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

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

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Access NCERT Solution for Class 11 MATHS Chapter 12- Introduction to Three Dimensional Geometry part-1

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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