3D Geometry Explained Part 1: Distance, Angles, and Planes
PUC Mathematics | 3D Geometry | Daniel Classes
Why 3D Geometry Breaks Intuition
In 2D, a line has one equation: y = 2x + 3. In 3D, the same “line” requires two equations. Why? Because in 3D, a single equation doesn’t pin down a line — it describes a plane. Two planes intersect at a line.
This is where most students get stuck. The formulas look similar to 2D, but the geometry has shifted. A point is still a point. But a line becomes more complex. And planes arrive as an entirely new object.
Here’s the core insight: 3D geometry is just 2D geometry with one more dimension. You already know how to find the distance between two points in 2D: √((x₂-x₁)² + (y₂-y₁)²). In 3D, you add one more term: √((x₂-x₁)² + (y₂-y₁)² + (z₂-z₁)²).
Once you see this pattern, 3D stops being mysterious.
Distance Between Two Points
2D formula (you know this): d = √((x₂-x₁)² + (y₂-y₁)²)
3D formula (just add z): d = √((x₂-x₁)² + (y₂-y₁)² + (z₂-z₁)²)
Example: Distance between (1, 2, 3) and (4, 6, 15):
d = √((4-1)² + (6-2)² + (15-3)²) = √(9 + 16 + 144) = √169 = 13
Angle Between Two Lines
In 3D, lines are represented by direction vectors. The angle between two lines is the angle between their direction vectors.
Formula: cos(θ) = (**d₁** · **d₂**) / (|**d₁**| × |**d₂**|)
Where **d₁** and **d₂** are direction vectors of the two lines.
Example: Find the angle between lines with direction vectors (1, 0, 0) and (1, 1, 0).
cos(θ) = (1×1 + 0×1 + 0×0) / (1 × √2) = 1/√2, so θ = 45°
The Equation of a Plane
General form: ax + by + cz + d = 0
Where (a, b, c) is the normal vector (perpendicular to the plane).
Example: The plane 2x + 3y + 4z = 12 has normal vector (2, 3, 4).
Finding the equation of a plane: You need either:
- Three non-collinear points, or
- A point and a normal vector, or
- A point and two direction vectors in the plane
Example: Find the equation of the plane through (1, 0, 0), (0, 1, 0), (0, 0, 1).
Solution: Substitute into ax + by + cz + d = 0 for each point. You get a = b = c = 1, d = -1. So the equation is x + y + z = 1.
Distance from a Point to a Plane
Formula: distance = |ax₀ + by₀ + cz₀ + d| / √(a² + b² + c²)
Where (x₀, y₀, z₀) is the point and ax + by + cz + d = 0 is the plane.
Example: Distance from (1, 2, 3) to the plane 2x + 3y + 4z – 5 = 0:
distance = |2(1) + 3(2) + 4(3) – 5| / √(4 + 9 + 16) = |2 + 6 + 12 – 5| / √29 = 15 / √29
How PUC Exams Test 3D Geometry
Type 1: Distance Between Points (2-3 marks)
Example: “Find the distance between (1, 2, 3) and (4, 6, 15).”
Solution: Apply the formula directly.
Type 2: Equation of a Plane (3-5 marks)
Example: “Find the equation of the plane through (1, 0, 0), (0, 1, 0), (0, 0, 1).”
Solution: Substitute into ax + by + cz + d = 0, solve for coefficients.
Type 3: Angle Between Lines (3-5 marks)
Example: “Find the angle between lines with direction vectors **d₁** = (1, 0, 0) and **d₂** = (1, 1, 0).”
Solution: Use the dot product formula.
The Five Common Mistakes
Mistake 1: Forgetting the Square Root
Example: Distance between (1, 2, 3) and (4, 6, 15).
Wrong: (4-1)² + (6-2)² + (15-3)² = 9 + 16 + 144 = 169 [stopping here]
Right: √169 = 13
Fix: Distance always takes a square root.
Mistake 2: Using Incorrect Plane Coefficients
Example: “Find the normal vector of 2x + 3y + 4z = 12.”
Wrong: “The normal is (2, 3, 12).” [Confusing the constant term]
Right: “The normal is (2, 3, 4).” [Only the coefficients of x, y, z]
Fix: The normal vector is (a, b, c) from ax + by + cz + d = 0.
Mistake 3: Forgetting the Denominator in Distance Formula
Example: Distance from (1, 2, 3) to 2x + 3y + 4z – 5 = 0.
Wrong: |2 + 6 + 12 – 5| = 15 [stopping, calling it the distance]
Right: 15 / √29 [dividing by the magnitude of the normal]
Fix: The distance formula divides by √(a² + b² + c²).
Mistake 4: Mixing Up Lines and Planes
Example: “In 3D, how many equations describe a line?”
Wrong: “One.”
Right: “Two (the intersection of two planes).”
Fix: In 3D: one equation = plane. Two equations = line. Three equations = point.
Mistake 5: Confusing Direction Vector with Normal Vector
Example: “A line has direction vector (2, 3, 4). What’s the normal to the plane it’s in?”
Wrong: “The normal is (2, 3, 4).” [Confusing the two]
Right: “The normal is perpendicular to (2, 3, 4).” [There are infinite perpendicular vectors]
Fix: Direction vector is along a line. Normal vector is perpendicular to a plane. They’re different concepts.
What to Do Next
To master 3D distances: Solve 10 problems finding distances between points.
To master plane equations: Find 10 plane equations from different combinations of points.
To master angles: Find angles between 8 pairs of lines.
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