RD Chapter 28- Introduction to 3D coordinate geometry Ex-28.1 |
RD Chapter 28- Introduction to 3D coordinate geometry Ex-28.2 |

**Answer
1** :

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 thelength of AD

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

The section formula is given as

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:

**Answer
2** :

Given:

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

By using the section formula,

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:

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).

**Answer
3** :

Given:

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

By using the section formula,

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:

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

**Answer
4** :

Given:

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

By using the section formula,

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:

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

**Answer
5** :

Given:

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

By using the section formula,

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:

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

5(k + 1) = 4

5k + 5 = 4

5k = 4 – 5

5k = – 1

k = -1/5

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

**Answer
6** :

Given:

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

By using the sectionformula,

Let C divides AB inratio k: 1

Three points arecollinear 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 ofC are:

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 thesame in all three cases.

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

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

**Answer
7** :

Given:

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

By using the sectionformula,

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

The coordinates of Cis given by,

P(-2, 3, 5) ismid-point of A(x_{1}, y_{1}, z_{1}) and B(x_{2},y_{2}, z_{2})

Then,

Q(4, -1, 7) ismid-point of B(x_{2}, y_{2}, z_{2}) and C(x_{3},y_{3}, z_{3})

Then,

R(6, 5, 3) ismid-point of A(x_{1}, y_{1}, z_{1}) and C(x_{3},y_{3}, z_{3})

Then,

Now solving for ‘x’terms

x_{1} + x_{2} =-4……………………(4)

x_{2} + x_{3} =8………………………(5)

x_{1} + x_{3} =12……………………(6)

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

x_{1} + x_{2} +x_{2} + x_{3} + x_{1} + x_{3} =8 + 12 – 4

2x_{1} +2x_{2} + 2x_{3} = 16

2(x_{1} +x_{2} + x_{3}) = 16

x_{1} + x_{2} +x_{3} = 8………………………(7)

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

x_{1} + x_{2} +x_{3} – x_{1} – x_{2} = 8 – (-4)

x_{3} =12

x_{1} + x_{2} +x_{3} – x_{2} – x_{3} = 8 – 8

x_{1} = 0

x_{1} + x_{2} +x_{3} – x_{1} – x_{3} = 8 – 12

x_{2} =-4

Now solving for ‘y’terms

y_{1} + y_{2} =6……………………(8)

y_{2} + y_{3} =-2……………………(9)

y_{1} + y_{3} =10……………………(10)

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

y_{1} + y_{2} +y_{2} + y_{3} + y_{1} + y_{3} =10 + 6 – 2

2y_{1} +2y_{2} + 2y_{3} = 14

2(y_{1} +y_{2} + y_{3}) = 14

y_{1} + y_{2} +y_{3} = 7………………………(11)

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

y_{1} + y_{2} +y_{3} – y_{1} – y_{2} = 7 – 6

y_{3} = 1

y_{1} + y_{2} +y_{3} – y_{2} – y_{3} = 7 – (-2)

y_{1} = 9

y_{1} + y_{2} +y_{3} – y_{1} – y_{3} = 7 – 10

y_{2} =-3

Now solving for ‘z’terms

z_{1} + z_{2} =10……………………(12)

z_{2} + z_{3} =14……………………(13)

z_{1} + z_{3} =6……………………(14)

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

z_{1} + z_{2} +z_{2} + z_{3} + z_{1} + z_{3} =6 + 14 + 10

2z_{1} +2z_{2} + 2z_{3} = 30

2(z_{1} +z_{2} + z_{3}) = 30

z_{1} + z_{2} +z_{3} = 15………………………(15)

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

z_{1} + z_{2} +z_{3} – z_{1} – z_{2} = 15 – 10

z_{3} = 5

z_{1} + z_{2} +z_{3} – z_{2} – z_{3} = 15 – 14

z_{1} = 1

z_{1} + z_{2} +z_{3} – z_{1} – z_{3} = 15 – 6

z_{2} = 9

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

**Answer
8** :

Given:

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

By using the distance formula,

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 byusing section formula is given as

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

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