# Chapter – 12: Introduction to Three-Dimensional Geometry

# EXERCISE 12.1

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

**Solution: –**

If a point is on the x-axis, then the coordinates of y and z are 0.

So the point is (x, 0, 0).

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

**Solution: –**

If a point is in XZ plane, then its y-co-ordinate is 0.

**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: –**

Here is the table which represents the octants:

Octants | I | II | III | IV | V | VI | VII | VIII |

x | + | – | – | + | + | – | – | + |

y | + | + | – | – | + | + | – | – |

z | + | + | + | + | – | – | – | – |

(i) (1, 2, 3)

Here x is positive, y is positive and z is positive.

So it lies in I octant.

(ii) (4, -2, 3)

Here x is positive, y is negative and z is positive.

So it lies in IV octant.

(iii) (4, -2, -5)

Here x is positive, y is negative and z is negative.

So it lies in VIII octant.

(iv) (4, 2, -5)

Here x is positive, y is positive and z is negative.

So it lies in V octant.

(v) (-4, 2, -5)

Here x is negative, y is positive and z is negative.

So it lies in VI octant.

(vi) (-4, 2, 5)

Here x is negative, y is positive and z is positive.

So it lies in II octant.

(vii) (-3, -1, 6)

Here x is negative, y is negative and z is positive.

So it lies in III octant.

(viii) (2, -4, -7)

Here x is positive, y is negative and z is negative.

So it lies in VIII octant.

**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) The x-axis and y-axis taken together determine a plane known as XY Plane.

(ii) The coordinates of points in the XY-plane are of the form (x, y, 0).

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

# EXERCISE 12.2

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

Let P be (2, 3, 5) and Q be (4, 3, 1)

By using the formula,

Distance PQ = **√**[(x_{2} – x_{1})^{2} + (y_{2} – y_{1})^{2} + (z_{2} – z_{1})^{2}]

So here,

x_{1} = 2, y_{1} = 3, z_{1} = 5

x_{2} = 4, y_{2} = 3, z_{2} = 1

Distance PQ = **√**[(4 – 2)^{2} + (3 – 3)^{2} + (1 – 5)^{2}]

= **√**[(2)^{2} + 0^{2} + (-4)^{2}]

= **√**[4 + 0 + 16]

**= √**20

= 2**√**5

∴ The required distance is 2**√**5 units.

**(ii) **(–3, 7, 2) and (2, 4, –1)

Let P be (– 3, 7, 2) and Q be (2, 4, – 1)

By using the formula,

Distance PQ = **√**[(x_{2} – x_{1})^{2} + (y_{2} – y_{1})^{2} + (z_{2} – z_{1})^{2}]

So here,

x_{1} = – 3, y_{1} = 7, z_{1} = 2

x_{2} = 2, y_{2} = 4, z_{2} = – 1

Distance PQ = **√**[(2 – (-3))^{2} + (4 – 7)^{2} + (-1 – 2)^{2}]

= **√**[(5)^{2} + (-3)^{2} + (-3)^{2}]

= **√**[25 + 9 + 9]

**= √**43

∴ The required distance is **√**43 units.

**(iii) **(–1, 3, – 4) and (1, –3, 4)

Let P be (– 1, 3, – 4) and Q be (1, – 3, 4)

By using the formula,

Distance PQ = **√**[(x_{2} – x_{1})^{2} + (y_{2} – y_{1})^{2} + (z_{2} – z_{1})^{2}]

So here,

x_{1} = – 1, y_{1} = 3, z_{1} = – 4

x_{2} = 1, y_{2} = – 3, z_{2} = 4

Distance PQ = **√**[(1 – (-1))^{2} + (-3 – 3)^{2} + (4 – (-4))^{2}]

= **√**[(2)^{2} + (-6)^{2} + (8)^{2}]

= **√**[4 + 36 + 64]

**= √**104

= 2**√**26

∴ The required distance is 2**√**26 units.

**(iv) **(2, –1, 3) and (–2, 1, 3)

Let P be (2, – 1, 3) and Q be (– 2, 1, 3)

By using the formula,

Distance PQ = **√**[(x_{2} – x_{1})^{2} + (y_{2} – y_{1})^{2} + (z_{2} – z_{1})^{2}]

So here,

x_{1} = 2, y_{1} = – 1, z_{1} = 3

x_{2} = – 2, y_{2} = 1, z_{2} = 3

Distance PQ = **√**[(-2 – 2)^{2} + (1 – (-1))^{2} + (3 – 3)^{2}]

= **√**[(-4)^{2} + (2)^{2} + (0)^{2}]

= **√**[16 + 4 + 0]

**= √**20

= 2**√**5

∴ The required distance is 2**√**5 units.

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

**Solution: –**

If three points are collinear, then they lie on a line.

Firstly let us calculate distance between the 3 points

i.e. PQ, QR and PR

Calculating PQ

P ≡ (– 2, 3, 5) and Q ≡ (1, 2, 3)

By using the formula,

Distance PQ = **√**[(x_{2} – x_{1})^{2} + (y_{2} – y_{1})^{2} + (z_{2} – z_{1})^{2}]

So here,

x_{1} = – 2, y_{1} = 3, z_{1} = 5

x_{2} = 1, y_{2} = 2, z_{2} = 3

Distance PQ = **√**[(1 – (-2))^{2} + (2 – 3)^{2} + (3 – 5)^{2}]

= **√**[(3)^{2} + (-1)^{2} + (-2)^{2}]

= **√**[9 + 1 + 4]

**= √**14

Calculating QR

Q ≡ (1, 2, 3) and R ≡ (7, 0, – 1)

By using the formula,

Distance QR = **√**[(x_{2} – x_{1})^{2} + (y_{2} – y_{1})^{2} + (z_{2} – z_{1})^{2}]

So here,

x_{1} = 1, y_{1} = 2, z_{1} = 3

x_{2} = 7, y_{2} = 0, z_{2} = – 1

Distance QR = **√**[(7 – 1)^{2} + (0 – 2)^{2} + (-1 – 3)^{2}]

= **√**[(6)^{2} + (-2)^{2} + (-4)^{2}]

= **√**[36 + 4 + 16]

**= √**56

= 2**√**14

Calculating PR

P ≡ (– 2, 3, 5) and R ≡ (7, 0, – 1)

By using the formula,

Distance PR = **√**[(x_{2} – x_{1})^{2} + (y_{2} – y_{1})^{2} + (z_{2} – z_{1})^{2}]

So here,

x_{1} = – 2, y_{1} = 3, z_{1} = 5

x_{2} = 7, y_{2} = 0, z_{2} = – 1

Distance PR = **√**[(7 – (-2))^{2} + (0 – 3)^{2} + (-1 – 5)^{2}]

= **√**[(9)^{2} + (-3)^{2} + (-6)^{2}]

= **√**[81 + 9 + 36]

**= √**126

= 3**√**14

Thus, PQ = **√**14, QR = 2**√**14 and PR = 3**√**14

So, PQ + QR = **√**14 + 2**√**14

= 3**√**14

= PR

∴ The points P, Q and R 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.**

**Solution: –**

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

Let us consider the points be

P(0, 7, –10), Q(1, 6, – 6) and R(4, 9, – 6)

If any 2 sides are equal, hence it will be an isosceles triangle

So firstly let us calculate the distance of PQ, QR

Calculating PQ

P ≡ (0, 7, – 10) and Q ≡ (1, 6, – 6)

By using the formula,

Distance PQ = **√**[(x_{2} – x_{1})^{2} + (y_{2} – y_{1})^{2} + (z_{2} – z_{1})^{2}]

So here,

x_{1} = 0, y_{1} = 7, z_{1} = – 10

x_{2} = 1, y_{2} = 6, z_{2} = – 6

Distance PQ = **√**[(1 – 0)^{2} + (6 – 7)^{2} + (-6 – (-10))^{2}]

= **√**[(1)^{2} + (-1)^{2} + (4)^{2}]

= **√**[1 + 1 + 16]

**= √**18

Calculating QR

Q ≡ (1, 6, – 6) and R ≡ (4, 9, – 6)

By using the formula,

Distance QR = **√**[(x_{2} – x_{1})^{2} + (y_{2} – y_{1})^{2} + (z_{2} – z_{1})^{2}]

So here,

x_{1} = 1, y_{1} = 6, z_{1} = – 6

x_{2} = 4, y_{2} = 9, z_{2} = – 6

Distance QR = **√**[(4 – 1)^{2} + (9 – 6)^{2} + (-6 – (-6))^{2}]

= **√**[(3)^{2} + (3)^{2} + (-6+6)^{2}]

= **√**[9 + 9 + 0]

**= √**18

Hence, PQ = QR

18 = 18

2 sides are equal

∴ PQR is an isosceles triangle.

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

Let the points be

P(0, 7, 10), Q(– 1, 6, 6) & R(– 4, 9, 6)

Firstly let us calculate the distance of PQ, OR and PR

Calculating PQ

P ≡ (0, 7, 10) and Q ≡ (– 1, 6, 6)

By using the formula,

Distance PQ = **√**[(x_{2} – x_{1})^{2} + (y_{2} – y_{1})^{2} + (z_{2} – z_{1})^{2}]

So here,

x_{1} = 0, y_{1} = 7, z_{1} = 10

x_{2} = – 1, y_{2} = 6, z_{2} = 6

Distance PQ = **√**[(-1 – 0)^{2} + (6 – 7)^{2} + (6 – 10)^{2}]

= **√**[(-1)^{2} + (-1)^{2} + (-4)^{2}]

= **√**[1 + 1 + 16]

**= √**18

Calculating QR

Q ≡ (1, 6, – 6) and R ≡ (4, 9, – 6)

By using the formula,

Distance QR = **√**[(x_{2} – x_{1})^{2} + (y_{2} – y_{1})^{2} + (z_{2} – z_{1})^{2}]

So here,

x_{1} = 1, y_{1} = 6, z_{1} = – 6

x_{2} = 4, y_{2} = 9, z_{2} = – 6

Distance QR = **√**[(4 – 1)^{2} + (9 – 6)^{2} + (-6 – (-6))^{2}]

= **√**[(3)^{2} + (3)^{2} + (-6+6)^{2}]

= **√**[9 + 9 + 0]

**= √**18

Calculating PR

P ≡ (0, 7, 10) and R ≡ (– 4, 9, 6)

By using the formula,

Distance PR = **√**[(x_{2} – x_{1})^{2} + (y_{2} – y_{1})^{2} + (z_{2} – z_{1})^{2}]

So here,

x_{1} = 0, y_{1} = 7, z_{1} = 10

x_{2} = – 4, y_{2} = 9, z_{2} = 6

Distance PR = **√**[(-4 – 0)^{2} + (9 – 7)^{2} + (6 – 10)^{2}]

= **√**[(-4)^{2} + (2)^{2} + (-4)^{2}]

= **√**[16 + 4 + 16]

**= √**36

Now,

PQ^{2} + QR^{2} = 18 + 18

= 36

= PR^{2}

By using converse of Pythagoras theorem,

∴ The given vertices P, Q & R are the vertices of a right – angled triangle at Q.

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

Let the points be: A(–1, 2, 1), B(1, –2, 5), C(4, –7, 8) & D(2, –3, 4)

ABCD can be vertices of parallelogram only if opposite sides are equal.

i.e. AB = CD and BC = AD

Firstly let us calculate the distance

Calculating AB

A ≡ (– 1, 2, 1) and B ≡ (1, – 2, 5)

By using the formula,

Distance AB = **√**[(x_{2} – x_{1})^{2} + (y_{2} – y_{1})^{2} + (z_{2} – z_{1})^{2}]

So here,

x_{1} = – 1, y_{1} = 2, z_{1} = 1

x_{2} = 1, y_{2} = – 2, z_{2} = 5

Distance AB = **√**[(1 – (-1))^{2} + (-2 – 2)^{2} + (5 – 1)^{2}]

= **√**[(2)^{2} + (-4)^{2} + (4)^{2}]

= **√**[4 + 16 + 16]

**= √**36

= 6

Calculating BC

B ≡ (1, – 2, 5) and C ≡ (4, – 7, 8)

By using the formula,

Distance BC = **√**[(x_{2} – x_{1})^{2} + (y_{2} – y_{1})^{2} + (z_{2} – z_{1})^{2}]

So here,

x_{1} = 1, y_{1} = – 2, z_{1} = 5

x_{2} = 4, y_{2} = – 7, z_{2} = 8

Distance BC = **√**[(4 – 1)^{2} + (-7 – (-2))^{2} + (8 – 5)^{2}]

= **√**[(3)^{2} + (-5)^{2} + (3)^{2}]

= **√**[9 + 25 + 9]

**= √**43

Calculating CD

C ≡ (4, – 7, 8) and D ≡ (2, – 3, 4)

By using the formula,

Distance CD = **√**[(x_{2} – x_{1})^{2} + (y_{2} – y_{1})^{2} + (z_{2} – z_{1})^{2}]

So here,

x_{1} = 4, y_{1} = – 7, z_{1} = 8

x_{2} = 2, y_{2} = – 3, z_{2} = 4

Distance CD = **√**[(2 – 4)^{2} + (-3 – (-7))^{2} + (4 – 8)^{2}]

= **√**[(-2)^{2} + (4)^{2} + (-4)^{2}]

= **√**[4 + 16 + 16]

**= √**36

= 6

Calculating DA

D ≡ (2, – 3, 4) and A ≡ (– 1, 2, 1)

By using the formula,

Distance DA = **√**[(x_{2} – x_{1})^{2} + (y_{2} – y_{1})^{2} + (z_{2} – z_{1})^{2}]

So here,

x_{1} = 2, y_{1} = – 3, z_{1} = 4

x_{2} = – 1, y_{2} = 2, z_{2} = 1

Distance DA = **√**[(-1 – 2)^{2} + (2 – (-3))^{2} + (1 – 4)^{2}]

= **√**[(-3)^{2} + (5)^{2} + (-3)^{2}]

= **√**[9 + 25 + 9]

**= √**43

Since AB = CD and BC = DA (given)

So, In ABCD both pairs of opposite sides are equal.

∴ ABCD is a parallelogram.

**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 (1, 2, 3) & B (3, 2, – 1)

Let point P be (x, y, z)

Since it is given that point P(x, y, z) is equal distance from point A(1, 2, 3) & B(3, 2, – 1)

i.e. PA = PB

Firstly let us calculate

Calculating PA

P ≡ (x, y, z) and A ≡ (1, 2, 3)

By using the formula,

Distance PA = **√**[(x_{2} – x_{1})^{2} + (y_{2} – y_{1})^{2} + (z_{2} – z_{1})^{2}]

So here,

x_{1} = x, y_{1} = y, z_{1} = z

x_{2} = 1, y_{2} = 2, z_{2} = 3

Distance PA = **√**[(1 – x)^{2} + (2 – y)^{2} + (3 – z)^{2}]

Calculating PB

P ≡ (x, y, z) and B ≡ (3, 2, – 1)

By using the formula,

Distance PB = **√**[(x_{2} – x_{1})^{2} + (y_{2} – y_{1})^{2} + (z_{2} – z_{1})^{2}]

So here,

x_{1} = x, y_{1} = y, z_{1} = z

x_{2} = 3, y_{2} = 2, z_{2} = – 1

Distance PB = **√**[(3 – x)^{2} + (2 – y)^{2} + (-1 – z)^{2}]

Since PA = PB

Square on both the sides, we get

PA^{2} = PB^{2}

(1 – x)^{2} + (2 – y)^{2} + (3 – z)^{2} = (3 – x)^{2} + (2 – y)^{2} + (– 1 – z)^{2}

(1 + x^{2} – 2x) + (4 + y^{2} – 4y) + (9 + z^{2} – 6z)

(9 + x^{2} – 6x) + (4 + y^{2} – 4y) + (1 + z^{2} + 2z)

– 2x – 4y – 6z + 14 = – 6x – 4y + 2z + 14

4x – 8z = 0

x – 2z = 0

∴ The required equation is x – 2z = 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 A (4, 0, 0) & B (– 4, 0, 0)

Let the coordinates of point P be (x, y, z)

Calculating PA

P ≡ (x, y, z) and A ≡ (4, 0, 0)

By using the formula,

Distance PA = **√**[(x_{2} – x_{1})^{2} + (y_{2} – y_{1})^{2} + (z_{2} – z_{1})^{2}]

So here,

x_{1} = x, y_{1} = y, z_{1} = z

x_{2} = 4, y_{2} = 0, z_{2} = 0

Distance PA = **√**[(4– x)^{2} + (0 – y)^{2} + (0 – z)^{2}]

Calculating PB

P ≡ (x, y, z) and B ≡ (– 4, 0, 0)

By using the formula,

Distance PB = **√**[(x_{2} – x_{1})^{2} + (y_{2} – y_{1})^{2} + (z_{2} – z_{1})^{2}]

So here,

x_{1} = x, y_{1} = y, z_{1} = z

x_{2} = – 4, y_{2} = 0, z_{2} = 0

Distance PB = **√**[(-4– x)^{2} + (0 – y)^{2} + (0 – z)^{2}]

Now it is given that:

PA + PB = 10

PA = 10 – PB

Square on both the sides, we get

PA^{2} = (10 – PB)^{2}

PA^{2} = 100 + PB^{2} – 20 PB

(4 – x)^{2} + (0 – y)^{2} + (0 – z)^{2}

100 + (– 4 – x)^{2} + (0 – y)^{2} + (0 – z)^{2} – 20 PB

(16 + x^{2} – 8x) + (y^{2}) + (z^{2})

100 + (16 + x^{2} + 8x) + (y^{2}) + (z^{2}) – 20 PB

20 PB = 16x + 100

5 PB = (4x + 25)

Square on both the sides again, we get

25 PB^{2} = 16x^{2} + 200x + 625

25 [(– 4 – x)^{2} + (0 – y)^{2} + (0 – z)^{2}] = 16x^{2} + 200x + 625

25 [x^{2} + y^{2} + z^{2} + 8x + 16] = 16x^{2} + 200x + 625

25x^{2} + 25y^{2} + 25z^{2} + 200x + 400 = 16x^{2} + 200x + 625

9x^{2} + 25y^{2} + 25z^{2} – 225 = 0

∴ The required equation is 9x^{2} + 25y^{2} + 25z^{2} – 225 = 0

# EXERCISE 12.3

**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 externally.**

**Solution: –**

Let the line segment joining the points P (-2, 3, 5) and Q (1, -4, 6) be PQ.

**(i)** 2: 3 internally

By using section formula,

We know that the coordinates of the point R which divides the line segment joining two points P (x_{1}, y_{1}, z_{1}) and Q (x_{2}, y_{2}, z_{2}) internally in the ratio m: n is given by:

Upon comparing we have

x_{1} = -2, y_{1} = 3, z_{1} = 5;

x_{2} = 1, y_{2} = -4, z_{2} = 6 and

m = 2, n = 3

So, the coordinates of the point which divides the line segment joining the points P (– 2, 3, 5) and Q (1, – 4, 6) in the ratio 2 : 3 internally is given by:

Hence, the coordinates of the point which divides the line segment joining the points (-2, 3, 5) and (1, -4, 6) is (-4/5, 1/5, 27/5)

**(ii)** 2: 3 externally

By using section formula,

We know that the coordinates of the point R which divides the line segment joining two points P (x_{1}, y_{1}, z_{1}) and Q (x_{2}, y_{2}, z_{2}) externally in the ratio m: n is given by:

Upon comparing we have

x_{1} = -2, y_{1} = 3, z_{1} = 5;

x_{2} = 1, y_{2} = -4, z_{2} = 6 and

m = 2, n = 3

So, the coordinates of the point which divides the line segment joining the points P (– 2, 3, 5) and Q (1, – 4, 6) in the ratio 2: 3 externally is given by:

∴ The co-ordinates of the point which divides the line segment joining the points (-2, 3, 5) and (1, -4, 6) is (-8, 17, 3).

**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 us consider Q divides PR in the ratio k: 1.

By using section formula,

We know that the coordinates of the point R which divides the line segment joining two points P (x_{1}, y_{1}, z_{1}) and Q (x_{2}, y_{2}, z_{2}) internally in the ratio m : n is given by:

Upon comparing we have,

x_{1} = 3, y_{1} = 2, z_{1} = -4;

x_{2} = 9, y_{2} = 8, z_{2} = -10 and

m = k, n = 1

So, we have

9k + 3 = 5 (k+1)

9k + 3 = 5k + 5

9k – 5k = 5 – 3

4k = 2

k = 2/4

= ½

Hence, the ratio in which Q divides PR is 1: 2.

**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 formed by joining the points P (-2, 4, 7) and Q (3, -5, 8) be PQ.

We know that any point on the YZ-plane is of the form (0, y, z).

So now, let R (0, y, z) divides the line segment PQ in the ratio k: 1.

Then,

Upon comparing we have,

x_{1} = -2, y_{1} = 4, z_{1} = 7;

x_{2} = 3, y_{2} = -5, z_{2} = 8 and

m = k, n = 1

By using the section formula,

We know that the coordinates of the point R which divides the line segment joining two points P (x_{1}, y_{1}, z_{1}) and Q (x_{2}, y_{2}, z_{2}) internally in the ratio m: n is given by:

So we have,

3k – 2 = 0

3k = 2

k = 2/3

Hence, the ratio in which the YZ-plane divides the line segment formed by joining the points (–2, 4, 7) and (3, –5, 8) is 2:3.

**4. Using section formula, show that the points A (2, –3, 4), B (–1, 2, 1) and C (0, 1/3, 2) are collinear.**

**Solution: –**

Let the point P divides AB in the ratio k: 1.

Upon comparing we have,

x_{1} = 2, y_{1} = -3, z_{1} = 4;

x_{2} = -1, y_{2} = 2, z_{2} = 1 and

m = k, n = 1

By using section formula,

We know that the coordinates of the point R which divides the line segment joining two points P (x_{1}, y_{1}, z_{1}) and Q (x_{2}, y_{2}, z_{2}) internally in the ratio m : n is given by:

So we have,

Now, we check if for some value of k, the point coincides with the point C.

Put (-k+2)/(k+1) = 0

-k + 2 = 0

k = 2

When k = 2, then (2k-3)/(k+1) = (2(2)-3)/(2+1)

= (4-3)/3

= 1/3

And, (k+4)/(k+1) = (2+4)/(2+1)

= 6/3

= 2

∴ C (0, 1/3, 2) is a point which divides AB in the ratio 2: 1 and is same as P.

Hence, A, B, C are collinear.

**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 A (x_{1}, y_{1}, z_{1}) and B (x_{2}, y_{2}, z_{2}) trisect the line segment joining the points P (4, 2, -6) and Q (10, -16, 6).

A divides the line segment PQ in the ratio 1: 2.

Upon comparing we have,

x_{1} = 4, y_{1} = 2, z_{1} = -6;

x_{2} = 10, y_{2} = -16, z_{2} = 6 and

m = 1, n = 2

By using the section formula,

We know that the coordinates of the point R which divides the line segment joining two points P (x_{1}, y_{1}, z_{1}) and Q (x_{2}, y_{2}, z_{2}) internally in the ratio m : n is given by:

So we have,

Similarly, we know that B divides the line segment PQ in the ratio 2: 1.

Upon comparing we have,

x_{1} = 4, y_{1} = 2, z_{1} = -6;

x_{2} = 10, y_{2} = -16, z_{2} = 6 and

m = 2, n = 1

By using the section formula,

We know that the coordinates of the point R which divides the line segment joining two points P (x_{1}, y_{1}, z_{1}) and Q (x_{2}, y_{2}, z_{2}) internally in the ratio m: n is given by:

So we have,

∴ The coordinates of the points which trisect the line segment joining the points P (4, 2, – 6) and Q (10, –16, 6) are (6, -4, -2) and (8, -10, 2).

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