Introduction To Three Dimensional Geometry

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

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

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

(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.

(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.

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

Distance PQ

Here,

x₁ = 2, y₁ = 3, z₁ = 5

x₂ = 4, y₂ = 3, z₂ = 1

Distance PQ

Thus, the required distance is units.

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

Distance PQ

Here,

x₁ = – 3, y₁ = 7, z₁ = 2

x₂ = 2, y₂ = 4, z₂ = – 1

Distance PQ

Thus, the required distance is units.

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

Distance PQ

Here,

x₁ = – 1, y₁ = 3, z₁ = – 4

x₂ = 1, y₂ = – 3, z₂ = 4

Distance PQ

Thus, the required distance is units.

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

Distance PQ

Here,

x₁ = 2, y₁ = – 1, z₁ = 3

x₂ = – 2, y₂ = 1, z₂ = 3

Distance PQ

Thus, the required distance is units.

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

Let first calculate distance between the 3 points

i.e. PQ, QR and PR

Calculating PQ

P ≡ (– 2, 3, 5) and Q ≡ (1, 2, 3)

Distance PQ

Here,

x₁ = – 2, y₁ = 3, z₁ = 5

x₂ = 1, y₂ = 2, z₂ = 3

Distance PQ

Calculating QR

Q ≡ (1, 2, 3) and R ≡ (7, 0, – 1)

Distance QR

Here,

x₁ = 1, y₁ = 2, z₁ = 3

x₂ = 7, y₂ = 0, z₂ = – 1

Distance QR

Calculating PR

P ≡ (– 2, 3, 5) and R ≡ (7, 0, – 1)

Distance PR

Here,

x₁ = – 2, y₁ = 3, z₁ = 5

x₂ = 7, y₂ = 0, z₂ = – 1

Distance PR

Thus, PQ = , QR = & PR =

So, PQ + QR = + = = PR

Thus, Points P, Q and R are collinear.

Let points be

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

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

Calculating PQ

P ≡ (0, 7, – 10) and Q ≡ (1, 6, – 6)

Distance PQ

Here,

x₁ = 0, y₁ = 7, z₁ = – 10

x₂ = 1, y₂ = 6, z₂ = – 6

Distance PQ

Calculating QR

Q ≡ (1, 6, – 6) and R ≡ (4, 9, – 6)

Distance QR

Here,

x₁ = 1, y₁ = 6, z₁ = – 6

x₂ = 4, y₂ = 9, z₂ = – 6

Distance QR

Since PQ = QR

∴ 2 sides are equal

Hence PQR is an isosceles triangle.

Let points be

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

Calculating PQ

P ≡ (0, 7, 10) and Q ≡ (– 1, 6, 6)

Distance PQ

Here,

x₁ = 0, y₁ = 7, z₁ = 10

x₂ = – 1, y₂ = 6, z₂ = 6

Distance PQ

Calculating QR

Q ≡ (– 1, 6, 6) and R ≡ (– 4, 9, 6)

Distance QR

Here,

x₁ = – 1, y₁ = 6, z₁ = 6

x₂ = – 4, y₂ = 9, z₂ = 6

Distance QR

Calculating PR

P ≡ (0, 7, 10) and R ≡ (– 4, 9, 6)

Distance PR

Here,

x₁ = 0, y₁ = 7, z₁ = 10

x₂ = – 4, y₂ = 9, z₂ = 6

Distance PR

Now,

PQ² + QR² = 18 + 18 = 36 = PR²

Hence, According to converse of pythagoras theorem,

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

Let 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 & BC = AD

Calculating AB

A ≡ (– 1, 2, 1) and B ≡ (1, – 2, 5)

Distance AB

Here,

x₁ = – 1, y₁ = 2, z₁ = 1

x₂ = 1, y₂ = – 2, z₂ = 5

Distance AB

Calculating BC

B ≡ (1, – 2, 5) and C ≡ (4, – 7, 8)

Distance BC

Here,

x₁ = 1, y₁ = – 2, z₁ = 5

x₂ = 4, y₂ = – 7, z₂ = 8

Distance BC

Calculating CD

C ≡ (4, – 7, 8) and D ≡ (2, – 3, 4)

Distance CD

Here,

x₁ = 4, y₁ = – 7, z₁ = 8

x₂ = 2, y₂ = – 3, z₂ = 4

Distance CD

Calculating DA

D ≡ (2, – 3, 4) and A ≡ (– 1, 2, 1)

Distance DA

Here,

x₁ = 2, y₁ = – 3, z₁ = 4

x₂ = – 1, y₂ = 2, z₂ = 1

Distance DA

Since AB = CD & BC = DA

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

Thus, ABCD is a parallelogram.

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

Calculating PA

P ≡ (x, y, z) and A ≡ (1, 2, 3)

Distance PA

Here,

x₁ = x, y₁ = y, z₁ = z

x₂ = 1, y₂ = 2, z₂ = 3

Distance PA

Calculating PB

P ≡ (x, y, z) and B ≡ (3, 2, – 1)

Distance PB

Here,

x₁ = x, y₁ = y, z₁ = z

x₂ = 3, y₂ = 2, z₂ = – 1

Since PA = PB

Squaring both sides, we get –

PA² = PB²

⇒ (1 – x)² + (2 – y)² + (3 – z)² = (3 – x)² + (2 – y)² + (– 1 – z)²

⇒ (1 + x² – 2x) + (4 + y² – 4y) + (9 + z² – 6z)

= (9 + x² – 6x) + (4 + y² - 4y) + (1 + z² + 2z)

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

⇒ 4x – 8z = 0

⇒ x – 2z = 0

This is the required equation.

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)

Distance PA

Here,

x₁ = x, y₁ = y, z₁ = z

x₂ = 4, y₂ = 0, z₂ = 0

Distance PA

Calculating PB

P ≡ (x, y, z) and B ≡ (– 4, 0, 0)

Distance PB

Here,

x₁ = x, y₁ = y, z₁ = z

x₂ = – 4, y₂ = 0, z₂ = 0

Distance PB

Given that –

PA + PB = 10

⇒ PA = 10 – PB

Squaring both sides, we get –

PA² = (10 – PB)²

⇒ PA² = 100 + PB² – 20 PB

⇒ (4 – x)² + (0 – y)² + (0 – z)²

= 100 + (– 4 – x)² + (0 – y)² + (0 – z)² – 20 PB

⇒ (16 + x² – 8x) + (y²) + (z²)

= 100 + (16 + x² + 8x) + (y²) + (z²) – 20 PB

⇒ 20 PB = 16x + 100

⇒ 5 PB = (4x + 25)

Squaring both sides again, we get –

⇒ 25 PB² = 16x² + 200x + 625

⇒ 25 [(– 4 – x)² + (0 – y)² + (0 – z)²] = 16x² + 200x + 625

⇒ 25 [x² + y² + z² + 8x + 16] = 16x² + 200x + 625

⇒ 25x² + 25y² + 25z² + 200x + 400 = 16x² + 200x + 625

⇒ 9x² + 25y² + 25z² – 225 = 0

This is the required equation.

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

(i) By Section Formula,

We know that the coordinates of the point R which divides the line segment joining two points P (x₁, y₁, z₁) and Q (x₂, y₂, z₂) internally in the ratio m : n is given by:

Comparing this information with the details given in the question, we have

x₁ = -2, y₁ = 3, z₁ = 5; x₂ = 1, y₂ = -4, z₂ = 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 .

(ii) By Section Formula,

We know that the coordinates of the point R which divides the line segment joining two points P (x₁, y₁, z₁) and Q (x₂, y₂, z₂) externally in the ratio m : n is given by:

Comparing this information with the details given in the question, we have

x₁ = -2, y₁ = 3, z₁ = 5; x₂ = 1, y₂ = -4, z₂ = 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:

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

Let Q divides PR in the ratio k : 1.

By Section Formula,

We know that the coordinates of the point R which divides the line segment joining two points P (x₁, y₁, z₁) and Q (x₂, y₂, z₂) internally in the ratio m : n is given by:

Comparing this information with the details given in the question, we have

x₁ = 3, y₁ = 2, z₁ = -4; x₂ = 9, y₂ = 8, z₂ = -10 and m = k, n = 1

So, we have

⇒ 9k + 3 = 5 (k+1) ⇒ 9k + 3 = 5k + 5 ⇒ 4k = 2

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

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).

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

Then,

Comparing this information with the details given in the question, we have

x₁ = -2, y₁ = 4, z₁ = 7; x₂ = 3, y₂ = -5, z₂ = 8 and m = k, n = 1

By section formula,

We know that the coordinates of the point R which divides the line segment joining two points P (x₁, y₁, z₁) and Q (x₂, y₂, z₂) internally in the ratio m : n is given by:

So, we have

⇒ 3k – 2 = 0 ⇒ 3k = 2

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.

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

Then,

Comparing this information with the details given in the question, we have

x₁ = 2, y₁ = -3, z₁ = 4; x₂ = -1, y₂ = 2, z₂ = 1 and m = k, n = 1

By section formula,

We know that the coordinates of the point R which divides the line segment joining two points P (x₁, y₁, z₁) and Q (x₂, y₂, z₂) internally in the ratio m : n is given by:

So, we have,

The coordinates of P =

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

Put

⇒ -k + 2 = 0 ⇒ k = 2

When k =2, then

And

Therefore, C is a point which divides AB in the ratio 2 : 1 and is same as P.

Hence, A, B, C are collinear.

Let A (x₁, y₁, z₁) and B (x₂, y₂, z₂) 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.

Then,

Comparing this information with the details given in the question, we have

x₁ = 4, y₁ = 2, z₁ = -6; x₂ = 10, y₂ = -16, z₂ = 6 and m = 1, n = 2

By section formula,

We know that the coordinates of the point R which divides the line segment joining two points P (x₁, y₁, z₁) and Q (x₂, y₂, z₂) internally in the ratio m : n is given by:

So, we have

The coordinates of A =

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

Then,

Comparing this information with the details given in the question, we have

x₁ = 4, y₁ = 2, z₁ = -6; x₂ = 10, y₂ = -16, z₂ = 6 and m = 2, n = 1

By section formula,

We know that the coordinates of the point R which divides the line segment joining two points P (x₁, y₁, z₁) and Q (x₂, y₂, z₂) internally in the ratio m : n is given by:

So, we have

The coordinates of B =

Hence, 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).

Given: ABCD is a parallelogram, with vertices A (3, -1, 2), B (1, 2, -4), C (-1, 1, 2).

⇒ x₁ = 3, y₁ = -1, z₁ = 2; x₂ = 1, y₂ = 2, z₂ = -4; x₃ = -1, y₃ = 1, z₃ = 2

Let the coordinates of the fourth vertex be D (x, y, z).

We know that the diagonals of a parallelogram bisect each other, so the mid points of AC and BD are equal, i.e. Midpoint of AC = Midpoint of BD ……….(i)

Now, by Midpoint Formula, we know that the coordinates of the mid-point of the line segment joining two points P (x₁, y₁, z₁) and Q (x₂, y₂, z₂) are .

So, we have

Coordinates of the midpoint of AC

Coordinates of the midpoint of BD

So, using (i), we have

⇒ 1 + x = 2, 2 + y = 0, -4 + z = 4

⇒ x = 1, y = -2, z = 8

Hence, the coordinates of the fourth vertex is D (1, -2, 8).

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

⇒ x₁ = 0, y₁ = 0, z₁ = 6; x₂ = 0, y₂ = 4, z₂ = 0; x₃ = 6, y₃ = 0, z₃ = 0

We know that the median is a line segment through a vertex of a triangle to the midpoint of the side opposite to the vertex.

So, let the medians of this triangle be AD, BE and CF corresponding to the vertices A, B and C respectively.

⇒ D, E and F are the midpoints of the sides BC, AC and AB respectively.

By Midpoint Formula, we know that the coordinates of the mid-point of the line segment joining two points P (x₁, y₁, z₁) and Q (x₂, y₂, z₂) are .

So, we have

The coordinates of D = (3, 2, 0)

The coordinates of E = (3, 0, 3)

And the coordinates of F = (0, 2, 3)

By Distance Formula, we know that the distance between two points P (x₁, y₁, z₁) and Q (x₂, y₂, z₂) is given by .

The lengths of the medians are:

So, the lengths of the medians of the given triangle are 7, and 7.

Given: The vertices of the triangle are P (2a, 2, 6), Q (-4, 3b, -10) and R (8, 14, 2c).

⇒ x₁ = 2a, y₁ = 2, z₁ = 6; x₂ = -4, y₂ = 3b, z₂ = -10; x₃ = 8, y₃ = 14, z₃ = 2c

We know that the coordinates of the centroid of the triangle, whose vertices are (x₁, y₁, z₁), (x₂, y₂, z₂) and (x₃, y₃, z₃), are .

So, the coordinates of the centroid of the triangle PQR are

Now, it is given that the origin (0, 0, 0) is the centroid.

So, we have

⇒ 2a +4 = 0, 3b + 16 = 0, 2c – 4 = 0

Let the point on y-axis be A (0, y, 0).

Then, it is given that the distance between the points A (0, y, 0) and P (3, -2, 5) is .

Now, by Distance Formula, we know that the distance between two points P (x₁, y₁, z₁) and Q (x₂, y₂, z₂) is given by .

So, the distance between the points A (0, y, 0) and P (3, -2, 5) is given by

Squaring both sides, we get

(-2 -y)² + 34 = 25 × 2

⇒ (-2 -y)² = 50 – 34

⇒ 4 + y² + (2 × -2 × -y) = 16

⇒ y² + 4y + 4 -16 = 0

⇒ y² + 4y – 12 = 0

⇒ y² + 6y – 2y – 12 = 0

⇒ y (y + 6) – 2 (y + 6) = 0

⇒ (y + 6) (y - 2) = 0

⇒ y = -6, y = 2

So, the points (0, 2, 0) and (0, -6, 0) are the required points on the y-axis.

Given: The coordinates of the points P (2, -3, 4) and Q (8, 0, 10).

_⇒x1 = 2, y₁ = -3, z₁ = 4; x₂ = 8, y₂ = 0, z₂ = 10

Let the coordinates of the required point be (4, y, z).

Now, let the point R (4, y, z) divides the line segment joining the points P (2, -3, 4) and Q (8, 0, 10) in the ratio k : 1.

Also, by Section Formula,

We know that the coordinates of the point R which divides the line segment joining two points P (x₁, y₁, z₁) and Q (x₂, y₂, z₂) internally in the ratio m : n is given by:

So, the coordinates of the point R are given by

So, we have

⇒ 8k + 2 = 4 (k + 1)

⇒ 8k + 2 = 4k + 4

⇒ 8k – 4k = 4 – 2

⇒ 4k = 2

Hence, the coordinates of the required point are (4, -2, 6).

Given: The points A (3, 4, 5) and B (-1, 3, -7)

⇒ x₁ = 3, y₁ = 4, z₁ = 5; x₂ = -1, y₂ = 3, z₂ = -7;

PA² + PB² = k² ……….(i)

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

Now, by Distance Formula, we know that the distance between two points P (x₁, y₁, z₁) and Q (x₂, y₂, z₂) is given by .

So,

And

Now, substituting these values in (i), we have

[(3 - x)² + (4 - y)² + (5 - z)²] + [(-1 - x)² + (3 - y)² + (-7 - z)²] = k²

⇒ [(9 + x² – 6x) + (16 + y² – 8y) + (25 + z² – 10z)] + [(1 + x² + 2x) + (9 + y² – 6y) + (49 + z² + 14z)] = k²

⇒ 9 + x² – 6x + 16 + y² – 8y + 25 + z² – 10z + 1 + x² + 2x + 9 + y² – 6y + 49 + z² + 14z = k²

⇒ 2x² + 2y² + 2z² - 4x - 14y + 4z + 109 = k²

⇒ 2x² + 2y² + 2z² - 4x - 14y + 4z = k² – 109

⇒ 2 (x² + y² + z² – 2x – 7y + 2z) = k² – 109

Hence, the required equation is .