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Introduction to Three Dimensional Geometry

Introduction to Three Dimensional Geometry

Study Action Plan (Sorted by Difficulty)
3D Distance-Based Shape Verification
Accuracy: 0%
Total Attempted1
Total Correct0
Total Incorrect1
Difficulty Level:
High
Centroid and Missing Coordinate Problems
Accuracy: 0%
Total Attempted2
Total Correct0
Total Incorrect2
Difficulty Level:
High
Collinearity of Points in 3D
Accuracy: 25%
Total Attempted4
Total Correct1
Total Incorrect3
Difficulty Level:
High
Octants and Sign of Coordinates
Accuracy: 50%
Total Attempted2
Total Correct1
Total Incorrect1
Difficulty Level:
Medium
Coordinates of a Point in Space
Accuracy: 80%
Total Attempted5
Total Correct4
Total Incorrect1
Difficulty Level:
Low
Three Dimensional Coordinate System
Accuracy: 93%
Total Attempted15
Total Correct14
Total Incorrect1
Difficulty Level:
Low
Distance Formula in 3D Space
Accuracy: 100%
Total Attempted3
Total Correct3
Total Incorrect0
Difficulty Level:
Low
Equation of a Set of Points
Accuracy: 100%
Total Attempted3
Total Correct3
Total Incorrect0
Difficulty Level:
Low
Points on Axes and Coordinate Planes
Accuracy: 100%
Total Attempted8
Total Correct8
Total Incorrect0
Difficulty Level:
Low

Learning Objectives

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Learning Objectives

  • Understand the structure and components of the Three Dimensional Coordinate System, including coordinate axes, planes, and octants.
  • Calculate the Coordinates of a Point in Space using perpendicular distances from the coordinate planes.
  • Derive and apply the Distance Formula in 3D Space to find the distance between two points.
  • Determine the Collinearity of Points in 3D using the distance formula.
  • Formulate the Equation of a Set of Points equidistant from two given points in three-dimensional space.
  • Identify the Octants and Sign of Coordinates by analyzing the signs of (x,y,z)(x, y, z) coordinates.
  • Recognize the forms of Points on Axes and Coordinate Planes, such as (x,0,0)(x,0,0), (0,y,0)(0,y,0), and (0,0,z)(0,0,z).
  • Verify geometric shapes like isosceles/right triangles and parallelograms using 3D Distance-Based Shape Verification.
  • Solve problems involving the Centroid and Missing Coordinate Problems by using centroid relations and coordinate geometry in 3D.

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Revision Notes & Summary

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Chapter Notes

Three Dimensional Coordinate System

  • Coordinate Axes and Planes: In three-dimensional space, three planes intersect at a point called the origin (OO), forming the xx, yy, and zz axes. These axes are mutually perpendicular.
  • Octants: The space is divided into eight parts known as octants, denoted as XOYZXOYZ, X′OYZX'OYZ, X′OY′ZX'OY'Z, XOY′ZXOY'Z, XOYZ′XOYZ', X′OYZ′X'OYZ', X′OY′Z′X'OY'Z', and XOY′Z′XOY'Z'.

Coordinates of a Point in Space

  • Definition: A point PP in space is represented by an ordered triplet (x,y,z)(x, y, z), which are the perpendicular distances from the YZYZ, ZXZX, and XYXY planes, respectively.
  • Location: To locate a point (x,y,z)(x, y, z), draw perpendiculars from PP to the coordinate planes, intersecting the axes at points that determine the coordinates.

Distance Formula in 3D Space

  • Formula: The distance PQPQ between two points P(x1,y1,z1)P(x_1, y_1, z_1) and Q(x2,y2,z2)Q(x_2, y_2, z_2) is given by: PQ=(x2−x1)2+(y2−y1)2+(z2−z1)2PQ = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}

Collinearity of Points in 3D

  • Condition: Points PP, QQ, and RR are collinear if the sum of the distances PQPQ and QRQR equals the distance PRPR.

Equation of a Set of Points

  • Equidistant Points: The equation of a set of points equidistant from two points A(x1,y1,z1)A(x_1, y_1, z_1) and B(x2,y2,z2)B(x_2, y_2, z_2) is derived from the condition PA=PBPA = PB.

Octants and Sign of Coordinates

  • Sign Table:
    Octantxxyyzz
    I+++
    II-++
    III--+
    IV+-+
    V++-
    VI-+-
    VII---
    VIII+--

Points on Axes and Coordinate Planes

  • Forms:
    • On xx-axis: (x,0,0)(x, 0, 0)
    • On yy-axis: (0,y,0)(0, y, 0)
    • On zz-axis: (0,0,z)(0, 0, z)
    • In XYXY-plane: (x,y,0)(x, y, 0)
    • In YZYZ-plane: (0,y,z)(0, y, z)
    • In ZXZX-plane: (x,0,z)(x, 0, z)

3D Distance-Based Shape Verification

  • Verification: Use the distance formula to verify geometric properties such as isosceles or right triangles, parallelograms, and rectangles.

Centroid and Missing Coordinate Problems

  • Centroid Formula: For a triangle with vertices A(x1,y1,z1)A(x_1, y_1, z_1), B(x2,y2,z2)B(x_2, y_2, z_2), C(x3,y3,z3)C(x_3, y_3, z_3), the centroid GG is given by: G=(x1+x2+x33,y1+y2+y33,z1+z2+z33)G = \left(\frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3}, \frac{z_1 + z_2 + z_3}{3}\right)
  • Finding Missing Coordinates: Use the centroid formula to solve for missing coordinates when the centroid is known.
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Practice Test – MCQs, True/False

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Multiple Choice Questions

A.

ABCDABCD is a rectangle.

B.

ABCDABCD is a parallelogram but not a rectangle.

C.

ABCDABCD is a square.

D.

ABCDABCD is not a parallelogram.
Correct Answer: B

Solution:

To show ABCDABCD is a parallelogram, we need to show opposite sides are equal. Calculating the distances, AB=CDAB = CD and BC=DABC = DA, confirming it is a parallelogram. However, the diagonals ACAC and BDBD are not equal, so it is not a rectangle.

Chapter Concept:

3D Distance-Based Shape Verification

A.

(-2, 3, 5)

B.

(4, -1, 3)

C.

(2, 3, -5)

D.

(3, 2, 1)
Correct Answer: D

Solution:

The first octant is defined by all coordinates being positive. Therefore, the point (3,2,1)(3, 2, 1) lies in the first octant.

Chapter Concept:

Coordinates of a Point in Space

A.

45\sqrt{45}

B.

50\sqrt{50}

C.

41\sqrt{41}

D.

55\sqrt{55}
Correct Answer: A

Solution:

Using the distance formula: PQ=(−4−1)2+(1+3)2+(2−4)2=25+16+4=45PQ = \sqrt{(-4-1)^2 + (1+3)^2 + (2-4)^2} = \sqrt{25 + 16 + 4} = \sqrt{45}.

Chapter Concept:

Distance Formula in 3D Space

A.

Yes, because AB+BC=ACAB + BC = AC

B.

No, because AB+BC≠ACAB + BC \neq AC

C.

Yes, because the vectors AB→\overrightarrow{AB} and BC→\overrightarrow{BC} are parallel.

D.

No, because the points do not lie on the same plane.
Correct Answer: A

Solution:

The points AA, BB, and CC are collinear because the sum of the distances ABAB and BCBC is equal to the distance ACAC.

Chapter Concept:

Collinearity of Points in 3D

A.

Both are zero

B.

Only y-coordinate is zero

C.

Only z-coordinate is zero

D.

Both are non-zero
Correct Answer: A

Solution:

A point on the x-axis has coordinates of the form (x,0,0)(x, 0, 0), meaning both y-coordinate and z-coordinate are zero.

Chapter Concept:

Equation of a Set of Points

A.

(0, 0, 0)

B.

(1, 1, 1)

C.

(x, y, z)

D.

(0, y, z)
Correct Answer: A

Solution:

The origin in a three-dimensional coordinate system is represented by the coordinates (0, 0, 0).

Chapter Concept:

Three Dimensional Coordinate System

A.

It is the point where the x, y, and z axes intersect.

B.

It is the point where the x and y axes intersect.

C.

It is the point where the y and z axes intersect.

D.

It is the point where the x and z axes intersect.
Correct Answer: A

Solution:

The origin is the point (0,0,0)(0, 0, 0) where the x, y, and z axes intersect in a three-dimensional coordinate system.

Chapter Concept:

Coordinates of a Point in Space

A.

x2+y2+z2=0x^2 + y^2 + z^2 = 0

B.

x+y+z=0x + y + z = 0

C.

x=2x = 2

D.

x=1x = 1
Correct Answer: C

Solution:

The equation of the set of points equidistant from (1,2,3)(1, 2, 3) and (3,2,−1)(3, 2, -1) is x=2x = 2.

Chapter Concept:

Distance Formula in 3D Space

A.

(1, 1, 2)

B.

(1, 1, 1)

C.

(0, 0, 0)

D.

(2, 2, 2)
Correct Answer: A

Solution:

Using the centroid formula G(x,y,z)=(x1+x2+x33,y1+y2+y33,z1+z2+z33)G(x, y, z) = \left(\frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3}, \frac{z_1 + z_2 + z_3}{3}\right), substitute G(1,1,1)G(1, 1, 1), A(3,−5,7)A(3, -5, 7), B(−1,7,−6)B(-1, 7, -6) to find C(1,1,2)C(1, 1, 2).

Chapter Concept:

Centroid and Missing Coordinate Problems

A.

The yy-coordinate is positive.

B.

The yy-coordinate is negative.

C.

The yy-coordinate is zero.

D.

The yy-coordinate is non-zero.
Correct Answer: C

Solution:

A point lying on the XZXZ-plane has its yy-coordinate equal to zero.

Chapter Concept:

Three Dimensional Coordinate System

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True or False

Correct Answer: False

Solution:

A point with coordinates (−3,1,2)(-3, 1, 2) lies in the second octant, as indicated by the signs of the coordinates: xx is negative, yy is positive, and zz is positive.

Chapter Concept :

Octants and Sign of Coordinates

Correct Answer: True

Solution:

In three-dimensional geometry, a point on the x-axis is represented by (x,0,0)(x, 0, 0), indicating that it lies on the x-axis with no displacement along the y or z axes.

Chapter Concept :

Points on Axes and Coordinate Planes

Correct Answer: True

Solution:

A point in the YZ-plane is represented by (0,y,z)(0, y, z), which signifies that the point does not have any displacement along the x-axis.

Chapter Concept :

Points on Axes and Coordinate Planes

Correct Answer: True

Solution:

The distance formula for two points in three-dimensional space is derived from the Pythagorean theorem applied in three dimensions, taking into account the differences in each coordinate direction.

Chapter Concept :

3D Distance-Based Shape Verification

Correct Answer: False

Solution:

A point on the x-axis has coordinates of the form (x,0,0)(x, 0, 0), meaning both yy and zz are zero.

Chapter Concept :

Coordinates of a Point in Space

Correct Answer: True

Solution:

In three-dimensional geometry, any point on the x-axis is represented by the coordinates (x,0,0)(x, 0, 0), where xx can be any real number and the yy and zz coordinates are zero.

Chapter Concept :

Points on Axes and Coordinate Planes

Correct Answer: True

Solution:

A point on the x-axis has no displacement along the y-axis or z-axis, hence its coordinates are (x,0,0)(x, 0, 0). This reflects zero distance from the YZ-plane.

Chapter Concept :

3D Distance-Based Shape Verification

Correct Answer: True

Solution:

The distance formula for two points in 3D space is derived by considering the diagonal of a rectangular parallelepiped formed by the projections of the points onto the coordinate planes, applying the Pythagorean theorem twice.

Chapter Concept :

Distance Formula in 3D Space

Correct Answer: True

Solution:

The distance formula in three-dimensional space is derived from the Pythagorean theorem and calculates the Euclidean distance between two points.

Chapter Concept :

Distance Formula in 3D Space

Correct Answer: True

Solution:

The signs of the coordinates (x,y,z)(x, y, z) determine the octant in which the point is located. Each octant corresponds to a unique combination of signs for the coordinates.

Chapter Concept :

Octants and Sign of Coordinates

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