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

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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.

(i)Let point P(x,y,z) divides the line segment joining the points(− 2, 3, 5) and (1,− 4,6) in the ratio 2:3 internally.Then,      x=mx2+nx1m+n =2(1)+3(−2)2+3=−45     y=my2+ny1m+n =2(−4)+3(3)2+3 =15      z=mz2+nz1m+n =2(6)+3(5)2+3 =275Therefore, the coordinates of the point P are (−45,15,275).(ii) Let point P (x,y,z) divides the line segment joining the points(− 2, 3, 5) and (1,− 4,6) in the ratio 2:3 externally.Then,      x=mx2−nx1m−n =2(1)−3(−2)2−3

=8−1=−8      y=my2−ny1m−n =2(−4)−3(3)2−3 =−17−1=17      z=mz2−nz1m−n =2(6)−3(5)2−3 =−3−1=3Therefore, the coordinates of the point P are (−8,17,3).

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.

Point Q(5, 4,−6) divides the line segment joining the pointsP(3, 2,− 4) and R(9, 8,−10) in the ratio k:1 internally.Then,     x=mx2+nx1m+n    5=k(9)+1(3)k+1 5(k+1)=9 k+3   5 k+5=9 k+3      5−3=9k−5k

⇒    2=4 k⇒   k=24⇒   k=12The required ratio is 12:1 i.e., 1:2.

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.

Plane YZ divides the line segment joining the pointsQ(−2, 4,  7)  and R(3,−5,  8) in the ratio k:1 internally.Then,let coordinates of point P in YZ plane: (0,y,z)    x=mx2+nx1m+n    0=k(3)+1(−2)k+1    0=3 k−2    2=3 k    k=23The required ratio is 23:1 i.e., 2:3.

Q.4

Using section formula, show that the points A(2,−3, 4), B(−1, 2, 1) and C(0,13,2) are collinear.

Ans.

The given points are A(2,−3, 4), B(−1, 2, 1) and C0,13, 2.Let a point P divides AB​in the ratio k:1.Hence, by the section formula, the coordinates of P ​are:mx2+nx1m+n,my2+ny1m+n,mz2+nz1m+nk−1+2k+1,k2−3k+1,k1+4k+1Now, we find the value of k at which P coincides with point C.So, by taking −k+2k+1=0⇒ k =2For k=2, the coordinates of P ​are: 0,13, 2Therefore, C0,13,2​​   is a point that divides AB ​externally in ratio 2:1.Hence, points A, B​ and C are collinear.

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.

coordinates of point A={1(10)+2(4)1+2,1(−16)+2(2)1+2,1(6)+2(− 6)1+2}=(10+83,−16+43,6−123)=(6,−4,−2)coordinates of point B={2(10)+1(4)1+2,2(−16)+1(2)1+2,2(6)+1(− 6)1+2}=(20+43,−32+23,12−63)=(8,−10,2)Thus, points (6,−4,−2) and (8,−10,2) trisect the line segment formed by the given points.

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