11. INTRODUCTION TO THREE DIMENSIONAL GEOMETRY

CBSE Class 11 Maths Chapter 11 Introduction to Three Dimensional Geometry

Chapter 11 . INTRODUCTION TO THREE DIMENSIONAL GEOMETRY

Class 11 Maths Chapter 11 . Introduction to three Dimensional Geometry Exercise 11.1 Solutions :

1. A point is on the -axis. What are its -coordinate and -coordinates?

Solution : The coordinate of the point on the x-axis is ().

[ Where x is the value of its x-coordinate, and the y and z coordinates are both zero.]
2. A point is in the -plane. What can you say about its -coordinate?

Solution : The y-coordinate of a point in the XZ-plane is always 0.

[Because , the coordinate of XZ-plane is . ]

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 : The name of octants are :

(1, 2, 3) : I octant (+, +, +)

(4, -2, 3) : IV octant (+, –, +)

(4, -2, -5) : VIII octant (+,–,–)

(4, 2, -5) : V octant (+, +, –)

(-4, 2, -5) : VI octant (–, +, –)

(-4, 2, 5) : II octant (–, +, +)

(-3, -1, 6) : III octant (–,–, +)

(-2, -4, -7) : VII octant (–,–, –)

4. Fill in the blanks:
(i) The -axis and -axis taken together determine a plane known as_______.
(ii) The coordinates of points in the -plane are of the form _______.
(iii) Coordinate planes divide the space into ______ octants.

Solution : (i) The x-axis and y-axis taken together determine a plane known as the XY-plane or the horizontal plane.

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

(iii) Coordinate planes divide the space into eight octants.

Class 11 Maths Chapter 11 . Introduction to three Dimensional Geometry Exercise 11.2 Solutions :

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) Given, the points are P(2, 3, 5) and Q(4, 3, 1) .

Here, ,

Using distance formula,

units .

(ii) Given, the points are P(–3, 7, 2) and Q(2, 4, –1) .
Here, ,

Using distance formula,

units .

(iii) Given, the points are P(–1, 3, – 4) and Q(1, –3, 4) .

Here, ,

Using distance formula,

units .

(iv) Given, the points are P(2, –1, 3) and Q(–2, 1, 3) .
Here, ,

Using distance formula,

units .

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

Solution : Given the three points are P(–2, 3, 5), Q(1, 2, 3) and R(7, 0, –1) .

Using distance formula, we have

units

Therefore, the points (–2, 3, 5), (1, 2, 3) and (7, 0, –1) are collinear.

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.

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

Solution : Given, P(0, 7, –10), Q(1, 6, – 6) and R(4, 9, – 6) are the vertices of a triangle .

Using distance formula, we have

units

So, (0, 7, –10), (1, 6, – 6) and (4, 9, – 6) are the vertices of an isosceles triangle. Verified .
(ii) Given, P(0, 7, 10), Q(–1, 6, 6) and R(– 4, 9, 6) are the vertices of a right angled triangle.

units

Therefore, P(0, 7, 10), Q(–1, 6, 6) and R(– 4, 9, 6) are the vertices of a right angled triangle PQR at Q .
(iii)Given, P (–1, 2, 1), Q(1, –2, 5), R(4, –7, 8) and S(2, –3, 4) are the vertices of a parallelogram.

units

and [opposite sides are equal.]

So, P (–1, 2, 1), Q(1, –2, 5), R(4, –7, 8) and S(2, –3, 4) are the vertices of a parallelogram PQRS .
4. Find the equation of the set of points which are equidistant from the points (1, 2, 3) and (3, 2, –1).

Solution : Let P ()is equidistant from the points A(1, 2, 3) and B(3, 2, –1).

= 0

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 the point be P() .

A/Q,

Class 11 Maths Chapter 11 . Introduction to three Dimensional Geometry Miscellaneous Exercise Solutions :

1. Three vertices of a parallelogram ABCD are A(3, – 1, 2), B (1, 2, – 4) and C (– 1, 1, 2). Find the coordinates of the fourth vertex.

Solution : Let, the coordinates of the fourth vertex be D() .

We know that diagonals of a parallelogram bisect each other .

So, the coordinate of the mid-point of AC = the coordinate of the mid-point of BD .

Therefore, the coordinates of the fourth vertex be (– 1 , – 2 , 8) .

2. Find the lengths of the medians of the triangle with vertices A (0, 0, 6), B (0,4, 0) and (6, 0, 0).

Solution : Given, the vertices of the triangle are A(0, 0, 6), B(0,4, 0) and C(6, 0, 0) .

Let, P , Q and R are the mid-points of the sides BC , AC and AB of the triangle ABC respectively .

So, the coordinate of the point P is

The coordinate of the point Q is

The coordinate of the point R is

Using distance formula , we have

units

uints

3. If the origin is the centroid of the triangle PQR with vertices P (, 2, 6), Q (– 4, , –10) and R(8, 14, ), then find the values of and .

Solution : [ Note : If A() , B() and C() are the vertices of triangle ABC , then coordinates of the centroid of the triangle is given by

, , ]

Here, , , ,

We have,

Therefore , the values of , and .
4.If A and B be the points (3, 4, 5) and (–1, 3, –7), respectively, find the equation of the set of points P such that , where is a constant.

Solution : let the point be P() .

A/Q ,