Three-dimensional Geometry

X and y coordinates of a point are its distance from the origin along or parallel to the horizontal x-axis and y-axis. To measure the x and y coordinates, you must move either to the left of the origin or to its right. In case of a point on the z-axis, you do not move to the right or to the left of the origin. Hence x and y coordinates are 0 for a point on the z-axis.

x-coordinate is the distance of a point from the origin parallel or along the x-axis. To measure the x coordinate, you must move either to the left of the origin or to its right. In case of a point lying on the yz-plane, you do not move to the right or to the left of the origin. Hence x coordinate is 0 for a point on the yz-plane.

Here the x, y, z coordinates of the point are 4, -3, 0. As the distance of point along the z-axis is 0, the plane in which the point lies is the xy-plane.

The position of a point in a octant is signified by the signs of the x, y, z coordinates.

Here is a table showing signs of the x, y, z coordinates in all the octants.

According to the table

(i) (-4, -1, -6) lies in octant VII

(ii) (2, 3, -4) lies in octant V

(iii) (-6, 5, -1) lies in octant VI

(iv) (4, -3, -2) lies in octant VIII

(v) (-1, -6, 5) lies in octant III

(vi) (4, 6, 8) lies in octant I

Formula:

The distance between two points (x₁,y₁,z₁) and (x₂,y₂,z₂) is given by

D=

(i) A(5, 1, 2) and B(4, 6, -1)

Here, (x₁,y₁,z₁) = (5, 1, 2)

(x₂,y₂,z₂)= (4, 6, -1)

Therefore,

D =

=

=

=

Distance between points A and B is .

(ii) P(1, -1, 3) and Q(2, 3, -5)

Here, (x₁,y₁,z₁)= (1, -1, 3)

(x₂,y₂,z₂)= (2, 3, -5)

Therefore,

D =

=

=

= = 9

Distance between points P and Q are 9 units.

(iii) R(1, -3, 4) and S(4, -2, -3)

Here, (x₁,y₁,z₁)= (1, -3, 4)

(x₂,y₂,z₂)= (4, -2, -3)

Therefore,

D =

=

=

=

Distance between points R and S is units.

(iv) C(9, -12, -8) and the origin

Coordinates of origin are (0, 0, 0)

Here, (x₁,y₁,z₁)= (9, -12, -8)

(x₂,y₂,z₂)= (0, 0, 0)

Therefore,

D =

=

=

= = 17

Distance between points C and origin is 17 units.

To prove: Points A, B, C form equilateral triangle.

Formula:

The distance between two points (x₁,y₁,z₁) and (x₂,y₂,z₂) is given by

D=

Here,

(x₁,y₁,z₁)= (1, -1, -5)

(x₂,y₂,z₂)= (3, 1,3)

(x₃,y₃,z₃)= (9, 1, -3)

Length AB =

=

=

=

= = 6

Length BC =

=

=

=

= = 6

Length AC =

=

=

=

= = 6

Hence, AB = BC = AC

Therefore, Points A, B, C make an equilateral triangle.

To prove: Points A, B, C form isosceles triangle.

Formula:

The distance between two points (x₁,y₁,z₁) and (x₂,y₂,z₂) is given by

D=

Here,

(x₁,y₁,z₁)= (4, 6, -3)

(x₂,y₂,z₂)= (0, 2, 3)

(x₃,y₃,z₃)= (-4, -4, -1)

Length AB =

=

=

=

= = 2

Length BC =

=

=

=

= = 2

Length AC =

=

=

=

=

Here, AB = BC

∴ vertices A, B, C forms an isosceles triangle.

To prove: Points A, B, C form isosceles triangle.

Formula:

The distance between two points (x₁,y₁,z₁) and (x₂,y₂,z₂) is given by

D=

Here,

(x₁,y₁,z₁)= (0, 1, 2)

(x₂,y₂,z₂)= (2, -1, 3)

(x₃,y₃,z₃)= (1, -3, 1)

Length AB =

=

=

=

=

Length BC =

=

=

=

=

Length AC =

=

=

=

=

Also, AB² + BC² = 9 + 9 = 18 = AC²

Therefore, points A, B, C forms an isosceles right-angled triangle.

To prove: Points A, B, C, D form square.

Formula:

The distance between two points (x₁,y₁,z₁) and (x₂,y₂,z₂) is given by

D=

Here,

(x₁,y₁,z₁)= (1, 1, 1)

(x₂,y₂,z₂)= (-2, 4, 1)

(x₃,y₃,z₃)= (-1, 5, 5)

(x₄,y₄,z₄)= (2, 2, 5)

Length AB =

=

=

=

=

Length BC =

=

=

=

=

Length CD =

=

=

=

=

Length AD =

=

=

=

=

Length AC =

=

=

=

=

Length BD =

=

=

=

=

Here, AB = BC = CD = AD

Also, AC = BD

This means all the sides are the same and diagonals are also equal.

Hence vertices A, B, C, D form a square.

To prove: Points A, B, C, D form parallelogram.

Formula:

The distance between two points (x₁,y₁,z₁) and (x₂,y₂,z₂) is given by

D=

Here,

(x₁,y₁,z₁)= (1, 2, 3)

(x₂,y₂,z₂)= (-1, -2, -1)

(x₃,y₃,z₃)= (2, 3, 2)

(x₄,y₄,z₄)= (4, 7, 6)

Length AB =

=

=

=

=

Length BC =

=

=

=

=

Length CD =

=

=

=

=

Length AD =

=

=

=

=

Length AC =

=

=

=

=

Length BD =

=

=

=

=

Here, AB = CD which are opposite sides of polygon.

BC = AD which are opposite sides of polygon.

Also the diagonals AC and BD are not equal in length.

Hence, the polygon is not a rectangle.

To prove: Points P, Q, R, S forms rectangle.

Formula:

The distance between two points (x₁,y₁,z₁) and (x₂,y₂,z₂) is given by

D=

Here,

(x₁,y₁,z₁)= (2, 3, 5)

(x₂,y₂,z₂)= (-4, 7, -7)

(x₃,y₃,z₃)= (-2, 1, -10)

(x₄,y₄,z₄)= (4, -3, 2)

Length PQ =

=

=

=

=

Length QR =

=

=

=

=

Length RS =

=

=

=

=

Length PS =

=

=

=

=

Length PR =

=

=

=

=

Length QS =

=

=

=

=

Here, PQ = RS which are opposite sides of polygon.

QR = PS which are opposite sides of polygon.

Also the diagonals PR = QS.

Hence, the polygon is a rectangle.

To prove: Points P, Q, R, S forms rhombus.

Formula:

The distance between two points (x₁,y₁,z₁) and (x₂,y₂,z₂) is given by

D=

Here,

(x₁,y₁,z₁)= (1, 3, 4)

(x₂,y₂,z₂)= (-1, 6, 10)

(x₃,y₃,z₃)= (-7, 4, 7)

(x₄,y₄,z₄)= (-5, 1, 1)

Length PQ =

=

=

=

=

Length QR =

=

=

=

=

Length RS =

=

=

=

=

Length PS =

=

=

=

=

Length PR =

=

=

=

=

Length QS =

=

=

=

=

Here, PQ = RS =QR = PS .

Also the diagonals PR ≠ QS.

Hence, the polygon is a rhombus as all sides are equal and diagonals are not equal.

To prove: D is circumcenter of triangle ABC

Let us consider D as circumcenter of triangle ABC.

∴ AD = BC = CD.

Formula:

The distance between two points (x₁,y₁,z₁) and (x₂,y₂,z₂) is given by

D=

Here,

(x₁,y₁,z₁)= (3, 2, -5)

(x₂,y₂,z₂)= (-3. 8, -5)

(x₃,y₃,z₃)= (-3, 2, 1)

(x₄,y₄,z₄) = (-1, 4, -3)

Length AD =

=

=

=

=

Length BD =

=

=

=

=

Length CD =

=

=

=

=

Hence, the condition is consistent.

Hence, D is circumcenter of triangle ABC.

To prove: the 3 points are collinear.

Formula:

The distance between two points (x₁,y₁,z₁) and (x₂,y₂,z₂) is given by

D=

Here,

(x₁,y₁,z₁)= (-2, 3, 5)

(x₂,y₂,z₂)= (1, 2, 3)

(x₃,y₃,z₃)= (7, 0, -1)

Length AB =

=

=

=

=

Length BC =

=

=

=

= = 2

Length AC =

=

=

=

= = 3

AB + BC = + 2 =3 = AC

Therefore A, B, C are collinear.

To prove: the 3 points are collinear.

Formula:

The distance between two points (x₁,y₁,z₁) and (x₂,y₂,z₂) is given by

D=

Here,

(x₁,y₁,z₁)= (3, -5, 1)

(x₂,y₂,z₂)= (-1, 0, 8)

(x₃,y₃,z₃)= (7, -10, -6)

Length AB =

=

=

=

= = 3

Length BC =

=

=

=

= = 6

Length AC =

=

=

=

= = 3

BA + BC = 3 + 3 =6 = BC

Therefore A, B, C are collinear.

To prove: the 3 points are collinear.

Formula:

The distance between two points (x₁,y₁,z₁) and (x₂,y₂,z₂) is given by

D=

Here,

Vertices should be R(-1, 4, -2)

The solution according is

(x₁,y₁,z₁)= (3, -2, 4)

(x₂,y₂,z₂)= (1, 1, 1)

(x₃,y₃,z₃)= (-1, 4,- 2)

Length PQ=

=

=

=

=

Length QR=

=

=

=

=

Length PR =

=

=

=

= = 2

PQ + QR= + =2 = PR

Therefore P, Q, R are collinear.

Consider, C(x,y,z) point equidistant from points A(-1, 2, 3) and B(3, 2, 1).

∴ AC = BC

Squaring both sides,

=

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

8x -4z = 0

Equation of curve is 8x-4z = 0

Consider, C(0,y,0) point which lies on y axis and is equidistant from points A(3, 1, 2) and B(5, 5, 2).

∴ AC = BC

Squaring both sides,

=

9+ y² -2y +1 + 4 = 25 + y² -10y +25 + 4

8y = 40

Y = 5

The point C is (0,5,0).

Consider, C(0,0,z) point which lies on z axis and is equidistant from points A(1, 5, 7) and B(5, 1, -4).

∴ AC = BC

Squaring both sides,

=

1 +25+ z² -14z +49 = 25 + 1 + z² +8z +16

-22z = -33

Z = 1.5

The point C is (0,0,1.5).

Consider, D(x,y,z) point equidistant from points A(a, 0, 0), B(0, b, 0), C(0, 0, c) and O(0, 0, 0).

∴ AD = 0D

Squaring both sides,

=

x² +2ax + a² + y² + z² = x² + y² + z²

a(2x-a) = 0

as a ≠ 0.

X = a/2

∴ BD = 0D

Squaring both sides,

=

x² + y^{2 +} 2by + b² + z² = x² + y² + z²

b(2y-b) = 0

as b ≠ 0.

y= b/2

∴ CD = 0D

Squaring both sides,

=

x² + y^{2 +} z² + 2cz + c² = x² + y² + z²

c(2z-c) = 0

as c ≠ 0.

z= c/2

Therefore, the pint D(a/2,b/2,c/2) is equidistant to points A(a, 0, 0), B(0, b, 0), C(0, 0, c) and O(0, 0, 0).

The general point on yz plane is D(0, y, z). Consider this point is equidistant to the points A(3, 2, -1), B(1, -1, 0) and C(2, 1, 2).

∴ AD = BD

Squaring both sides,

=

9+ y² -4y + 4 + z² + 2z + 1 = 1+ y² +2y + 1+ z²

-6y + 2z + 12 = 0 ….(1)

Also, AD = CD

Squaring both sides,

=

9+ y² -4y + 4 + z² + 2z + 1 = 4+ y² - 2y + 1+ z² – 4z + 4

-2y + 6z + 5 = 0 ….(2)

Simultaneously solving equation (1) and (2) we get

Y= 31/16, z = -3/16

The point which is equidistant to the points A(3, 2, -1), B(1, -1, 0) and C(2, 1, 2) is (0, 31/16, -3/16).

The general point on xy plane is D(x, y, 0). Consider this point is equidistant to the points A(2, 0, 3), B(0, 3, 2) and C(0, 0, 1).

∴ AD = BD

Squaring both sides,

=

X² – 4x +4 + y² + 9 = X² + y² -6y + 9 + 4

-4x = -6y ….(1)

Also, AD = CD

Squaring both sides,

=

X² – 4x +4 + y² + 9 = X² + y² +1

-4x = -12 ….(2)

Simultaneously solving equation (1) and (2) we get

X = 3, y = 2.

The point which is equidistant to the points A(2, 0, 3), B(0, 3, 2) and C(0, 0, 1) is (3, 2, 0).

The coordinates of point R that divides the line segment joining points P (x₁, y₁, z₁)

and Q (x₂, y₂, z₂) in the ratio m: n are

Point A( 3, 2, 5 ) and B( -4, 2, -2 ), m and n are 4 and 3 respectively.

Using the above formula, we get,

( -1,2,1), is the point which divides the two points in ratio 4 : 3.

The coordinates of point R that divides the line segment joining points P (x₁, y₁, z₁)

and Q (x₂, y₂, z₂) in the ratio m: n are

Point A( 2, 1, -3 ) and B( 5, -8, 3 ), m and n are 2 and 1 respectively.

Using the above formula, we get,

( 4,-5, 1), is the point of trisection of the segment AB.

The coordinates of point R that divides the line segment joining points P (x₁, y₁, z₁)

and Q (x₂, y₂, z₂) externally in the ratio m: n are

Point A( -2, 4, 7 ) and B( 3, -5, 8 ), m and n are 2 and 1 respectively.

Using the above formula, we get,

= ( 8,-14,9), is the point that divides the two point A and B externally in the ratio 2:1.

Let the ratio be k:1 in which point R divides point P and point Q.

Using , we get,

Here m and n are k and 1. The point which this formula gives is already given, i.e. R(5,4,-6) and the joining points are P(3, 2, -4) and Q(9, 8, -10).

Taking any point and finding the value of k, we get

5k + 5 = 9k + 3

4k = 2

Therefore, the ratio be 1:2.

Let the ratio be k:1 in which point R divides point P and point Q.

Using , we get,

Here m and n are k and 1. The point which this formula gives is already given, i.e. R(5,9,-14) and the joining points are P(2, -3, 4) and Q(3, 1, -2).

Taking any point and finding the value of k, we get

5k + 5 = 3k + 2

2k = -3

Since, the ratio is -3:2. hence the division is external division.

The external division ratio is 3:2.

Let the plane XZ divides the points A(-1, -3, 4) and B(4, 2, -1) in ratio k:1.

Hence, using section formula (, we get

On XZ plane, Y co- ordinate of every point be zero, therefore

2k – 3 = 0

The ratio is 3:2 in XZ plane which divides the line joined from points A and B.

Let the plane XY divides the points A(3,4,1) and B(5, 1, 6) in ratio k:1.

Hence, using section formula (, we get

On XY plane, Z co- ordinate of every point be zero, therefore

6k + 1 = 0

The ratio is 1:6 externally in XZ plane which divides the line joined from points A and B.

Let the plane x – 2y + 3z = 5 divides the join of A(3, -5, 4) and B(2, 3, -7) in ratio k:1.

The point which will come by section formula will be in the plane. Putting that in the plane equation will give the point coordinates. The points are A(3, -5, 4) and B(2, 3, -7).

Using section formula, , we get

Putting this point in the plane equation, we get

2k + 3 – 6k + 10 – 21k + 12 = 5k + 5

-25k + 25 = 5k + 5

-30k = -20

the ratio is 2:3. And the point of intersection of the plane and the line is

The given co-ordinates: A(3, 2, 0), B(5, 3, 2) and C(-9, 6, -3)

Now, AB = 3

Also, AC

Now, we have,

By the property of internal angle bisector,

Therefore,

Applying the section formula, we get,

D(x, y, z)

D(x, y, z)

Answer.

the vertices of the parallelogram be A(3, 4, -3), B(7, 10, -3) and C(5, -2, 7), and the fourth coordinate be D(a,b,c).

the property of parallelogram is the diagonal bisect each other. Therefore,

diagonal AC and BD will bisect each other, and the bisecting point will be equal to the two diagnals. By using section formula, we get

, where m and n are 1 and 1.

Finding the coordinate of mid point of diagonal AC,

= (4,1,2)

Now, Finding the coordinate of mid point of diagonal BD,

Equating the two mid points, we get

. Thus,

Therefore,

8 = a + 7

a = 1

b + 10 = 2

b = -8

and

c = 7

therefore the point is ( 1, -8, 7).

Since the centroid of a triangle

The points are A(2, -4, 3) and B(3, -1, -2), and its centroid is (1, 0, 3). And let its third vertex C(a,b,c).

Using the formula, we get

Equating it with the coordinates of centroid, we get

a = -2

b = -5

and,

c = 7

therefore, the point is (-2,-5,7)

Since, centroid of a triangle is found by

The points are A(a,1,3) and B(-2, b, -5), and its centroid is (0, 0, 0) and its third vertex C(4,7,c).

Using the formula, we get

Equating it with the coordinates of centroid, we get

a = -2

b = -8

and,

c = 2

therefore, a= -2, b= -8, c=2.

The midpoints of the sides of a triangle are (1, 5, -1), (0, 4, -2) and (2, 3, 4).

Let its vertices be A(x_1,y₁,z₁), B(x₂,y₂,z₂), C(x₃,y₃,z₃) .

The mid point of AB is (1,5,-1), therefore

x₁ +x₂ = 2………..eq.1

y₁ + y₂ = 10……eq.2

z₁ + z₂ = -2……..eq.3

Mid point of AC is (2,3,4), therefore

x₁ +x₃ = 4………..eq.4

y₁ + y₃ = 6……eq.5

z₁ + z₃ = 8……..eq.6

Mid point of BC is (0,4,-2), therefore

x₂ +x₃ = 0………..eq.7

y₃ + y₂ = 8……eq.8

z₃ + z₂ = -4……..eq.9

now, adding the equations 1,4 and 7, and divide it by two we get,

x₁ + x₂ + x₃ = 3

now subtracting 1, 4, 7 individually, we get

x₁ = 3, x₂ = -1 and x₃ = 1

now, adding the equations 2,5 and 8, and divide it by two we get,

y₁ + y₂ + y₃ = 12

now subtracting 1, 4, 7 individually, we get

y₁ = 4, y₂ = 6 and y₃ = 2

now, adding the equations 3,6 and 9, and divide it by two we get,

z₁ + z₂ + z₃ = 1

now subtracting 1, 4, 7 individually, we get

z₁ = 5, z₂ = -7 and z₃ = 3

therefore, the coordinates are A(3,4,5), B(-1,6,-7) and C(1,2,3).