NCERT Solutions for Class 11 Maths Chapter 12 Introduction to Three Dimensional Geometry are part of NCERT Solutions for Class 11 Maths. Here we have given NCERT Solutions for Class 11 Maths Chapter 12 Introduction to Three Dimensional Geometry.

Board |
CBSE |

Textbook |
NCERT |

Class |
Class 11 |

Subject |
Maths |

Chapter |
Chapter 12 |

Chapter Name |
Introduction to Three Dimensional Geometry |

Exercise |
Ex 12.1, Ex 12.2, Ex 12.3 |

Number of Questions Solved |
14 |

Category |
NCERT Solutions |

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

### Chapter 12 Introduction to Three Dimensional Geometry Exercise 12.1

**Question 1.**

A point is on the x-axis.What are its y-coordinate and z-coordinate?

**Solution:**

The coordinates of any point on the x-axis will be (x, 0, 0). Thus y-coordinate and z-coordinate of the point are zero.

**Question 2.**

A point is in the XZ-plane. What can you say about its y-coordinate?

**Solution:**

The coordinates of any point in XZ-plane will be (x, 0, z). Thus y-coordinate of the point is zero.

**Question 3.**

Name the octants in which the following points lie:

(1, 2, 3), (4, -2, 3), (4, -2, -5), (4, 2, -5), (-4, 2, -5), (-4, 2, 5), (-3, -1, 6), (2, -4, -7)

**Solution:**

Point (1, 2, 3) lies in Octant I.

Point (4, -2, 3) lies in Octant IV.

Point (4, -2, -5) lies in Octant VIII.

Point (4, 2, -5) lies in Octant V.

Point (- 4, 2, -5) lies in Octant VI.

Point (- 4, 2, 5) lies in Octant II.

Point (- 3, -1, 6) lies in Octant III.

Point (2, – 4, -7) lies in Octant VIII.

**Question 4.**

**Fill in the blanks:**

**(i)** The x-axis and y-axis taken together determine a plane known as ______

**(ii)** The coordinates of points in the XY-plane are of the form _______

**(iii)** Coordinate planes divide the space into ______ octants.

**Solution:**

**(i)** XY-plane

**(ii)** (x, y, 0)

**(iii)** Eight

### Chapter 12 Introduction to Three Dimensional Geometry Exercise 12.2

**Question 1.**

**Find the distance between the following pairs of points:**

**(i)** (2, 3, 5) and (4, 3, 1)

**(ii)** (-3, 7, 2) and (2, 4, -1)

**(iii)** (-1, 3, -4) and (1, -3, 4)

**(iv)** (2, -1, 3) and (-2, 1, 3)

**Solution:**

**(i) The distance PQ between the points P(2, 3, 5) and Q(4, 3, 1) is**

=

**(ii) The distance PQ between the points P(-3, 7, 2) and Q(2, 4, -1) is**

**(iii) The distance PQ between the points P(-1, 3, -4) and Q(1, -3, 4) is**

**(iv) The distance PQ between the points P(2, -1, 3) and Q(-2, 1, 3) is**

**Question 2.**

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

**Solution:**

Let A(-2, 3, 5), B(1, 2, 3) and C(7, 0, -1) be three given points.

Now AC = AB + BC

Thus, points A, B and C are collinear.

**Question 3.**

**Verify the following:**

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

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

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

**Solution:**

**(i)** Let A(0, 7, -10), B(l, 6, -6) and C(4, 9, -6) be three vertices of triangle ABC. Then

Now, AB = BC

Thus, ABC is an isosceles triangle.

**(ii)** Let A(0, 7,10), B(-l, 6, 6) and C(-A, 9, 6) be three vertices of triangle ABC. Then

Now, AC^{2} = AB^{2} + BC^{2}

Thus, ABC is a right angled triangle.

**(iii)** Let A(-1, 2, 1), B(1, -2, 5) and C(4, -7, 8) and D(2, -3,4) be four vertices of quadrilateral ABCD. Then

Now AB = CD, BC = AD and AC ≠ BD

Thus A, B, C and D are vertices of a parallelogram ABCD.

**Question 4.**

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

**Solution:**

Let A(x, y, z) be any point which is equidistant from points B(1, 2, 3) and C(3, 2, -1).

**Question 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.

**Solution:**

Let P(x, y, z) be any point.

### Chapter 12 Introduction to Three Dimensional Geometry Exercise 12.3

**Question 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 enternally

**Solution:**

**(i)** Let P(x, y, z) be any point which divides the line segment joining the points A(-2, 3, 5) and B(1, -4, 6) in the ratio 2 : 3 internally.

**(ii)** Let P(x, y, z) be any point which divides the line segment joining the points 71 (-2, 3, 5) and B(1, -4, 6) in the ratio 2 : 3 externally. Then

**Question 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.

**Solution:**

Let Q(5, 4, -6) divides the line segment joining the points P(3, 2, -4) and R(9, 8, -10) in the ratio k : 1 internally.

**Question 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).

**Solution:**

Let the line segment joining the points A(-2, 4, 7) and B(3, -5, 8) be divided by the YZ -plane at a point C in the ratio k : 1.

**Question 4.**

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

**Solution:**

Let the points A(2, -3, 4), B(-l, 2,1) and C be the given points. Let the point P divides AB in the ratio k : 1. Then coordinates of P are

Let us examine whether for some value of k, the point P coincides with point C.

AB internally in the ratio 2:1. Hence A, B, C are collinear.

**Question 5.**

Find the coordinates of the points which trisect the line segment joining the points P(4, 2, -6) and Q(10, -16, 6).

**Solution:**

Let R and S be two points which trisect the line segment PQ.

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