# NCERT Solutions for Class 11 Maths Chapter 12 Introduction to Three Dimensional Geometry (Ex 12.2) Exercise 12.2

Mathematics is a world of shapes and figures. To some, it seems like a battle to be fought using digits. Students feel a pang of panic upon hearing these heavy words like Geometry, Trigonometry etc. It is evident when it comes to Class 11, that the aura of Mathematics is overwhelming. Hereby, easy-to-go digits and concepts take a big turn, and a wide range of concepts are added to this world of Mathematics.

Chapter 12 of Mathematics by NCERT for Class 11, deals with Geometry. The chapter is entitled ‘Introduction to Three Dimensional Geometry’. It deals with the concepts of distance formula, coordinates of a point on a plane, and so on. Students find it an uphill task to solve the answers to the questions of this chapter.

The solutions can now be found on Extramarks under NCERT Solutions for Class 11 Maths Chapter 12 Exercise 12.2.These questions are critically analysed by experts to determine the frequency with which they appear in the exam, and they can be found at the link NCERT Solutions for Class 11 Maths Chapter 12 Exercise 12.2.

## NCERT Solutions for Class 11 Maths Chapter 12 Introduction to Three Dimensional Geometry (Ex 12.2) Exercise 12.2

Solving the bulky exercises of Mathematics is not an easy task for everyone. Some students do it out of passion and with full vigour. Others find it exciting.Not only this, but finding the prim and proper solutions to these giants, and verifying one’s solutions, is another obstacle that a student has to face. The process becomes a lot easier with Extramarks. It provides NCERT solutions to these answers with the help of the expertise of great masters of the subject. For Class 11, Introduction to Three Dimensional Geometry, which is Chapter 12 in the NCERT book, the solutions for Introduction to Three Dimensional Geometry Exercise 12.2 are available on Extramarks under NCERT Solutions for Class 11 Maths Chapter 12 Exercise 12.2. Class 11 Maths Exercise 12.2 deals with the questions related to the distance formula. Answers to this question can be found in a variety of sources. The NCERT Solutions for Class 11 Maths Chapter 12 Exercise 12.2. provide not only answers to these questions or exercises but also teach the method of representing a solution in the exam. Solving the problem in an appropriate way fetches more marks for students. These NCERT Solutions for Class 11 Maths Chapter 12 Exercise 12.2 are available in PDF format. The NCERT Solutions for Class 11 Maths Chapter 12 Exercise 12.2 can be viewed by students in online as well as offline mode. However, to reduce screen time, they can download these NCERT Solutions for Class 11 Maths Chapter 12 Exercise 12.2 in a PDF format from Extramarks. By taking help from these NCERT Solutions for Class 11 Maths Chapter 12 Exercise 12.2 or the similar notes available on Extramarks, students can polish their representation skills for the exam and can attempt the paper with full confidence.

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

In the exam of Mathematics, the correct representation and solution of the question are all that matter. Students of Class 11 have a strange fear regarding answer solving in the Mathematics exam. Chapter 12 of the NCERT book related to Three-dimensional Geometry creates a fuss in their mind regarding Mathematics as a subject. It deals with section formula, distance formula, coordinates of a point etc. The solutions to the main exercises and the miscellaneous exercises are available on Extramarks. The solutions for the 2nd exercise of Chapter 12, i.e., 12.2, can be found under NCERT Solutions for Class 11 Maths Chapter 12 Exercise 12.2. These NCERT Solutions for Class 11 Maths Chapter 12 Exercise 12.2 are updated and revised at frequent intervals by the Mathematics experts of the Extramarks team. The story doesn’t end here. The past years’ papers for Class 11 Mathematics Chapter 12 are available on NCERT Solutions for Class 11 Maths Chapter 12 Exercise 12.2. It would be a great exercise if students could attempt these past years’ papers on a regular basis.

### NCERT Solutions for Class 11 Maths Chapter 12 Introduction to Three Dimensional Geometry Exercise 12.2

A point in space has three coordinates. Students in Class 11 are taught the fundamentals of geometry in Chapter 11 of Maths NCERT.This chapter deals with the concepts of geometry, such as coordinate axes and coordinate planes in three dimensions, coordinates of a point, the distance between two points, and the section formula, in-depth. In total, there are three exercises in the chapter, which are suffused with a good quantity and quality of questions.. Therefore, the NCERT Solutions for Class 11 Maths Chapter 12 Exercise 12.2 should be used by students alongside their preparation.

Extramarks assists students in better understanding the concept.At the end of the chapter, some miscellaneous exercises are available that contain additional questions from the entire chapter. Extramarks solutions, such as those for Class 11 Maths Chapter 12 Exercise 12.2, can also be used by students.

Students are advised to go through the solutions to these questions, which are available on Extramarks. The NCERT Solutions for Class 11 Maths Chapter 12 Exercise 12.2 help students clear even the slightest doubt in the exercise. Apart from this, the solutions for other exercises in the same chapter are also available on the website in the form of a PDF. These NCERT Solutions for Class 11 Maths Chapter 12 Exercise 12.2 also reveal the most recent trend in exam questions.Past years’ papers available on Extramarks make the path of exam preparation more smooth, as the analysis of the same displays the sort of questions expected in the exam.

Some major benefits of using Extramarks for accessing the NCERT Solutions for Class 11 Maths Chapter 12 Exercise 12.2 are:

Firstly, the solutions for NCERT or CBSE Mathematics for Class 11 are available to them within their comfort zone and as per their convenience.

Secondly, these NCERT Solutions for Class 11 Maths Chapter 12 Exercise 12.2 are available in the form of a PDF which is available both on the website and mobile app of Extramarks. These NCERT Solutions for Class 11 Maths Chapter 12 Exercise 12.2 are also available in offline mode by downloading.This helps the students prepare effectively for their exams. The PDF for NCERT Solutions for Class 11 Maths Chapter 12 Exercise 12.2 is also available.

Also, students have the advantage of customising their learning journey by making their own study plan to review each topic at their own pace without any sort of time constraints.

Q.1 Find the distance between the following pairs of points:

(i) (2, 3, 5) and (4, 3, 1)
(ii) (–3, 7, 2) and (2, 4, –1)
(iii) (–1, 3, – 4) and (1, –3, 4)
(iv) (2, –1, 3) and (–2, 1, 3).

Ans.

$\begin{array}{l}\text{Distance between two points}\left({\mathrm{x}}_{1},{\mathrm{y}}_{1},{\mathrm{z}}_{1}\right)\text{and}\left({\mathrm{x}}_{2},{\mathrm{y}}_{2},{\mathrm{z}}_{2}\right),\\ \mathrm{d}=\sqrt{{\left({\mathrm{x}}_{2}-{\mathrm{x}}_{1}\right)}^{2}+{\left({\mathrm{y}}_{2}-{\mathrm{y}}_{1}\right)}^{2}+{\left({\mathrm{z}}_{2}-{\mathrm{z}}_{1}\right)}^{2}}\\ \left(\mathrm{i}\right)\left(2,\text{}3,\text{}5\right)\text{}\mathrm{and}\text{}\left(4,\text{}3,\text{}1\right)\\ \mathrm{d}=\sqrt{{\left(4-2\right)}^{2}+{\left(3-3\right)}^{2}+{\left(1-5\right)}^{2}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\sqrt{4+0+16}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\sqrt{20}\text{\hspace{0.17em}}\mathrm{unit}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=2\sqrt{5}\text{\hspace{0.17em}}\mathrm{unit}\end{array}$

$\begin{array}{l}\left(\mathrm{ii}\right)\left(–3,\text{}7,\text{}2\right)\text{}\mathrm{and}\text{}\left(2,\text{}4,\text{}–1\right)\\ \mathrm{d}=\sqrt{{\left(2+3\right)}^{2}+{\left(4-7\right)}^{2}+{\left(2+1\right)}^{2}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\sqrt{25+9+9}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\sqrt{43}\text{\hspace{0.17em}}\mathrm{unit}\\ \left(\mathrm{iii}\right)\text{\hspace{0.17em}}\left(-1,\text{}3,-\text{\hspace{0.17em}}4\right)\text{}\mathrm{and}\text{}\left(1,-3,\text{}4\right)\\ \mathrm{d}=\sqrt{{\left(1+1\right)}^{2}+{\left(-3-3\right)}^{2}+{\left(4+4\right)}^{2}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\sqrt{4+36+64}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\sqrt{104}\text{\hspace{0.17em}}\mathrm{unit}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=2\sqrt{26}\text{\hspace{0.17em}}\mathrm{unit}\\ \left(\mathrm{iv}\right)\left(2,-1,3\right)\text{}\mathrm{and}\text{}\left(-2,1,3\right)\\ \mathrm{d}=\sqrt{{\left(-2-2\right)}^{2}+{\left(1+1\right)}^{2}+{\left(3-3\right)}^{2}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\sqrt{16+4+0}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\sqrt{20}\text{\hspace{0.17em}}\mathrm{unit}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=2\sqrt{5}\text{\hspace{0.17em}}\mathrm{unit}\end{array}$

Q.2 Show that the points (–2, 3, 5), (1, 2, 3) and (7, 0, –1) are collinear.

Ans.

$\begin{array}{l}\mathrm{Let}\text{three points be A}\left(-2,3,5\right),\mathrm{B}\left(\text{1},\text{2},\text{3}\right)\text{and}\mathrm{}\text{\hspace{0.17em}}\mathrm{C}\left(7,0,–1\right).\mathrm{Then},\\ \mathrm{AB}=\sqrt{{\left(1+2\right)}^{2}+{\left(2-3\right)}^{2}+{\left(3-5\right)}^{2}}\\ =\sqrt{9+1+4}\\ =\sqrt{14}\\ =3.741\end{array}$

$\begin{array}{l}\mathrm{BC}=\sqrt{{\left(7-1\right)}^{2}+{\left(0-2\right)}^{2}+{\left(-1-3\right)}^{2}}\\ =\sqrt{36+4+16}\\ =\sqrt{56}\\ =14.966\\ \mathrm{CA}=\sqrt{{\left(-2-7\right)}^{2}+{\left(0-3\right)}^{2}+{\left(-1-5\right)}^{2}}\\ =\sqrt{81+9+36}\\ =\sqrt{126}\\ =11.225\\ \therefore \mathrm{AB}+\mathrm{CA}=3.741+11.225\\ =14.966=\mathrm{BC}\\ \mathrm{Thus},\text{the points A, B and C are collinear.}\end{array}$

Q.3 Verify the following:
(i) (0, 7, –10), (1, 6, – 6) and (4, 9, – 6) are the vertices of an isosceles triangle.
(ii) (0, 7, 10), (–1, 6, 6) and (– 4, 9, 6) are the vertices of a right angled triangle.
(iii) (–1, 2, 1), (1, –2, 5), (4, –7, 8) and (2, –3, 4) are the vertices of a parallelogram.

Ans.

$\begin{array}{l}\left(\text{i}\right)\text{ Let three points be A}\left(\text{0, 7,–10}\right)\text{,B}\left(\text{1, 6,– 6}\right)\text{ and C}\left(\text{4, 9,– 6}\right)\text{.}\\ \text{Then,}\\ \text{ AB = }\sqrt{{\left(\text{1-0}\right)}^{\text{2}}\text{+}{\left(\text{6-7}\right)}^{\text{2}}\text{+}{\left(\text{-\hspace{0.17em}6+10}\right)}^{\text{2}}}\\ \text{ = }\sqrt{\text{1+1+16}}\\ \text{ = }\sqrt{\text{18}}\\ \text{ BC = }\sqrt{{\left(\text{4-1}\right)}^{\text{2}}\text{+}{\left(\text{9-6}\right)}^{\text{2}}\text{+}{\left(\text{-\hspace{0.17em}6+6}\right)}^{\text{2}}}\end{array}$ $\begin{array}{l}\text{}\end{array}$

$\begin{array}{l}\text{ = }\sqrt{\text{9+9+0}}\\ \text{ = }\sqrt{\text{18}}\\ \text{ CA = }\sqrt{{\left(\text{4-0}\right)}^{\text{2}}\text{+}{\left(\text{9-7}\right)}^{\text{2}}\text{+}{\left(\text{-\hspace{0.17em}6+10}\right)}^{\text{2}}}\\ \text{ = }\sqrt{\text{16+4+16}}\\ \text{ = }\sqrt{\text{36}}\\ \text{Since, AB = BC = }\sqrt{\text{18}}\text{.}\\ \text{So, }\left(\text{0,7,–10}\right)\text{,}\left(\text{1,6,– 6}\right)\text{ and }\left(\text{4,9,– 6}\right)\text{ are the vertices of}\\ \text{an isosceles triangle.}\\ \left(\text{ii}\right)\text{ Let three points be A}\left(\text{0, 7, 10}\right)\text{,B}\left(\text{–1, 6, 6}\right)\text{ and C}\left(\text{– 4, 9, 6}\right)\text{. }\\ \text{Then,}\\ \text{ AB = }\sqrt{{\left(\text{-1-0}\right)}^{\text{2}}\text{+}{\left(\text{6-7}\right)}^{\text{2}}\text{+}{\left(\text{6-10}\right)}^{\text{2}}}\\ \text{ = }\sqrt{\text{1+1+16}}\\ \text{ = }\sqrt{\text{18}}\\ \text{ BC = }\sqrt{{\left(\text{-\hspace{0.17em}4+1}\right)}^{\text{2}}\text{+}{\left(\text{9-6}\right)}^{\text{2}}\text{+}{\left(\text{\hspace{0.17em}6-6}\right)}^{\text{2}}}\\ \text{ = }\sqrt{\text{9+9+0}}\\ \text{ = }\sqrt{\text{18}}\\ \text{ CA = }\sqrt{{\left(\text{-\hspace{0.17em}4-0}\right)}^{\text{2}}\text{+}{\left(\text{9-7}\right)}^{\text{2}}\text{+}{\left(\text{\hspace{0.17em}6-10}\right)}^{\text{2}}}\\ \text{ = }\sqrt{\text{16+4+16}}\\ \text{ = }\sqrt{\text{36}}\\ \text{Since,}\\ {\text{AB}}^{\text{2}}{\text{+BC}}^{\text{2}}\text{ = 18+18}\\ \text{ = 36}\\ {\text{ = CA}}^{\text{2}}\end{array}$

$\begin{array}{l}\text{Therefore, \hspace{0.17em}​}\left(\text{0, 7, 10}\right)\text{,}\left(\text{–1, 6, 6}\right)\text{ and }\left(\text{– 4, 9, 6}\right)\text{ are the \hspace{0.17em}}\\ \text{vertices of a right angled triangle.}\\ \left(\text{iii}\right)\text{ Let four points be A}\left(\text{–1, 2, 1}\right)\text{, B}\left(\text{1,–2, 5}\right)\text{, C}\left(\text{4,–7, 8}\right)\text{}\\ \text{and D}\left(\text{2,–3,4}\right)\text{. Then,}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}AB = }\sqrt{{\left(\text{1+1}\right)}^{\text{2}}\text{+}{\left(\text{-2-2}\right)}^{\text{2}}\text{+}{\left(\text{5-1}\right)}^{\text{2}}}\\ \text{= }\sqrt{\text{4+16+16}}\\ \text{= }\sqrt{\text{36}}\text{=6}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}BC = }\sqrt{{\left(\text{4-1}\right)}^{\text{2}}\text{+}{\left(\text{-7+2}\right)}^{\text{2}}\text{+}{\left(\text{8-5}\right)}^{\text{2}}}\\ \text{= }\sqrt{\text{9+25+9}}\\ \text{= }\sqrt{\text{43}}\\ \text{\hspace{0.17em}\hspace{0.17em}CD = }\sqrt{{\left(\text{2-4}\right)}^{\text{2}}\text{+}{\left(\text{-3+7}\right)}^{\text{2}}\text{+}{\left(\text{4-8}\right)}^{\text{2}}}\\ \text{= }\sqrt{\text{4+16+16}}\\ \text{= }\sqrt{\text{36}}\\ \text{\hspace{0.17em}\hspace{0.17em}DA = }\sqrt{{\left(\text{-1-2}\right)}^{\text{2}}\text{+}{\left(\text{2+3}\right)}^{\text{2}}\text{+}{\left(\text{1-4}\right)}^{\text{2}}}\\ \text{= }\sqrt{\text{9+25+9}}\\ \text{= }\sqrt{\text{43}}\\ \text{Since, opposite sides AB and CD, BC and DA are equal.}\\ \text{So, ABCD is a parallelogram.}\\ \text{Therefore, }\left(\text{–1,2,1}\right)\text{,}\left(\text{1,–2,5}\right)\text{,}\left(\text{4,–7,8}\right)\text{ and }\left(\text{2,–3,4}\right)\text{}\\ \text{are the vertices of a parallelogram.}\end{array}$

Q.4 Find the equation of the set of points which are equidistant from the points (1, 2, 3) and (3, 2, –1).

Ans.

$\begin{array}{l}\mathrm{Let}\text{the point P}\left(\mathrm{x},\mathrm{y},\mathrm{z}\right)\text{are equidistant from the points Q}\left(\text{1},\text{2},\text{3}\right)\text{}\\ \text{and R}\left(\text{3},\text{2},\text{}–\text{1}\right).\text{Then,}\\ \mathrm{PQ}=\sqrt{{\left(\mathrm{x}-1\right)}^{2}+{\left(\mathrm{y}-2\right)}^{2}+{\left(\mathrm{z}-3\right)}^{2}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\sqrt{{\mathrm{x}}^{2}-2\mathrm{x}+1+{\mathrm{y}}^{2}-4\mathrm{y}+4+{\mathrm{z}}^{2}-6\mathrm{z}+9}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\sqrt{{\mathrm{x}}^{2}-2\mathrm{x}+{\mathrm{y}}^{2}-4\mathrm{y}+{\mathrm{z}}^{2}-6\mathrm{z}+14}\\ \mathrm{PR}=\sqrt{{\left(\mathrm{x}-3\right)}^{2}+{\left(\mathrm{y}-2\right)}^{2}+{\left(\mathrm{z}+1\right)}^{2}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\sqrt{{\mathrm{x}}^{2}-6\mathrm{x}+9+{\mathrm{y}}^{2}-4\mathrm{y}+4+{\mathrm{z}}^{2}+2\mathrm{z}+1}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\sqrt{{\mathrm{x}}^{2}-6\mathrm{x}+{\mathrm{y}}^{2}-4\mathrm{y}+{\mathrm{z}}^{2}+2\mathrm{z}+14}\\ \mathrm{According}\text{to given condition,}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} PQ}=\text{PR}\\ ⇒{\text{\hspace{0.17em}PQ}}^{\text{2}}={\text{PR}}^{2}\\ {\mathrm{x}}^{2}-2\mathrm{x}+{\mathrm{y}}^{2}-4\mathrm{y}+{\mathrm{z}}^{2}-6\mathrm{z}+14\\ ={\mathrm{x}}^{2}-6\mathrm{x}+{\mathrm{y}}^{2}-4\mathrm{y}+{\mathrm{z}}^{2}+2\mathrm{z}+14\\ 6\mathrm{x}-2\mathrm{x}-2\mathrm{z}-6\mathrm{z}=0\\ 4\mathrm{x}-8\mathrm{z}=0\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{x}-2\mathrm{z}=0,\text{which is the required equation of the set of points}\\ \text{which are equidistant from the points}\left(1,2,3\right)\text{and}\left(3,2,-1\right).\end{array}$

Q.5 Find the equation of the set of points P, the sum of whose distances from A (4, 0, 0) and B (– 4, 0, 0) is equal to 10.

Ans.

$\mathrm{Let}\text{the sum of distances of point P}\left(\mathrm{x},\mathrm{y},\mathrm{z}\right)\text{from the points Q}\left(\text{4},\text{}0,\text{}0\right)\text{and R}\left(\text{}–\text{4},\text{}0,\text{}0,\right)\text{is 1}0.\text{Then}$ $\begin{array}{l}\mathrm{PQ}=\sqrt{{\left(\mathrm{x}-4\right)}^{2}+{\left(\mathrm{y}-0\right)}^{2}+{\left(\mathrm{z}-0\right)}^{2}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\sqrt{{\mathrm{x}}^{2}-8\mathrm{x}+16+{\mathrm{y}}^{2}+{\mathrm{z}}^{2}}\\ \mathrm{PR}=\sqrt{{\left(\mathrm{x}+4\right)}^{2}+{\left(\mathrm{y}-0\right)}^{2}+{\left(\mathrm{z}-0\right)}^{2}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\sqrt{{\mathrm{x}}^{2}+8\mathrm{x}+16+{\mathrm{y}}^{2}+{\mathrm{z}}^{2}}\\ \mathrm{According}\text{to given condition,}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} PQ}+\text{PR}=\text{10}\\ ⇒{\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}PQ}}^{\text{2}}={\left(10-\text{PR}\right)}^{2}\\ ⇒{\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}PQ}}^{\text{2}}=100-20\text{\hspace{0.17em}}\mathrm{PR}+{\mathrm{PR}}^{2}\\ 20\text{\hspace{0.17em}}\mathrm{PR}-100={\mathrm{x}}^{2}+8\mathrm{x}+16+{\mathrm{y}}^{2}+{\mathrm{z}}^{2}-{\mathrm{x}}^{2}+8\mathrm{x}-16-{\mathrm{y}}^{2}-{\mathrm{z}}^{2}\\ 20\text{\hspace{0.17em}}\mathrm{PR}-100=16\mathrm{x}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}20\text{\hspace{0.17em}}\mathrm{PR}=100+16\mathrm{x}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}5\text{\hspace{0.17em}}\mathrm{PR}=25+4\mathrm{x}\\ \mathrm{Squarring}\text{both sides, we get}\\ \text{\hspace{0.17em}}{\left(5\text{\hspace{0.17em}}\mathrm{PR}\right)}^{2}={\left(25+4\mathrm{x}\right)}^{2}\\ \text{\hspace{0.17em}\hspace{0.17em}}25\text{\hspace{0.17em}}{\mathrm{PR}}^{2}=625+200\mathrm{x}+16{\mathrm{x}}^{2}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}25\left({\mathrm{x}}^{2}+8\mathrm{x}+16+{\mathrm{y}}^{2}+{\mathrm{z}}^{2}\right)=625+200\mathrm{x}+16{\mathrm{x}}^{2}\\ 25{\mathrm{x}}^{2}+200\mathrm{x}+400+25{\mathrm{y}}^{2}+25{\mathrm{z}}^{2}=625+200\mathrm{x}+16{\mathrm{x}}^{2}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}9{\mathrm{x}}^{2}+25{\mathrm{y}}^{2}+25{\mathrm{z}}^{2}-225=0\end{array}$

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In today’s world of technology, the entire world of education has become digital. Due to this digitalisation, a number of e-learning platforms are available online which cater to the need of students in getting NCERT Solutions for Class 11 Maths. This comes as a boon to them. However, the same blessing can become a bane as there are many fraud sites which provide wrong solutions to students in order to mint money. Thus, students are advised to opt only for those platforms that are known for their authenticity and provide cent- percent accurate and cross-verified solutions. Students of Class 11, who are looking for NCERT Solutions for Class 11 Maths Chapter 12 Exercise 12.2, are advised to visit Extramarks to find the most authentic solutions to the exercise. The NCERT Solutions for Class 11 Maths Chapter 12 Exercise 12.2, available at Extramarks, contain the link to other exercises as well. Here one can find the perfect and scrutinised NCERT Solutions for Class 11 Maths Chapter 12 Introduction to Three Dimensional Geometry and other chapters under NCERT Solutions for Class 11 Maths Chapter 12 Exercise 12.2. Prepared by experts in Mathematics, who are well versed in the guidelines issued by NCERT and the pattern of the exam. These solutions under NCERT Solutions for Class 11 Maths Chapter 12 Exercise 12.2 are highly effective in scoring very high marks in the exams.

### 2. How should students use the NCERT Solutions for Class 11 Maths Chapter 12 Exercise 12.2 provided by Extramarks?

For the basic concepts, the students should follow the NCERT Mathematics book as recommended by CBSE. This will lay the foundation of their knowledge. The solutions for the same and for Chapter 12 of the book can be accessed under NCERT Solutions for Class 11 Maths Chapter 12 Exercise 12.2. Apart from this, taking multiple tests and practising can be done to fortify the concepts in a better and more accomplished way. Practising the questions from these sources can be proved helpful in the preparation.

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