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

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

Free PDF download of NCERT Solutions for Class 11 Maths Chapter 12 Exercise 12.3 (Ex 12.3) and all chapter exercises at one place prepared by expert teacher as per NCERT (CBSE) books guidelines. Class 11 Maths Chapter 12 Introduction to Three Dimensional Geometry Exercise 12.3 Questions with Solutions to help you to revise complete Syllabus and Score More marks. Register and get all exercise solutions in your emails.


Class:

NCERT Solutions for Class 11

Subject:

Class 11 Maths

Chapter Name:

Chapter 12 - Introduction to Three Dimensional Geometry

Exercise:

Exercise - 12.3

Content-Type:

Text, Videos, Images and PDF Format

Academic Year:

2024-25

Medium:

English and Hindi

Available Materials:

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  • Important Questions

  • Revision Notes

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Competitive Exams after 12th Science

Access NCERT Solution for Class 11 MATHS Chapter 12- Introduction to Three Dimensional Geometry

Exercise 12.3

Refer to pages 1-3 for exercise 12.3 in the PDF

1. Find the coordinates of the point which divides the line segment joining the points \[\left( { - 2,3,5} \right)\] and \[\left( {1, - 4,6} \right)\] in the ratio.

(i) $2:3$ internally

Ans: The coordinates of point R that divides the line segment joining points \[P\left( {{x_1},{y_1},{z_1}} \right)\] and \[Q\left( {{x_2},{y_2},{z_2}} \right)\] internally in the ratio m:n are 

\[\left( {\dfrac{{m{x_2} + n{x_1}}}{{m + n}},\dfrac{{m{y_2} + n{y_2}}}{{m + n}},\dfrac{{m{z_1} + n{z_1}}}{{m + n}}} \right)\]

Let \[R\left( {x,y,z} \right)\] be the point that divides the line segment joining points \[\left( { - 2,3,5} \right)\]and \[\left( {1, - 4,6} \right)\] internally in the ratio 2:3 .

\[x = \dfrac{{2\left( 1 \right) + 3\left( { - 2} \right)}}{{2 + 3}},y = \dfrac{{2\left( { - 4} \right) + 3\left( 3 \right)}}{{2 + 3}},z = \dfrac{{2\left( 6 \right) + 3\left( 5 \right)}}{{2 + 3}}\]

\[i.e.,x = \dfrac{{ - 4}}{5},y = \dfrac{1}{5}\]and\[z = \dfrac{{27}}{5}\]

Thus, the coordinates of the required point are \[\left( {\dfrac{{ - 4}}{5},\dfrac{1}{5},\dfrac{{27}}{5}} \right)\]

(ii) $2:3$ externally

Ans: The coordinates of point R that divides the line segment joining points \[P\left( {{x_1},{y_1},{z_1}} \right)\] and \[Q\left( {{x_2},{y_2},{z_2}} \right)\]externally in the ratio m: n are 

\[\left( {\dfrac{{m{x_2} - n{x_1}}}{{m - n}},\dfrac{{m{y_2} - n{y_2}}}{{m - n}},\dfrac{{m{z_1} - n{z_1}}}{{m - n}}} \right)\]

Let \[R\left( {x,y,z} \right)\]be the point that divides the line segment joining points $\left( { - 2,3,5} \right)$ and $\left( {1, - 4,6} \right)$ externally in the ratio $2:3$. 

\[x = \dfrac{{2\left( 1 \right) - 3\left( { - 2} \right)}}{{2 - 3}},y = \dfrac{{2\left( { - 4} \right) - 3\left( 3 \right)}}{{2 - 3}},z = \dfrac{{2\left( 6 \right) - 3\left( 5 \right)}}{{2 - 3}}\]

\[i.e.,x =  - 8,y = 17\]and\[z = 3\]

Thus, the coordinates of the required point are \[\left( { - 8,17,3} \right)\].


2. Given that \[P\left( {3,2, - 4} \right),Q = \left( {5,4, - 6} \right)\] and \[R = \left( {9,8, - 10} \right)\] are collinear. Find the ratio in which Q divides PR.

Ans: Let point \[Q = \left( {5,4, - 6} \right)\]divide the line segment joining points \[P\left( {3,2, - 4} \right)\]\[R = \left( {9,8, - 10} \right)\] in the ratio $k:1$. 

Therefore, by section formula,

\[\left( {5,4, - 6} \right) = \left( {\dfrac{{k\left( 9 \right) + 3}}{{k + 1}},\dfrac{{k\left( 8 \right) + 2}}{{k + 1}},\dfrac{{k\left( { - 10} \right) - 4}}{{k + 1}}} \right)\]

\[ \Rightarrow \dfrac{{9k + 3}}{{k + 1}} = 5\]

\[ \Rightarrow 9k + 3 = 5k + 2\]

\[ \Rightarrow 4k = 2\]

\[ \Rightarrow k = \dfrac{1}{2}\]

Thus, point Q divides PR in the ratio $1:2$.


3. Find the ratio in which the YZ-plane divides the line segment formed by joining the points \[\left( { - 2,4,7} \right)\] and \[\left( {3, - 5,8} \right)\]

Ans: Let the YZ plane divide the line segment joining points \[\left( { - 2,4,7} \right)\] and \[\left( {3, - 5,8} \right)\] in the ratio $k:1$. 

Hence, by section formula, the coordinates of point of intersection are given by 

\[\left( {\dfrac{{k\left( 3 \right) - 2}}{{k + 1}},\dfrac{{k\left( { - 5} \right) + 4}}{{k + 1}},\dfrac{{k\left( 8 \right) + 7}}{{k + 1}}} \right)\]

On the YZ plane, the $x$ -coordinate of any point is zero. 

\[ \Rightarrow \dfrac{{3k - 2}}{{k + 1}} = 0\]

\[ \Rightarrow 3k - 2 = 0\]

\[ \Rightarrow k = \dfrac{2}{3}\]

Thus, the YZ plane divides the line segment formed by joining the given points in the ratio $2:3$.


4. Using section formula, show that the \[A\left( {2, - 3,4} \right),B\left( { - 1,2,1} \right)\] and \[C\left( {0,\dfrac{1}{3},2} \right)\] are collinear. 

Ans: The given points are \[A\left( {2, - 3,4} \right),B\left( { - 1,2,1} \right)\] and \[C\left( {0,\dfrac{1}{3},2} \right)\]

Let P be a point that divides AB in the ratio $k:1$.

Hence, by section formula, the coordinates of P are given by

\[\left( {\dfrac{{k\left( { - 1} \right) + 2}}{{k + 1}},\dfrac{{k\left( 2 \right) - 3}}{{k + 1}},\dfrac{{k\left( 1 \right) + 4}}{{k + 1}}} \right)\]

Now, we find the value of k at which point P coincides with point C.

By taking

\[\dfrac{{ - k + 2}}{{k + 1}} = 0\]

,we obtain $k = 2$.

For $k = 2$, the coordinates of point P are\[\left( {0,\dfrac{1}{3},2} \right)\]

i.e., \[\left( {0,\dfrac{1}{3},2} \right)\]is a point that divides AB externally in the ratio $2:1$ and is the same as point P.

5. Find the coordinates of the points which trisect the line segment joining the points \[P\left( {4,2, - 6} \right)\]and \[Q\left( {10, - 16,6} \right)\].

Ans: Let A and B be the points that trisect the line segment joining points \[P\left( {4,2, - 6} \right)\] and \[Q\left( {10, - 16,6} \right)\].

\[P\left( {4,2, - 6} \right)\]\[Q\left( {10, - 16,6} \right)\]

(Image Will Be Updated Soon)

Point A divides PQ in the ratio $1:2$. Therefore, by section formula, the coordinates of point A are given by 

\[\left( {\dfrac{{1\left( {10} \right) + 2\left( 4 \right)}}{{1 + 2}},\dfrac{{1\left( { - 16} \right) + 2\left( 2 \right)}}{{1 + 2}},\dfrac{{1\left( 6 \right) + 2\left( { - 6} \right)}}{{1 + 2}}} \right) = \left( {6, - 4, - 2} \right)\]

Point B divides PQ in the ratio $2:1$ . Therefore, by section formula, the coordinates of point B are given by 

\[\left( {\dfrac{{2\left( {10} \right) + 1\left( 4 \right)}}{{2 + 1}},\dfrac{{2\left( { - 16} \right) + 1\left( 2 \right)}}{{2 + 1}},\dfrac{{2\left( 6 \right) + 1\left( { - 6} \right)}}{{2 + 1}}} \right) = \left( {8, - 10,2} \right)\]

Thus, $\left( {6, - 4, - 2} \right)$ and $\left( {8, - 10,2} \right)$ are the points that trisect the line segment joining points $P\left( {4,2, - 6} \right)$ and $Q\left( {10, - 16,6} \right)$.

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

Opting for the NCERT solutions for Ex 12.3 Class 11 Maths is considered as the best option for the CBSE students when it comes to exam preparation. This chapter consists of many exercises. Out of which we have provided the Exercise 12.3 Class 11 Maths NCERT solutions on this page in PDF format. You can download this solution as per your convenience or you can study it directly from our website/ app online.

Vedantu in-house subject matter experts have solved the problems/ questions from the exercise with the utmost care and by following all the guidelines by CBSE. Class 11 students who are thorough with all the concepts from the Maths textbook and quite well-versed with all the problems from the exercises given in it, then any student can easily score the highest possible marks in the final exam. With the help of this Class 11 Maths Chapter 12 Exercise 12.3 solutions, students can easily understand the pattern of questions that can be asked in the exam from this chapter and also learn the marks weightage of the chapter. So that they can prepare themselves accordingly for the final exam.

Besides these NCERT solutions for Class 11 Maths Chapter 12 Exercise 12.3, there are plenty of exercises in this chapter which contain innumerable questions as well. All these questions are solved/answered by our in-house subject experts as mentioned earlier. Hence all of these are bound to be of superior quality and anyone can refer to these during the time of exam preparation. In order to score the best possible marks in the class, it is really important to understand all the concepts of the textbooks and solve the problems from the exercises given next to it.

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FAQs on NCERT Solutions for Class 11 Maths Chapter 12: Introduction to Three Dimensional Geometry - Exercise 12.3

1. Where can we find the NCERT Solutions for Class 11 Maths Chapter 12 Introduction to Three Dimensional Geometry (Ex 12.3) Exercise 12.3 in free PDF format?

NCERT Solutions for Class 11 Maths Chapter 12 Exercise 12.3 (Ex 12.3) and all chapter exercises at one time produced by qualified teachers in agreement with NCERT (CBSE) book guidelines in Vedantu are available for free download in PDF format. Questions and answers for Class 11 Math Chapter 12 Introduction to Three-Dimensional Geometry Exercise 12.3 are provided to assist you in reviewing the entire syllabus and achieving higher scores. Register to receive all Vedantu workout solutions in your inbox.

2. Are the NCERT Solutions for Class 11 Maths Chapter 12 Introduction to Three Dimensional Geometry (Ex 12.3) Exercise 12.3 beneficial?

Yes definitely, NCERT Solutions for Class 11 Maths Chapter 12 Introduction to Three Dimensional Geometry (Ex 12.3). For CBSE students preparing for exams, using the NCERT solutions for Ex 12.3 Class 11 Math is thought to be the best alternative. There are numerous exercises in this chapter. We offer Exercise 12.3 Class 11 Maths NCERT solutions. You can study this answer straight from the Vedantu website or mobile app, or you can download it as needed.

3. Why should I choose Vedantu's NCERT Solutions for Class 11 Mathematics Exercise 12.3: Introduction to Three-Dimensional Geometry?

If you have the correct research material, receiving all A's in math shouldn't be too difficult. For a topic like mathematics, studying just the chapters is never enough; therefore, students need to practice the corresponding exercise questions. NCERT Solutions have been developed by Vedantu's knowledgeable staff in accordance with CBSE standards. The solutions are correctly and thoroughly stated. Utilizing quick-response strategies will help you finish your exam on time, increase your scores, and become ready for competitive exams.

4. How many questions are there in the NCERT Solutions for Class 11 Maths Subject Chapter 12, Introduction to Three Dimensional Geometry (Ex 12.3) Exercise 12.3?

There are 5 problems in all in Exercise 12.3 of Chapter 12's Introduction to Three-Dimensional Geometry. NCERT questions can be viewed as helpful tools because they are filled with techniques that will enable students to get better results. For the purpose of assisting students in their exam preparation, Vedantu offers NCERT Solutions for all the chapters of Class 11 math at no charge. On Vedantu's official website.

5. How can I practice Exercise 12.3 from Chapter 11 of NCERT Class 11 Maths, Introduction to Three-Dimensional Geometry?

For students taking any board test or competitive exam, the constant review is important during the preparation phase. After finishing each chapter, move on to the corresponding exercises to practice solving the corresponding problems. Up to your final exam, frequently practice such problems. You can refer to the revision notes for Class 11 Math written by the subject-matter specialists at Vedantu, which will help you remember the equations and particulars of Chapter 11 Introduction to Three-Dimensional Geometry.