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# NCERT Solutions for Class 11 Maths Chapter 11 - Introduction to Three-Dimensional Geometry Exercise 11.2

Last updated date: 18th Sep 2024
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## NCERT Solutions for Maths Chapter 11 Exercise 11.2 Class 11 - FREE PDF Download

NCERT Solutions for Class 11 Maths Chapter 11 Introduction to Three Dimensional Geometry Exercise 11.2 provides students with comprehensive step-by-step solutions to problems related to direction cosines, direction ratios, and equations of lines in space as per the updated Class 11 Maths Syllabus. These solutions help students strengthen their understanding of Three Dimensional geometry concepts and easily approach complex problems. Designed to support effective exam preparation, this resource is ideal for mastering the fundamental concepts of Three Dimensional geometry.

Table of Content
1. NCERT Solutions for Maths Chapter 11 Exercise 11.2 Class 11 - FREE PDF Download
2. Glance on NCERT Solutions Maths Chapter 11 Exercise 11.2 Class 11 | Vedantu
3. Access NCERT Solutions for Maths Class 11 Chapter 11 - Introduction to Three Dimensional Geometry
3.1Exercise 11.2
4. Class 11 Maths Chapter 11: Exercises Breakdown
5. CBSE Class 11 Maths Chapter 11 Other Study Materials
6. Chapter-Specific NCERT Solutions for Class 11 Maths
FAQs

## Glance on NCERT Solutions Maths Chapter 11 Exercise 11.2 Class 11 | Vedantu

• Exercise 11.2 of Class 11 Maths Chapter 11 Introduction to Three-Dimensional Geometry helps students practice and master the basic concepts of three-dimensional geometry, which is essential for higher-level mathematics.

• This exercise requires students to use the distance formula to calculate the distance between two given points with 3D coordinates. This helps in understanding the spatial relationship between points in three dimensions.

• Some questions ask students to determine whether given points are collinear by checking if the sum of the distances between consecutive points equals the total distance between the endpoints.

• Students are asked to find the coordinates of a point dividing a line segment between two points in a given ratio using the section formula. These questions may involve internal or external divisions.

• Some questions involve determining the equation of a set of points that are equidistant from given points or satisfy certain conditions related to distances in 3D space.

Competitive Exams after 12th Science

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

### Exercise 11.2

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

1. $(2, 3, 5)$ and $(4, 3, 1)$

2. $(-3, 7, 2)$ and $(2, 4, -1)$

3. $(-1, 3, -4)$ and $(1, -3, 4)$

4. $(2, -1, 3)$ and $(-2, 1, 3)$

Ans: (i) $(2, 3, 5)$ and $(4, 3, 1)$:

We use the distance formula between two points $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$:

$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}$

$d = \sqrt{(4 - 2)^2 + (3 - 3)^2 + (1 - 5)^2} = \sqrt{(2)^2 + (0)^2 + (-4)^2} = \sqrt{4 + 16} = \sqrt{20} = 2\sqrt{5}$

(ii) $(-3, 7, 2)$ and $(2, 4, -1)$:

$d = \sqrt{(2 + 3)^2 + (4 - 7)^2 + (-1 - 2)^2} = \sqrt{(5)^2 + (-3)^2 + (-3)^2} = \sqrt{25 + 9 + 9} = \sqrt{43}$

(iii) $(-1, 3, -4)$ and $(1, -3, 4)$:

$d = \sqrt{(1 + 1)^2 + (-3 - 3)^2 + (4 + 4)^2} = \sqrt{(2)^2 + (-6)^2 + (8)^2} = \sqrt{4 + 36 + 64} = \sqrt{104} = 2\sqrt{26}$

(iv) $(2, -1, 3)$ and $(-2, 1, 3)$:

$d = \sqrt{(-2 - 2)^2 + (1 + 1)^2 + (3 - 3)^2} = \sqrt{(-4)^2 + (2)^2 + (0)^2} = \sqrt{16 + 4} = \sqrt{20} = 2\sqrt{5}$

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

Ans: To prove collinearity, we check if the distances between the points satisfy $AB + BC = AC$.

- Distance between $A(-2, 3, 5)$ and $B(1, 2, 3)$:

$AB = \sqrt{(1 + 2)^2 + (2 - 3)^2 + (3 - 5)^2} = \sqrt{(3)^2 + (-1)^2 + (-2)^2} = \sqrt{9 + 1 + 4} = \sqrt{14}$

- Distance between $B(1, 2, 3)$ and $C(7, 0, -1)$:

$BC = \sqrt{(7 - 1)^2 + (0 - 2)^2 + (-1 - 3)^2} = \sqrt{(6)^2 + (-2)^2 + (-4)^2} = \sqrt{36 + 4 + 16} = \sqrt{56}$

- Distance between $A(-2, 3, 5)$ and $C(7, 0, -1)$:

$AC = \sqrt{(7 + 2)^2 + (0 - 3)^2 + (-1 - 5)^2} = \sqrt{(9)^2 + (-3)^2 + (-6)^2} = \sqrt{81 + 9 + 36} = \sqrt{126}$

Since $AB + BC \neq AC$, the points are not collinear.

3. Verify the following:

(i) Show that $(0, 7, -10)$, $(1, 6, -6)$, and $(4, 9, -6)$ are vertices of an isosceles triangle.

Ans:

We check if two sides of the triangle are equal:

- Distance between $(0, 7, -10)$ and $(1, 6, -6)$:

$d = \sqrt{(1 - 0)^2 + (6 - 7)^2 + (-6 + 10)^2} = \sqrt{1 + 1 + 16} = \sqrt{18}$

- Distance between $(1, 6, -6)$ and $(4, 9, -6)$:

$d = \sqrt{(4 - 1)^2 + (9 - 6)^2 + (-6 + 6)^2} = \sqrt{9 + 9 + 0} = \sqrt{18}$

- Distance between $(0, 7, -10)$ and $(4, 9, -6)$:

$d = \sqrt{(4 - 0)^2 + (9 - 7)^2 + (-6 + 10)^2} = \sqrt{16 + 4 + 16} = \sqrt{36}$

Since two sides are equal, the triangle is isosceles.

(ii) Show that $(0, 7, 10)$, $(-1, 6, 6)$, and $(4, 9, 6)$ are the vertices of a right-angled triangle.

Ans: To verify if the triangle is right-angled, we check if the square of the longest side equals the sum of the squares of the other two sides.

- Distance between $(0, 7, 10)$ and $(-1, 6, 6)$:

$d_1 = \sqrt{(-1 - 0)^2 + (6 - 7)^2 + (6 - 10)^2} = \sqrt{(-1)^2 + (-1)^2 + (-4)^2} = \sqrt{1 + 1 + 16} = \sqrt{18}$

- Distance between $(-1, 6, 6)$ and $(4, 9, 6)$:

$d_2 = \sqrt{(4 + 1)^2 + (9 - 6)^2 + (6 - 6)^2} = \sqrt{(5)^2 + (3)^2 + (0)^2} = \sqrt{25 + 9 + 0} = \sqrt{34}$

- Distance between $(0, 7, 10)$ and $(4, 9, 6)$:

$d_3 = \sqrt{(4 - 0)^2 + (9 - 7)^2 + (6 - 10)^2} = \sqrt{(4)^2 + (2)^2 + (-4)^2} = \sqrt{16 + 4 + 16} = \sqrt{36} = 6$

Now check if the Pythagorean theorem holds:

$d_1^2 + d_2^2 = 18 + 34 = 52$

$d_3^2 = 6^2 = 36$

Since $d_1^2 + d_2^2 \neq d_3^2$, the triangle is not right-angled based on this configuration.

(iii) Show that $(-1, 2, 1)$, $(1, -2, 5)$, $(4, -7, 8)$, and $(2, -3, 4)$ are the vertices of a parallelogram.

Ans: To verify if the points form a parallelogram, we need to show that the diagonals bisect each other, i.e., the midpoints of both diagonals should coincide.

- Midpoint of diagonal $(-1, 2, 1)$ and $(4, -7, 8)$:

$M_1 = \left(\frac{-1 + 4}{2}, \frac{2 - 7}{2}, \frac{1 + 8}{2}\right) = \left(\frac{3}{2}, \frac{-5}{2}, \frac{9}{2}\right)$

- Midpoint of diagonal $(1, -2, 5)$ and $(2, -3, 4)$:

$M_2 = \left(\frac{1 + 2}{2}, \frac{-2 - 3}{2}, \frac{5 + 4}{2}\right) = \left(\frac{3}{2}, \frac{-5}{2}, \frac{9}{2}\right)$

Since the midpoints are the same, the diagonals bisect each other, proving that the points form a parallelogram.

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

Ans: Let the coordinates of the point $P(x, y, z)$ be equidistant from the points $(1, 2, 3)$ and $(3, 2, -1)$. Then,

$\text{Distance from } P(x, y, z) \text{ to } (1, 2, 3) = \text{Distance from } P(x, y, z) \text{ to } (3, 2, -1)$

Using the distance formula:

$\sqrt{(x - 1)^2 + (y - 2)^2 + (z - 3)^2} = \sqrt{(x - 3)^2 + (y - 2)^2 + (z + 1)^2}$

Squaring both sides:

$(x - 1)^2 + (y - 2)^2 + (z - 3)^2 = (x - 3)^2 + (y - 2)^2 + (z + 1)^2$

Expanding both sides:

$(x^2 - 2x + 1) + (y^2 - 4y + 4) + (z^2 - 6z + 9) = (x^2 - 6x + 9) + (y^2 - 4y + 4) + (z^2 + 2z + 1)$

Simplifying:

$x^2 - 2x + 1 + y^2 - 4y + 4 + z^2 - 6z + 9 = x^2 - 6x + 9 + y^2 - 4y + 4 + z^2 + 2z + 1$

Cancel the common terms:

$-2x + 1 - 6z + 9 = -6x + 9 + 2z + 1$

Simplifying further:

$-2x - 6z + 10 = -6x + 2z + 10$

Bringing all terms to one side:

$4x - 8z = 0$

Thus, the equation of the set of points is:

$x = 2z$

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:This is the equation of an ellipse with foci at points $A(4, 0, 0)$ and $B(-4, 0, 0)$, and the sum of the distances from these points is equal to the constant 10. The standard equation of an ellipse with foci at $(a, 0, 0)$ and $(-a, 0, 0)$, and the sum of the distances equal to $2c$, is:

$2a = 10, \quad \Rightarrow a = 5, \quad b^2 = a^2 - f^2$

Here, the distance between the foci is 8, so $f = 4$. Thus:

$b^2 = 5^2 - 4^2 = 25 - 16 = 9, \quad b = 3$

Therefore, the equation of the ellipse is:

$\frac{x^2}{25} + \frac{y^2}{9} = 1$

## Conclusion

The NCERT Solutions for Class 11 Maths Chapter 11: Introduction to Three Dimensional Geometry, Exercise 11.2 offer a clear and structured approach to mastering key concepts like distance and section formulas in 3D space. By working through these questions, students will solidify their understanding of spatial relationships and improve their problem-solving skills. This exercise is fundamental in building a foundation for advanced topics in geometry and vector analysis. Using Vedantu’s solutions, students can easily grasp these concepts and be better prepared for their exams.

## Class 11 Maths Chapter 11: Exercises Breakdown

 Exercises Number of Questions Exercise 11.1 4 Questions and Solutions Miscellaneous Exercise 4 Questions and Solutions

## Chapter-Specific NCERT Solutions for Class 11 Maths

Given below are the Chapter-wise NCERT Solutions for Class 11 Maths. Go through these chapter-wise solutions to be thoroughly familiar with the concepts.

## FAQs on NCERT Solutions for Class 11 Maths Chapter 11 - Introduction to Three-Dimensional Geometry Exercise 11.2

1. What topics are covered in NCERT Solutions for Class 11 Maths Chapter 11, Exercise 11.2?

The exercise focuses on the distance formula, section formula, and finding the midpoint of a line segment in three-dimensional geometry.

2. How does the distance formula work in 3D geometry?

The distance formula in 3D geometry calculates the distance between two points $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$ using:

$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}$

3. What is the section formula in Three-Dimensional Geometry?

The section formula helps to find the coordinates of a point dividing a line segment joining two points in a specific ratio. For points $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$ divided in the ratio $m:n$, the coordinates are:

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

4. How can I verify if three points are collinear in 3D geometry?

To verify collinearity, calculate the distances between consecutive points. If the sum of the two smaller distances equals the total distance between the endpoints, the points are collinear.

5. What is the midpoint formula in three-dimensional geometry?

The midpoint of a line segment joining two points $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$ is given by:

$\left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}, \frac{z_1 + z_2}{2}\right)$

6. How does solving Exercise 11.2 help in understanding 3D geometry?

Exercise 11.2 builds a strong understanding of spatial relationships in three dimensions by focusing on fundamental concepts like distances and points, which are essential for more complex topics like planes and vectors.

7. What type of questions are asked in Exercise 11.2 of Chapter 11?

The questions involve calculating distances between points, verifying collinearity, finding midpoints, and using the section formula to divide line segments in a given ratio.

8. Why is the section formula important in 3D geometry?

The section formula helps determine how a line segment is divided internally or externally, which is crucial in solving problems involving the division of space in 3D geometry.

9. Are there any real-life applications of the distance formula in 3D geometry?

Yes, the distance formula is widely used in physics, engineering, and computer graphics to calculate distances between objects in three-dimensional space.

10. How can Vedantu’s NCERT solutions help in mastering Exercise 11.2?

Vedantu’s NCERT solutions provide step-by-step explanations, helping students to understand and solve complex problems easily, thereby improving their performance in exams.