Introduction to Three-Dimensional Geometry

From 2D to 3D: Understanding the Coordinate System

Imagine your familiar 2D Cartesian plane (the XY-plane) lying flat on a table. Now, imagine a new line, the z-axis, passing through the origin (0,0) and standing perpendicular to the table. This is the essence of a three-dimensional coordinate system.

We now have three mutually perpendicular axes: the x-axis, the y-axis, and the z-axis. Their point of intersection is the Origin (0, 0, 0). These axes define three coordinate planes:

1. XY-plane: The plane containing the x and y axes. Here, the z-coordinate is always zero (z = 0).
2. YZ-plane: The plane containing the y and z axes. Here, the x-coordinate is always zero (x = 0).
3. XZ-plane: The plane containing the x and z axes. Here, the y-coordinate is always zero (y = 0).

Just as the x and y axes divide a plane into four quadrants, these three planes divide space into eight regions called octants. The sign of the coordinates (x, y, z) of a point determines which octant it lies in. For example, any point in the first octant has all positive coordinates (+, +, +). A point with coordinates (-, +, +) lies in the second octant.

Key Formulas in 3D Geometry

Distance Formula

The distance between two points P(x₁, y₁, z₁) and Q(x₂, y₂, z₂) in space is given by the formula: d = √[(x₂ - x₁)² + (y₂ - y₁)² + (z₂ - z₁)²]. This is a direct extension of the Pythagorean theorem into three dimensions.

Section Formula (Internal Division)

The coordinates of a point R(x, y, z) that divides the line segment joining P(x₁, y₁, z₁) and Q(x₂, y₂, z₂) internally in the ratio m:n are: x = (mx₂ + nx₁)/(m + n), y = (my₂ + ny₁)/(m + n), z = (mz₂ + nz₁)/(m + n).

Section Formula (External Division)

The coordinates of a point R(x, y, z) that divides the line segment joining P(x₁, y₁, z₁) and Q(x₂, y₂, z₂) externally in the ratio m:n are: x = (mx₂ - nx₁)/(m - n), y = (my₂ - ny₁)/(m - n), z = (mz₂ - nz₁)/(m - n).

Mid-point Formula

As a special case of internal division where m=n=1, the coordinates of the midpoint of the line segment joining P(x₁, y₁, z₁) and Q(x₂, y₂, z₂) are: ((x₁ + x₂)/2, (y₁ + y₂)/2, (z₁ + z₂)/2).

Worked Examples: Applying 3D Formulas

• Example 1: Using the Distance Formula Find the distance between the points A(2, 5, -1) and B(-3, 1, 4). Solution: Step 1: Identify the coordinates. Here, (x₁, y₁, z₁) = (2, 5, -1) and (x₂, y₂, z₂) = (-3, 1, 4). Step 2: Apply the distance formula: d = √[(x₂ - x₁)² + (y₂ - y₁)² + (z₂ - z₁)²]. AB = √[(-3 - 2)² + (1 - 5)² + (4 - (-1))²] Step 3: Calculate the result. AB = √[(-5)² + (-4)² + (5)²] AB = √[25 + 16 + 25] AB = √66 Final Answer: The distance between points A and B is √66 units.
• Example 2: Using the Section Formula (Internal Division) Find the coordinates of the point which divides the line segment joining P(1, -2, 3) and Q(3, 4, -5) internally in the ratio 2:3. Solution: Step 1: Identify the coordinates and the ratio. P(x₁, y₁, z₁) = (1, -2, 3) Q(x₂, y₂, z₂) = (3, 4, -5) Ratio m:n = 2:3 Step 2: Apply the section formula for each coordinate. x = (mx₂ + nx₁)/(m + n) = (23 + 31)/(2 + 3) = (6 + 3)/5 = 9/5 y = (my₂ + ny₁)/(m + n) = (24 + 3(-2))/(2 + 3) = (8 - 6)/5 = 2/5 z = (mz₂ + nz₁)/(m + n) = (2(-5) + 33)/(2 + 3) = (-10 + 9)/5 = -1/5 Final Answer: The coordinates of the required point are (9/5, 2/5, -1/5).
• Example 3: Checking for Collinearity Show that the points A(-2, 3, 5), B(1, 2, 3), and C(7, 0, -1) are collinear. Solution: Step 1: Find the distances between each pair of points using the distance formula. AB = √[(1 - (-2))² + (2 - 3)² + (3 - 5)²] = √[3² + (-1)² + (-2)²] = √(9 + 1 + 4) = √14 BC = √[(7 - 1)² + (0 - 2)² + (-1 - 3)²] = √[6² + (-2)² + (-4)²] = √(36 + 4 + 16) = √56 = 2√14 AC = √[(7 - (-2))² + (0 - 3)² + (-1 - 5)²] = √[9² + (-3)² + (-6)²] = √(81 + 9 + 36) = √126 = 3√14 Step 2: Check if the sum of two smaller distances equals the largest distance. AB + BC = √14 + 2√14 = 3√14 Step 3: Compare with the largest distance. Since AB + BC = AC, the points lie on the same straight line. Final Answer: The points A, B, and C are collinear.

Exam Traps and Key Points

1. Master the Octant Signs: A common source of error is mixing up the signs of the coordinates for each octant. Create a quick table during your exam revision to memorize them. Remember: Octant I is (+,+,+), Octant II is (-,+,+), and so on.

2. Points on Axes and Planes: This is a favourite for MCQs.

• Any point on the x-axis is of the form (x, 0, 0).
• Any point on the y-axis is of the form (0, y, 0).
• Any point on the z-axis is of the form (0, 0, z).
• Any point in the XY-plane is of the form (x, y, 0).
• Any point in the YZ-plane is of the form (0, y, z).
• Any point in the XZ-plane is of the form (x, 0, z).

3. Ratio Questions: When a question asks you to find the ratio in which a line segment is divided by one of the coordinate planes (e.g., the YZ-plane), assume the ratio is k:1. Use the section formula to find the coordinates of the point of division in terms of k. Since the point lies on the YZ-plane, its x-coordinate must be 0. Set the expression for the x-coordinate to zero and solve for k.

Practice Questions with Solutions

• Q: In which octant does the point (-3, 1, -2) lie? A: Step 1: Analyze the sign of each coordinate. The x-coordinate is -3 (negative). The y-coordinate is 1 (positive). The z-coordinate is -2 (negative). Step 2: Match the sign pattern (-, +, -) to the corresponding octant. Octant I: (+,+,+) Octant II: (-,+,+) Octant III: (-,-,+) Octant IV: (+,-,+) Octant V: (+,-,-) Octant VI: (-,-,-) Octant VII: (-,+,-) Octant VIII: (+,+,-) Final answer: The pattern (-, +, -) corresponds to the VII (seventh) octant.
• Q: Show that the points P(0, 7, 10), Q(–1, 6, 6) and R(–4, 9, 6) form an isosceles triangle. A: Step 1: Use the distance formula to find the lengths of the three sides PQ, QR, and PR. PQ = √[(-1-0)² + (6-7)² + (6-10)²] = √[(-1)² + (-1)² + (-4)²] = √(1+1+16) = √18 = 3√2. QR = √[(-4-(-1))² + (9-6)² + (6-6)²] = √[(-3)² + 3² + 0²] = √(9+9+0) = √18 = 3√2. PR = √[(-4-0)² + (9-7)² + (6-10)²] = √[(-4)² + 2² + (-4)²] = √(16+4+16) = √36 = 6. Step 2: Compare the lengths of the sides. We observe that PQ = QR = 3√2. Final answer: Since two sides of the triangle (PQ and QR) are equal in length, the triangle PQR is an isosceles triangle.
• Q: Find the ratio in which the YZ-plane divides the line segment formed by joining the points (–2, 4, 7) and (3, –5, 8). A: Step 1: Let the YZ-plane divide the line segment joining A(–2, 4, 7) and B(3, –5, 8) at a point P in the ratio k : 1. Step 2: Use the section formula to find the coordinates of P. The coordinates of P are given by ((k3 + 1(-2))/(k+1), (k(-5) + 14)/(k+1), (k8 + 17)/(k+1)). P = ((3k-2)/(k+1), (-5k+4)/(k+1), (8k+7)/(k+1)). Step 3: Use the property of the YZ-plane. Since the point P lies on the YZ-plane, its x-coordinate must be 0. Therefore, (3k-2)/(k+1) = 0. 3k - 2 = 0 => 3k = 2 => k = 2/3. Final answer: The required ratio is k : 1, which is (2/3) : 1 or 2 : 3.
• Q: Find the coordinates of a point on the y-axis which is equidistant from the points P(3, 1, 2) and Q(5, 5, 2). A: Step 1: Let the point on the y-axis be A. Any point on the y-axis has coordinates of the form (0, y, 0). Step 2: Set up the condition that A is equidistant from P and Q. This means AP = AQ, or AP² = AQ². AP² = (3-0)² + (1-y)² + (2-0)² = 9 + (1-y)² + 4 = 13 + (1-y)² AQ² = (5-0)² + (5-y)² + (2-0)² = 25 + (5-y)² + 4 = 29 + (5-y)² Step 3: Solve the equation AP² = AQ² for y. 13 + (1-y)² = 29 + (5-y)² 13 + 1 - 2y + y² = 29 + 25 - 10y + y² 14 - 2y = 54 - 10y 10y - 2y = 54 - 14 8y = 40 y = 5 Final answer: The required point on the y-axis is (0, 5, 0).

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