Three-Dimensional Geometry

While Class 11 introduced 3D geometry in Cartesian coordinates, this chapter leverages chapter-10-vector-algebra to make 3D geometry elegant and powerful. Rather than memorizing formulas, we use vectors to describe lines and planes as dynamic objects. A line becomes a starting point plus a direction; a plane becomes a point with a normal vector. This vector approach unifies concepts, simplifies calculations, and extends naturally to higher dimensions. Understanding 3D geometry is essential for physics (motion, fields), engineering (design, optimization), and advanced mathematics.

Direction Cosines and Direction Ratios

For a line in 3D, direction ratios (a, b, c) indicate the direction. They're proportional to components of a direction vector.

Direction cosines are the cosines of angles the line makes with coordinate axes:

cos(α) = a/√(a² + b² + c²)
cos(β) = b/√(a² + b² + c²)
cos(γ) = c/√(a² + b² + c²)

Important property: cos²(α) + cos²(β) + cos²(γ) = 1 (sum of squares of direction cosines equals 1)

Building from Class 11: Class 11 introduced these concepts; now we use them to describe lines and planes vectorially.

Equations of a Line in 3D

A line passing through point P(x₀, y₀, z₀) with direction ratios (a, b, c) can be expressed:

Vector form: r = r₀ + td, where r₀ = (x₀, y₀, z₀), d = (a, b, c), and t is a parameter

Parametric form:

x = x₀ + at
y = y₀ + bt
z = z₀ + ct

Symmetric form: (x - x₀)/a = (y - y₀)/b = (z - z₀)/c (if a, b, c ≠ 0)

Example: Line through (1, 2, 3) with direction (2, -1, 4):

• Vector: r = (1, 2, 3) + t(2, -1, 4)
• Parametric: x = 1 + 2t, y = 2 - t, z = 3 + 4t
• Symmetric: (x - 1)/2 = (y - 2)/(-1) = (z - 3)/4

Equations of a Plane in 3D

A plane is determined by a point P(x₀, y₀, z₀) and a normal vector n = (A, B, C) perpendicular to the plane.

Vector form: n · (r - r₀) = 0

Cartesian form: A(x - x₀) + B(y - y₀) + C(z - z₀) = 0

General form: Ax + By + Cz + D = 0 (expanding and simplifying)

Normal vector: From the equation Ax + By + Cz + D = 0, the normal is n = (A, B, C)

Example: Plane through (1, 2, 3) with normal (2, -1, 4):

• 2(x - 1) - 1(y - 2) + 4(z - 3) = 0
• 2x - y + 4z - 12 = 0

Angle Between Two Lines

For lines with direction vectors d₁ = (a₁, b₁, c₁) and d₂ = (a₂, b₂, c₂):

cos(θ) = |**d₁** · **d₂**| / (||**d₁**|| ||**d₂**||)

(Use absolute value to get acute angle)

Perpendicularity: Lines are perpendicular if d₁ · d₂ = 0

Parallelism: Lines are parallel if d₁ and d₂ are proportional (same direction)

Angle Between Two Planes

For planes with normal vectors n₁ = (A₁, B₁, C₁) and n₂ = (A₂, B₂, C₂):

cos(θ) = |**n₁** · **n₂**| / (||**n₁**|| ||**n₂**||)

Perpendicularity: Planes are perpendicular if n₁ · n₂ = 0

Parallelism: Planes are parallel if n₁ and n₂ are proportional

Distance Calculations

Distance from point P to line (through Q with direction d):

Distance = ||(**QP**) × **d**|| / ||**d**||

(Use cross product to find perpendicular distance)

Distance from point P to plane Ax + By + Cz + D = 0:

Distance = |Ax₀ + By₀ + Cz₀ + D| / √(A² + B² + C²)

(Numerator is plane equation evaluated at P; denominator normalizes the normal vector)

Distance between two parallel planes A₁x + B₁y + C₁z + D₁ = 0 and A₂x + B₂y + C₂z + D₂ = 0 (with proportional normals):

Pick any point on the first plane, use point-to-plane distance formula.

Line-Plane Intersection

A line r = r₀ + td intersects plane Ax + By + Cz + D = 0 when:

A(x₀ + ta) + B(y₀ + tb) + C(z₀ + tc) + D = 0

Solve for t; if solution exists, find intersection point by substituting t back.

Special cases:

• No solution if d ⊥ n (line parallel to plane)
• Infinitely many solutions if line lies in plane

Connections to Other Topics

• chapter-10-vector-algebra: Vectors describe directions and normals; dot and cross products for angles and distances
• chapter-06-application-of-derivatives: Tangent lines and planes use similar vector concepts
• chapter-08-application-of-integrals: Line and surface integrals over curves and planes in 3D
• chapter-04-determinants: Determinants in scalar triple product for volume

Socratic Questions

1. Explain why a line in 3D requires both a point and a direction vector, while a line in 2D can be specified by a single equation like y = mx + b. What extra information does the direction vector provide in 3D?

1. For a plane with equation Ax + By + Cz + D = 0, the normal vector is n = (A, B, C). Why must the normal be perpendicular to the plane? How does this relate to the dot product of the normal with any vector lying in the plane?

1. Two planes are parallel if their normals are proportional. Two planes are perpendicular if their normals are perpendicular (dot product = 0). Explain why these conditions make geometric sense.

1. To find the distance from a point P to a plane, we use the formula |Ax₀ + By₀ + Cz₀ + D| / √(A² + B² + C²). Why is the denominator √(A² + B² + C²) present? What does it represent, and why is the division necessary?

1. A line is given parametrically as r = (1, 2, 3) + t(2, -1, 4). How would you find where this line intersects the plane 2x + y - z + 5 = 0? What would it mean geometrically if you found no solution?