# NCERT Solutions for Class 11 Maths Chapter 12 Introduction to Three Dimensional Geometry (Ex 12.3) Exercise 12.3

Geometry is not a topic to be ignored because it is all around us. “Geo” stands for “Earth,” and “metry” stands for “measurement.”Thus, it basically means the measurement of the earth. The space-time continuum of Einstein’s Theory of Relativity is where one can find the most abundant use of three-dimensional geometry.To solve a problem involving a specific shape, one needs to be well versed in how its dimensions will be measured or recognised. Thus, being aware of the relevance of the topic, it has been included in the syllabus of Class 11 Mathematics.

This chapter includes some important concepts related to measurement, like the section formula, the distance formula, etc.

The questions framed by the NCERT for the same concepts have been distributed in three exercises. The solutions for all the exercises and Ex 12.3 Class11 are available on Extramarks. These NCERT Solutions For Class 11 Maths Chapter 12 Exercise 12.3, have been prepared by the top experts in the field of Mathematics and are upgraded from time to time. Under this, the solution to each and every question of Exercise 12.3 is available. The style of attempting the questions in the exam has also been discussed. Students who know the formulae but are falling behind due to a lack of understanding of the correct step sequence to take can improve their grades by using the NCERT Solutions for Class 11 Maths Chapter 12 Exercise 12.3 on Extramarks. NCERT Solutions for Class 11 Maths Chapter 12 Introduction to Three Dimensional Geometry (Ex 12.3) Exercise 12.3

In the digitalised world of the modern era, the role of the internet and technology in the lives of students has been enhanced by leaps and bounds. There was a time when, for every solution, a student needed to scan the entire library to find the correct solutions. Sometimes, the answers to the entire exercise werenot available at all. Fortunately, if they came to find them, they had to compromise on the quality of the solutions and representation.

This is not the case anymore. It is now possible for users to access the best and most complete solutions to their desired questions, whether these are related to Mathematics, Science or any other subject.

Extramarks is a destination where the best quality NCERT Solutions such as Class 11 Maths Chapter 12 Exercise 12.3 prepared by highly qualified subject experts are available. For Mathematics, solutions to every chapter and exercise are made available to the students. The NCERT Solutions For Class 11 Maths Chapter 12 Exercise 12.3 related to Three-dimensional geometry can also be found on Extramarks. These NCERT Solutions For Class 11 Maths Chapter 12 Exercise 12.3 can be accessed in PDF format on Extramarks.

The answers to Class 11 Maths Chapter 12 Exercise 12.3 questions can be found by scrolling through the PDF online.However, if, for some reason, students cannot spend much time on screen, then these NCERT Solutions For Class 11 Maths Chapter 12 Exercise 12.3 can be availed of by them in the form of a hard copy as well. These NCERT Solutions for Class 11 Maths Chapter 12 Exercise 12.3 are available on Extramarks’ website as well as their mobile application.

Students only have to click on the “Download” button to download the PDF of NCERT Solutions For Class 11 Maths Chapter 12 Exercise 12.3 and to get a print of the same. With this, one can compile a set of supreme notes in one place, avoiding any last-minute panic before exams. Also, going through the important and expected questions in NCERT Solutions For Class 11 Maths Chapter 12 Exercise 12.3 boosts the preparation. These NCERT Solutions for Class 11 Maths, Chapter 12, Exercise 12.3, also show how to perfectly represent a solution in the exam.Access NCERT Solution for Class 11 MATHS Chapter 12- Introduction to Three Dimensional Geometry

Extramarks is a learning resource that assists students of all levels throughout their educational journey.For every class and subject, the extraordinary quality of solutions can be accessed on the website of Extramarks. It provides NCERT Solutions Class 12, NCERT Solutions Class 11, NCERT Solutions Class 10 and NCERT Solutions Class 9 for multiple subjects. For Class 11, the NCERT solutions to the lengthy chapter, i.e., Chapter 12 of Mathematics – ‘Introduction to Three Dimensional Geometry’ are available on the website of Extramarks.  In today’s world, when lakhs of students take the same exam, presenting the same information in a unique and effective manner can be extremely beneficial.NCERT Solutions for Class 11 Maths Chapter 12 Exercise 12.3 can help with this aspect of preparation.To boost one’s preparation to the maximum level, and to gain marks better than the competition, students can go through these NCERT Solutions For Class 11 Maths Chapter 12 Exercise 12.3 with full dedication. The availability of past years’ papers under NCERT Solutions For Class 11 Maths Chapter 12 Exercise 12.3 adds to the zeal for preparation. A detailed analysis of the questions that appeared in the exam from Class 11 Maths Exercise 12.3 is also available on the same link, NCERT Solutions For Class 11 Maths Chapter 12 Exercise 12.3. Students are encouraged to visit the Extramarks website to prepare thoroughly for the exams and earn more points.

## NCERT Solutions for Class 11 Maths Chapter 12 Introduction to Three Dimensional Geometry Exercise 12.3

The concepts of Geometry are indeed somewhat daunting and difficult to comprehend, but it is important to remember that with practise the concepts can be grasped. This has been proven by the team at Extramarks. It offers the solutions to Ex 12.3 Class 11 entitled – Introduction to Three Dimensional Geometry in the easiest possible way. The NCERT Solutions For Class 11 Maths Chapter 12 Exercise 12.3 have been prepared and are polished frequently in such a way that makes practising Mathematics innovative. The concepts of section formula, the coordinates of a point, the distance between two distinct points, collinearity of points, etc., have been described in depth under NCERT Solutions For Class 11 Maths Chapter 12 Exercise 12.3. Students can practice the questions and verify the aptness of their solutions by cross-verifying the same on NCERT Solutions For Class 11 Maths Chapter 12 Exercise 12.3.

Q.1 Find the coordinates of the point which divides the line segment joining the points (– 2, 3, 5) and (1, – 4, 6) in the ratio

(i) 2 : 3 internally, (ii) 2 : 3 externally.

Ans.

$\begin{array}{l}\left(\mathrm{i}\right)\mathrm{Let}\text{point P}\left(\mathrm{x},\mathrm{y},\mathrm{z}\right)\text{divides the line segment joining the points}\\ \left(-\text{\hspace{0.17em}2},\text{3},\text{5}\right)\text{and}\left(\text{1},-\text{\hspace{0.17em}4},\text{6}\right)\text{in the ratio 2:3 internally.Then,}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{x}=\frac{{\mathrm{mx}}_{2}+{\mathrm{nx}}_{1}}{\mathrm{m}+\mathrm{n}}\\ =\frac{2\left(1\right)+3\left(-2\right)}{2+3}=\frac{-4}{5}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{y}=\frac{{\mathrm{my}}_{2}+{\mathrm{ny}}_{1}}{\mathrm{m}+\mathrm{n}}\\ =\frac{2\left(-4\right)+3\left(3\right)}{2+3}\\ =\frac{1}{5}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{z}=\frac{{\mathrm{mz}}_{2}+{\mathrm{nz}}_{1}}{\mathrm{m}+\mathrm{n}}\\ =\frac{2\left(6\right)+3\left(5\right)}{2+3}\\ =\frac{27}{5}\\ \mathrm{Therefore},\text{the coordinates of the point P are}\left(-\frac{4}{5},\frac{1}{5},\frac{27}{5}\right).\\ \left(\mathrm{ii}\right)\mathrm{Let}\text{point P}\left(\mathrm{x},\mathrm{y},\mathrm{z}\right)\text{divides the line segment joining the points}\\ \left(-\text{\hspace{0.17em}2},\text{3},\text{5}\right)\text{and}\left(\text{1},-\text{\hspace{0.17em}4},\text{6}\right)\text{in the ratio 2:3 externally.Then,}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{x}=\frac{{\mathrm{mx}}_{2}-{\mathrm{nx}}_{1}}{\mathrm{m}-\mathrm{n}}\\ =\frac{2\left(1\right)-3\left(-2\right)}{2-3}\end{array}$ $\begin{array}{l}\end{array}$

$\begin{array}{l}=\frac{8}{-1}=-8\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{y}=\frac{{\mathrm{my}}_{2}-{\mathrm{ny}}_{1}}{\mathrm{m}-\mathrm{n}}\\ =\frac{2\left(-4\right)-3\left(3\right)}{2-3}\\ =\frac{-17}{-1}=17\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{z}=\frac{{\mathrm{mz}}_{2}-{\mathrm{nz}}_{1}}{\mathrm{m}-\mathrm{n}}\\ =\frac{2\left(6\right)-3\left(5\right)}{2-3}\\ =\frac{-3}{-1}=3\\ \mathrm{Therefore},\text{the coordinates of the point P are}\left(-8,17,3\right).\end{array}$

Q.2 Given that P (3, 2, – 4), Q (5, 4, – 6) and R (9, 8, –10) are collinear. Find the ratio in which Q divides PR.

Ans.

$\begin{array}{l}\mathrm{P}\text{oint}\mathrm{Q}\left(5,\text{}4,-6\right)\text{divides the line segment joining the points}\\ \mathrm{P}\left(3,\text{}2,-\text{\hspace{0.17em}}4\right)\mathrm{}\text{and}\mathrm{R}\left(9,\text{}8,-10\right)\text{in the ratio k:1 internally.Then,}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{x}=\frac{{\mathrm{mx}}_{2}+{\mathrm{nx}}_{1}}{\mathrm{m}+\mathrm{n}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}5=\frac{\mathrm{k}\left(9\right)+1\left(3\right)}{\mathrm{k}+1}\\ 5\left(\mathrm{k}+1\right)=9\text{\hspace{0.17em}}\mathrm{k}+3\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}5\text{\hspace{0.17em}}\mathrm{k}+5=9\text{\hspace{0.17em}}\mathrm{k}+3\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}5-3=9\mathrm{k}-5\mathrm{k}\end{array}$

$\begin{array}{l}⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}2=4\text{\hspace{0.17em}}\mathrm{k}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{k}=\frac{2}{4}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{k}=\frac{1}{2}\\ \mathrm{The}\text{required ratio is}\frac{1}{2}:1\text{i.e., 1:2.}\end{array}$

Q.3 Find the ratio in which the YZ-plane divides the line segment formed by joining the points (–2, 4, 7) and (3, –5, 8).

Ans.

$\begin{array}{l}\mathrm{Plane}\text{\hspace{0.17em}}\mathrm{YZ}\text{divides the line segment joining the points}\\ \mathrm{Q}\left(-2,\text{}4,\text{\hspace{0.17em}\hspace{0.17em}}7\right)\mathrm{}\text{\hspace{0.17em}and}\mathrm{R}\left(3,-5,\text{\hspace{0.17em}\hspace{0.17em}}8\right)\text{in the ratio k:1 internally.Then,}\\ \text{let coordinates of point P in YZ plane:}\left(0,\mathrm{y},\mathrm{z}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{x}=\frac{{\mathrm{mx}}_{2}+{\mathrm{nx}}_{1}}{\mathrm{m}+\mathrm{n}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}0=\frac{\mathrm{k}\left(3\right)+1\left(-2\right)}{\mathrm{k}+1}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}0=3\text{\hspace{0.17em}}\mathrm{k}-2\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}2=3\text{\hspace{0.17em}}\mathrm{k}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{k}=\frac{2}{3}\\ \mathrm{The}\text{required ratio is}\frac{2}{3}:1\text{i.e., 2:3.}\end{array}$

Q.4

$\begin{array}{l}\mathbf{\text{Using section formula, show that the points A(2,}}\mathbf{-}\mathbf{\text{3, 4), B(}}\mathbf{-}\mathbf{\text{1, 2, 1)}}\\ \mathbf{\text{and C}}\left(\text{0,}\frac{\text{1}}{\text{3}}\text{,2}\right)\mathbf{\text{are collinear.}}\end{array}$

Ans.

$\begin{array}{l}\text{The given points are A(2,}-\text{3, 4), B(}-\text{1, 2, 1) and C}\left(\text{0,}\frac{\text{1}}{\text{3}}\text{, 2}\right)\text{.}\\ \text{Let a point P divides AB​in the ratio k:1.}\\ \text{Hence, by the section formula, the coordinates of P ​are:}\\ \left(\frac{{\text{mx}}_{\text{2}}{\text{+nx}}_{\text{1}}}{\text{m+n}}\text{,}\frac{{\text{my}}_{\text{2}}{\text{+ny}}_{\text{1}}}{\text{m+n}}\text{,}\frac{{\text{mz}}_{\text{2}}{\text{+nz}}_{\text{1}}}{\text{m+n}}\right)\\ \left(\frac{\text{k}\left(-\text{1}\right)\text{+2}}{\text{k+1}}\text{,}\frac{\text{k}\left(\text{2}\right)-\text{3}}{\text{k+1}}\text{,}\frac{\text{k}\left(\text{1}\right)\text{+4}}{\text{k+1}}\right)\\ \text{Now, we find the value of k at which P coincides with point C.}\\ \text{So, by taking}\frac{-\text{k+2}}{\text{k+1}}=0\\ ⇒\text{k =2}\\ \text{For k=2, the coordinates of P ​are:}\left(\text{0,}\frac{\text{1}}{\text{3}}\text{, 2}\right)\\ \text{Therefore, C}\left(\text{0,}\frac{\text{1}}{\text{3}}\text{,2}\right)\text{​​ \hspace{0.17em}\hspace{0.17em}is a point that divides AB ​externally in ratio 2:1.}\\ \text{Hence, points A, B​ and C are collinear.}\end{array}$

Q.5 Find the coordinates of the points which trisect the line segment joining the points P(4, 2, – 6) and Q(10, –16, 6).

Ans.
Let points A and B trisect the line segment joining the points P(4, 2, – 6) and Q(10, –16, 6). Point A divides the line segment in the ratio of 1:2 and point B divides the line segment in the ratio of 2:1.

$\begin{array}{l}\mathrm{coordinates}\text{of point A}=\left\{\frac{1\left(10\right)+2\left(4\right)}{1+2},\frac{1\left(-16\right)+2\left(2\right)}{1+2},\frac{1\left(6\right)+2\left(-\text{\hspace{0.17em}}6\right)}{1+2}\right\}\\ =\left(\frac{10+8}{3},\frac{-16+4}{3},\frac{6-12}{3}\right)\\ =\left(6,-4,-2\right)\\ \mathrm{coordinates}\text{of point B}=\left\{\frac{2\left(10\right)+1\left(4\right)}{1+2},\frac{2\left(-16\right)+1\left(2\right)}{1+2},\frac{2\left(6\right)+1\left(-\text{\hspace{0.17em}}6\right)}{1+2}\right\}\\ =\left(\frac{20+4}{3},\frac{-32+2}{3},\frac{12-6}{3}\right)\\ =\left(8,-10,2\right)\\ \mathrm{Thus},\text{points}\left(6,-4,-2\right)\text{and}\left(8,-10,2\right)\text{trisect the line segment}\\ \text{formed by the given points.}\end{array}$

## Please register to view this section

### 1. What are the benefits of going through NCERT Solutions For Class 11 Maths Chapter 12 Exercise 12.3 while preparing for the exams?

Extramarks has the best subject matter experts to prepare the solutions for NCERT textbooks. The NCERT Solutions For Class 11 Maths Chapter 12 Exercise 12.3 are beneficial to the students in many aspects, whether it is about the quality of content, important questions or past years’ papers, everything is scrutinised with extraneous efforts to make the students secure high marks. The availability of NCERT Solutions For Class 11 Maths Chapter 12 Exercise 12.3 in blended mode is an additional benefit.

### 2. Does Extramarks provide NCERT solutions for Mathematics only for Class 11 or other classes too?

The range of solutions on Extramarks varies from primary to upper primary and from secondary to senior secondary classes. The primary classes are available under the links NCERT Solutions Class 1, NCERT Solutions Class 2, NCERT Solutions Class 3, NCERT Solutions Class 4 and NCERT Solutions Class 5. Apart from these, solutions for classes 6 to 8 are available on NCERT Solutions Class 6, NCERT Solutions Class 7 and NCERT Solutions Class 8. These solutions can be accessed by students whenever required.

### 3. What points to keep in mind while attempting the exam of Mathematics to make the solutions more effective?

To make the solutions representable in the exams, students should follow the following tips:

• Revise the concepts thoroughly and practise the questions from NCERT Solutions For Class 11 Maths Chapter 12 Exercise 12.3.
• Write the solutions to the questions asked in a step-by-step method.
• Write the units of measurement wherever required.
• Clearly represent the formula before solving the question.
• Draw graphs wherever necessary.