 6) and Q (10, –16, 6).

The point which trisects the line segment are the points which divide PQ in either 1:2 or 2:1

Let R (x,y,z) be the point which divides Line segment PR in ratio 1:2

Now, according to the section formula

Let S be the point which divides the Line segment PQ in ratio 2:1

So, The point S according to section formula is

Hence the points which trisect the line segment AB are (6,-4,-2) and (8,-10,2).

NCERT solutions for class 11 maths chapter 12 introduction to three dimensional geometry-Miscellaneous Exercise

Given

Three vertices of the parallelogram, ABCD are:

A(3, – 1, 2), B (1, 2, – 4) and C (– 1, 1, 2).

Let the fourth vertice be (a, b, c )

Now, As we know the concept that diagonal of the parallelogram bisect each other,

Hence here in parallelogram ABCD, the midpoint of the line segment AC will be equal to the midpoint of the line segment BD.

So,

On comparing both points, we get

Hence the Fourth vertice of the is (1,-2,8)

Given,

Three vertices of the triangle,A (0, 0, 6), B (0,4, 0)and C(6, 0, 0).

Now,

Let D be the midpoint of the AB, E be the midpoint of the BC and F be the midpoint of the AC.

Vertice of the D =

Vertices of E =

Vertices of the F =

Now, Medians of the triangle are CD, AE, and BF

So the lengths of the medians are:

Hence lengths of the median are .

Given,

Triangle PQR with vertices P (2a, 2, 6),  Q (– 4, 3b, –10) and R(8, 14, 2c),

Now, As we know,

The centroid of  a triangle is given by

Where coordinates of vertices of the triangle are

Since Centroid of the triangle, PQR is origin =(0,0,0)

On equating both coordinates, we get

Let the point  Q be (0,y,0)

Now Given

The distance of point Q From point P =

So,

The coordinates of the required point is (0,2,0) or (0,-6,0).

Let R divides PQ in the ratio k: 1.

The coordinates of the point R are given   by

now, as given the x coordinate of the R is 4.

So,

Hence the coordinates of R are:

Given Points,

A (3, 4, 5) and B  (–1, 3, –7),

Let the coordinates of point P be (x,y,z)

Now,

Given condition :

And

Now, Given Condition

Hence Equation of the set of the point P is

.

## NCERT solutions for class 11 mathematics

 chapter-1 NCERT solutions for class 11 maths chapter 1 Sets chapter-2 Solutions of NCERT for class 11 chapter 2 Relations and Functions chapter-3 CBSE NCERT solutions for class 11 chapter 3 Trigonometric Functions chapter-4 NCERT solutions for class 11 chapter 4 Principle of Mathematical Induction chapter-5 Solutions of NCERT for class 11 chapter 5 Complex Numbers and Quadratic equations chapter-6 CBSE NCERT solutions for class 11 maths chapter 6 Linear Inequalities chapter-7 NCERT solutions for class 11 maths chapter 7 Permutation and Combinations chapter-8 Solutions of NCERT for class 11 maths chapter 8 Binomial Theorem chapter-9 CBSE NCERT solutions for class 11 maths chapter 9 Sequences and Series chapter-10 NCERT solutions for class 11 maths chapter 10 Straight Lines chapter-11 Solutions of NCERT for class 11 maths chapter 11 Conic Section chapter-12 CBSE NCERT solutions for class 11 maths chapter 12 Introduction to Three Dimensional Geometry chapter-13 NCERT solutions for class 11 maths chapter 13 Limits and Derivatives chapter-14 Solutions of NCERT for class 11 maths chapter 14 Mathematical Reasoning chapter-15 CBSE NCERT solutions for class 11 maths chapter 15 Statistics chapter-16 NCERT solutions for class 11 maths chapter 16 Probability

## NCERT solutions for class 11- Subject wise

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There are some important formulas from NCERT solutions for class 11 maths chapter 12 introduction to three dimensional geometry which you should remember after studying this chapter.

• Distance between two points  is

• The coordinates of the point R which divides the line segment joining two points  internally  in the ratio m : n

• The coordinates of the point R which divides the line segment joining two points  externally in the ratio m : n

• Distance between origin  is

• The coordinates of the mid-point of the line segment joining two points  is

• The coordinates of the centroid of the triangle, whose vertices are

Tip- All the  3D geometry formulas are very similar to 2D geometry, so try to relate it with 2D geometry it will very easy for you to remember the formulas. You should write the formulas which you are using in solving problems. There are 6 exercises in the miscellaneous exercise. You should solve that also to get command on this chapter.  In CBSE NCERT solutions for class 11 maths chapter 12 introduction to three dimensional geometry, you will get solutions of miscellaneous exercise too.