RD Sharma Solutions for Class 11 Chapter 28 - Introduction to Three Dimensional Coordinate Geometry Exercise 28.3

In Exercise 28.3 of Chapter 28, we shall discuss some problems related to section formulae. From the exam point of view, solutions are formulated by our subject expert faculty team that contain explanations in a simple language, which is very helpful for students to understand the problems. Students can improve their problem-solving ability by practising the problems provided in the RD Sharma Class 11 Maths Solutions which is very beneficial during their exam preparation. Students can download the pdf from the available links given below.

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Access answers to RD Sharma Solutions for Class 11 Maths Exercise 28.3 Chapter 28 – Introduction to Three Dimensional Coordinate Geometry

ExERCISE 28.3 PAGE NO: 28.19

1. The vertices of the triangle are A(5, 4, 6), B(1, -1, 3) and C(4, 3, 2). The internal bisector of angle A meets BC at D. Find the coordinates of D and the length AD.

Solution:

Given:

The vertices of the triangle are A (5, 4, 6), B (1, -1, 3) and C (4, 3, 2).

By using the formulas let us find the coordinates of D and the length of AD

The distance between any two points (a, b, c) and (m, n, o) is given by,

RD Sharma Solutions for Class 11 Maths Chapter 28 – image 41

The section formula is given as

RD Sharma Solutions for Class 11 Maths Chapter 28 – image 42

RD Sharma Solutions for Class 11 Maths Chapter 28 – image 43

AB : AC = 5:3

BD: DC = 5:3

So, m = 5 and n = 3


B(1, -1, 3) and C(4, 3, 2)

Coordinates of D using section formula:

RD Sharma Solutions for Class 11 Maths Chapter 28 – image 44

RD Sharma Solutions for Class 11 Maths Chapter 28 – image 45

RD Sharma Solutions for Class 11 Maths Chapter 28 – image 46

2. A point C with z-coordinate 8 lies on the line segment joining the points A(2, -3, 4) and B(8, 0, 10). Find the coordinates.

Solution:

Given:

The points A (2, -3, 4) and B (8, 0, 10)

By using the section formula,

RD Sharma Solutions for Class 11 Maths Chapter 28 – image 47

Let Point C(x, y, 8), and C divides AB in ratio k: 1

So, m = k and n = 1

A(2, -3, 4) and B(8, 0, 10)

Coordinates of C are:

RD Sharma Solutions for Class 11 Maths Chapter 28 – image 48

On comparing we get,

[10k + 4] / [k + 1] = 8

10k + 4 = 8(k + 1)

10k + 4 = 8k + 8

10k – 8k = 8 – 4

2k = 4

k = 4/2

= 2

Here C divides AB in ratio 2:1

x = [8k + 2] / [k + 1]

= [8(2) + 2] / [2 + 1]

= [16 + 2] / [3]

= 18/3

= 6

y = -3 / [k + 1]

= -3 / [2 + 1]

= -3 / 3

= -1

∴ The Coordinates of C are (6, -1, 8).

3. Show that the three points A(2, 3, 4), B(-1, 2, -3) and C(-4, 1, -10) are collinear and find the ratio in which C divides AB.

Solution:

Given:

The points A (2, 3, 4), B (-1, 2, -3) and C (-4, 1, -10)

By using the section formula,

RD Sharma Solutions for Class 11 Maths Chapter 28 – image 49

Let C divides AB in ratio k: 1

Three points are collinear if the value of k is the same for x, y and z coordinates.

So, m = k and n = 1

A(2, 3, 4), B(-1, 2, -3) and C(-4, 1, -10)

Coordinates of C are:

RD Sharma Solutions for Class 11 Maths Chapter 28 – image 50

On comparing we get,

[-k + 2] / [k + 1] = -4

-k + 2 = -4(k + 1)

-k + 2 = -4k – 4

4k – k = – 2 – 4

3k = -6

k = -6/3

= -2

[2k + 3] / [k + 1] = 1

2k + 3 = k + 1

2k – k = 1 – 3

k = – 2

[-3k + 4] / [k + 1] = -10

-3k + 4 = -10(k + 1)

-3k + 4 = -10k – 10

-3k + 10k = -10 – 4

7k = -14

k = -14/7

= -2

The value of k is the same in all three cases.

So, A, B and C are collinear [as k = -2]

∴We can say that, C divides AB externally in ratio 2:1

4. Find the ratio in which the line joining (2, 4, 5) and (3, 5, 4) is divided by the yz-plane.

Solution:

Given:

The points (2, 4, 5) and (3, 5, 4)

By using the section formula,

RD Sharma Solutions for Class 11 Maths Chapter 28 – image 51

We know X coordinate is always 0 on yz-plane

So, let Point C(0, y, z), and let C divide AB in ratio k: 1

Then, m = k and n = 1

A(2, 4, 5) and B(3, 5, 4)

The coordinates of C are:

RD Sharma Solutions for Class 11 Maths Chapter 28 – image 52

On comparing we get,

[3k + 2] / [k + 1] = 0

3k + 2 = 0(k + 1)

3k + 2 = 0

3k = – 2

k = -2/3

∴We can say that, C divides AB externally in ratio 2: 3

5. Find the ratio in which the line segment joining the points (2, -1, 3) and (-1, 2, 1) is divided by the plane x + y + z = 5.

Solution:

Given:

The points(2, -1, 3) and (-1, 2, 1)

By using the section formula,

RD Sharma Solutions for Class 11 Maths Chapter 28 – image 53

Let C(x, y, z) be any point on the given plane and C divides AB in ratio k: 1

Then, m = k and n = 1

A(2, -1, 3) and B(-1, 2, 1)


Coordinates of C are:

RD Sharma Solutions for Class 11 Maths Chapter 28 – image 54

On comparing we get,

[-k + 2] / [k + 1] = x

[2k – 1] / [k + 1] = y

[-k + 3] / [k + 1] = z

We know that x + y + z = 5

RD Sharma Solutions for Class 11 Maths Chapter 28 – image 55

5(k + 1) = 4

5k + 5 = 4

5k = 4 – 5

5k = – 1

k = -1/5

∴We can say that, the plane divides AB externally in the ratio 1:5

6. If the points A(3, 2, -4), B(9, 8, -10) and C(5, 4, -6) are collinear, find the ratio in which C divided AB.

Solution:

Given:

The points A (3, 2, -4), B (9, 8, -10) and C (5, 4, -6)

By using the section formula,

RD Sharma Solutions for Class 11 Maths Chapter 28 – image 56

Let C divides AB in ratio k: 1

Three points are collinear if the value of k is the same for x, y and z coordinates.

Then, m = k and n = 1

A(3, 2, -4), B(9, 8, -10) and C(5, 4, -6)

Coordinates of C are:

RD Sharma Solutions for Class 11 Maths Chapter 28 – image 57

On comparing we get,

[9k + 3] / [k + 1] = 5

9k + 3 = 5(k + 1)

9k + 3 = 5k + 5

9k – 5k = 5 – 3

4k = 2

k = 2/4

= ½

[8k + 2] / [k + 1] = 4

8k + 2 = 4(k + 1)

8k + 2 = 4k + 4

8k – 4k = 4 – 2

4k = 2

k = 2/4

= ½

[-10k – 4] / [k + 1] = -6

-10k – 4 = -6(k + 1)

-10k – 4 = -6k – 6

-10k + 6k = 4 – 6

-4k = -2

k = -2/-4

= ½

The value of k is the same in all three cases.

So, A, B and C are collinear [as, k = ½]

∴We can say that, C divides AB externally in ratio 1:2

7. The mid-points of the sides of a triangle ABC are given by (-2, 3, 5), (4, -1, 7) and (6, 5, 3). Find the coordinates of A, B and C.

Solution:

Given:

The mid-points of the sides of a triangle ABC is given as (-2, 3, 5), (4, -1, 7) and (6, 5, 3).

By using the section formula,

RD Sharma Solutions for Class 11 Maths Chapter 28 – image 58

We know the mid-point divides side in the ratio of 1:1.

The coordinates of C is given by,

RD Sharma Solutions for Class 11 Maths Chapter 28 – image 59

P(-2, 3, 5) is mid-point of A(x1, y1, z1) and B(x2, y2, z2)

Then,

RD Sharma Solutions for Class 11 Maths Chapter 28 – image 60

Q(4, -1, 7) is mid-point of B(x2, y2, z2) and C(x3, y3, z3)

Then,

RD Sharma Solutions for Class 11 Maths Chapter 28 – image 61

R(6, 5, 3) is mid-point of A(x1, y1, z1) and C(x3, y3, z3)

Then,

RD Sharma Solutions for Class 11 Maths Chapter 28 – image 62

Now solving for ‘x’ terms

x1 + x2 = -4……………………(4)

x2 + x3 = 8………………………(5)

x1 + x3 = 12……………………(6)

By adding equation (4), (5), (6)

x1 + x2 + x2 + x3 + x1 + x3 = 8 + 12 – 4

2x1 + 2x2 + 2x3 = 16

2(x1 + x2 + x3) = 16

x1 + x2 + x3 = 8………………………(7)

Now, subtract equation (4), (5) and (6) from equation (7) separately:

x1 + x2 + x3 – x1 – x2 = 8 – (-4)

x3 = 12


x1 + x2 + x3 – x2 – x3 = 8 – 8

x1 = 0


x1 + x2 + x3 – x1 – x3 = 8 – 12

x2 = -4

Now solving for ‘y’ terms

y1 + y2 = 6……………………(8)

y2 + y3 = -2……………………(9)

y1 + y3 = 10……………………(10)


By adding equation (8), (9) and (10) we get,

y1 + y2 + y2 + y3 + y1 + y3 = 10 + 6 – 2

2y1 + 2y2 + 2y3 = 14

2(y1 + y2 + y3) = 14

y1 + y2 + y3 = 7………………………(11)


Now, subtract equation (8), (9) and (10) from equation (11) separately:

y1 + y2 + y3 – y1 – y2 = 7 – 6

y3 = 1


y1 + y2 + y3 – y2 – y3 = 7 – (-2)

y1 = 9


y1 + y2 + y3 – y1 – y3 = 7 – 10

y2 = -3

Now solving for ‘z’ terms

z1 + z2 = 10……………………(12)

z2 + z3 = 14……………………(13)

z1 + z3 = 6……………………(14)


By adding equation (12), (13) and (14) we get,

z1 + z2 + z2 + z3 + z1 + z3 = 6 + 14 + 10

2z1 + 2z2 + 2z3 = 30

2(z1 + z2 + z3) = 30

z1 + z2 + z3 = 15………………………(15)


Now, subtract equation (8), (9) and (10) from equation (11) separately:

z1 + z2 + z3 – z1 – z2 = 15 – 10

z3 = 5


z1 + z2 + z3 – z2 – z3 = 15 – 14

z1 = 1


z1 + z2 + z3 – z1 – z3 = 15 – 6

z2 = 9

∴Thevertices of sides of a triangle ABC are A(0, 9, 1) B(-4,-3, 9) and C(12, 1, 5).

8. A(1, 2, 3), B(0, 4, 1), C(-1, -1, -3) are the vertices of a triangle ABC. Find the point in which the bisector of the angle ∠BAC meets BC.

Solution:

Given:

The vertices of a triangle are A (1, 2, 3), B (0, 4, 1), C (-1, -1, -3)

By using the distance formula,

RD Sharma Solutions for Class 11 Maths Chapter 28 – image 63

RD Sharma Solutions for Class 11 Maths Chapter 28 – image 64

So, AB/AC = 3/7

AB: AC = 3:7

BD: DC = 3:7

Then, m = 3 and n = 7

B(0, 4, 1) and C(-1, -1, -3)


Coordinates of D by using section formula is given as

RD Sharma Solutions for Class 11 Maths Chapter 28 – image 65

∴The coordinates of D are (-3/10, 5/2, -1/5).

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