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

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

Edited By Ravindra Pindel | Updated on Jul 12, 2022 04:56 PM IST

In the previous exercises, you have already learned about the coordinate axes, coordinate planes, coordinate of a point in the three-dimensional space, the distance between two points in the three-dimensional space. In the NCERT solutions for Class 11 Maths chapter 12 exercise 12.3, you will learn about an important concept called section formula. In two-dimensional geometry, you have learned about the section formula to find the coordinates of a point dividing a line segment in a given ratio internally.

In Class 11 NCERT book Maths chapter 12 exercise 12 you will learn the same concept to find coordinates of a point dividing a line segment in a given ratio internally in three-dimensional space. The section formula for three-dimensional space is very similar to the section formula for two-dimensional space. If you have a good command of the two-dimensional space, it won't take much effort from you to get command on the exercise 12.3 Class 11 Maths. You can go through the Class 11 Maths chapter 12 exercise 12.3 solutions to get conceptual clarity. If you are looking for NCERT Solutions, you can click on the given link where you will get NCERT solutions at one place.

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

The line segment joining the points A (– 2, 3, 5) and B(1, – 4, 6)

Let point P(x,y,z) be the point that divides the line segment AB internally in the ratio 2:3.

Now, As we know by section formula, The coordinate of the point P which divides line segment $A(x_1,y_1,z_1)$ And $B(x_2,y_2,z_2)$ in ratio m:n is

$\left (\frac{mx_2+nx_1}{m+n},\frac{my_2+ny_1}{m+n},\frac{mz_2+nz_1}{m+n} \right )$

Now the point that divides A (– 2, 3, 5) and B(1, – 4, 6) in ratio 2:3 is

$\left (\frac{2(1)+3(-2)}{2+3},\frac{2(-4)+3(3)}{2+3},\frac{2(6)+3(5)}{2+3} \right )=\left ( \frac{-4}{5},\frac{1}{5},\frac{27}{5} \right )$

Hence required point is

$\left ( \frac{-4}{5},\frac{1}{5},\frac{27}{5} \right )$.

he line segment joining the points A (– 2, 3, 5) and B(1, – 4, 6)

Let point P(x,y,z) be the point that divides the line segment AB externally in the ratio 2:3.

Now, As we know by section formula , The coordinate of the point P which divides line segment $A(x_1,y_1,z_1)$ And $B(x_2,y_2,z_2)$ externally in ratio m:n is

$\left (\frac{mx_2-nx_1}{m-n},\frac{my_2-ny_1}{m-n},\frac{mz_2-nz_1}{m-n} \right )$

Now the point that divides A (– 2, 3, 5) and B(1, – 4, 6) externally in ratio 2:3 is

$\left (\frac{2(1)-3(-2)}{2-3},\frac{2(-4)-3(3)}{2-3},\frac{2(6)-3(5)}{2-3} \right )=\left ( -8,17,3\right )$

Hence required point is

. $( -8,17,3)$

Given Three points,

P (3, 2, – 4), Q (5, 4, – 6) and R (9, 8, –10)

Let point Q divides PR internally in the ratio $\lambda:1$

Now,

According to the section formula , The point Q in terms of P,Q and $\lambda$ is:

$\left ( \frac{9\lambda+3}{\lambda+1},\frac{8\lambda+2 }{\lambda +1},\frac{-10\lambda-4}{\lambda+1}\right )=(5,4,-6)$

$\frac{9\lambda+3}{\lambda+1}=5$

${9\lambda+3}=5(\lambda+1)$

${9\lambda+3}=5\lambda+5$

$4\lambda =2$

$\lambda =\frac{1}{2}$

Hence, point Q divides PR in ratio 1:2.

Given,

two points A(–2, 4, 7) and B(3, –5, 8)

Let Y-Z plane divides AB in $\lambda:1$

So, According to the section formula, the point which divides AB in $\lambda:1$ is

$\left ( \frac{3\lambda-2}{\lambda+1},\frac{-5\lambda +4}{\lambda+1},\frac{8\lambda +7}{\lambda+1} \right )$

Since this point is in YZ plane, x coordinate of this point will be zero.

So,

$\frac{3\lambda-2}{\lambda+1}=0$

${3\lambda-2}=0$

$\lambda = \frac{2}{3}$

Hence YZ plane divides Line segment AB in a ratio 2:3.

Given,

three points, A (2, –3, 4), B (–1, 2, 1) and C (0, 1/3, 2)

Let a point P divides Line segment AB in the ratio $\lambda:1$

SO, according to the section formula, the point P will be

$\left (\frac{-\lambda +2}{\lambda+1},\frac{2\lambda-3}{\lambda+1},\frac{\lambda+4}{\lambda+1} \right )$

Now, let's compare this point P with point C.

$\frac{-\lambda +2}{\lambda+1},=0$

$\lambda=2$

From here, we see that for $\lambda=2$, point C divides the line segment AB in ratio 2:1. Since point C divides the line segment AB, it lies in the line joining A and B and Hence they are colinear.

Given,

two points P (4, 2, – 6) and Q (10, –16, 6).

The point which trisects the line segment are the points which divide PQ in either 1:2 or 2:1

Let R (x,y,z) be the point which divides Line segment PR in ratio 1:2

Now, according to the section formula

$(x,y,z)=\left ( \frac{10+2(4)}{1+2},\frac{-16+2(2)}{1+2},\frac{6-2(6)}{1+2} \right )=(6,-4,-2)$

Let S be the point which divides the Line segment PQ in ratio 2:1

So, The point S according to section formula is

$(x,y,z)=\left ( \frac{(2)10+(4)}{1+2},\frac{2(-16)+(2)}{1+2},\frac{(2)6-(6)}{1+2} \right )=(8,-10,2)$

Hence the points which trisect the line segment AB are (6,-4,-2) and (8,-10,2).

More About NCERT Solutions for Class 11 Maths Chapter 12 Exercise 12.3:-

Class 11 Maths chapter 12 exercise 12.3 consists of questions related to finding the coordinates of the point which divides the line segment in the given ratio internally and externally. Also, you will get questions related to checking the collinearity of three-point in the three-dimensional space. You can go through the derivation of the section formula given in NCERT book before Class 11 Maths chapter 12 exercise 12.3 to get more clarity.

Benefits of NCERT Solutions for Class 11 Maths Chapter 12 Exercise 12.3:-

• Exercise 12.3 Class 11 Maths is short, concise, and very similar to the section formula in two-dimensional geometry.
• You can try to solve problems from the NCERT syllabus Class 11 Maths chapter 12 exercise 12.3.
• You can take help from Class 11 Maths chapter 12 exercise 12.3 solutions if you find difficulties while solving NCERT problems.

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1. Do we need buy NCERT solutions book for Class 11 Maths ?

No, you don't need to buy any book for NCERT solutions for Class 11 Maths.

3. What are the important topics in the CBSE Class 11 Maths chapter 12 ?

Coordinate axes, coordinate planes, coordinates of a point in three-dimensional space, the distance between two points in three dimensions, section formula are important topics in the CBSE Class 11 Maths chapter 12.

4. How many chapters are there in the CBSE Class 11 Maths ?

There are 16 chapters starting from sets to probability in the CBSE Class 11 Maths.

5. Which is the best for CBSE Class 11 Maths?

As most of the questions in the CBSE exam come directly from the NCERT, it is the best book for CBSE Class 11 Maths.

6. Which are the high weighatge units in the CBSE Class 11 Maths ?

Sets, Functions, and Algebra are the high weightage unit in the CBSE Class 11 Maths.

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