Mathematics — Std 12
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Three Dimensional Geometry

Ch. 6Std 12

Easy Overview

You've spent years doing geometry in two dimensions — plotting points on x and y axes, finding distances, slopes, and equations of lines. But the world we live in is three-dimensional. This chapter is your passport to working in 3D space. Every point now has three coordinates (x, y, z), and lines have direction in all three dimensions. This is the geometry behind 3D movies, GPS navigation, flight paths, architectural blueprints, and even the way your phone tracks its orientation in space. We start with the basics of the 3D coordinate system. There are three mutually perpendicular axes — X, Y, and Z — meeting at the origin. The space is divided into 8 octants (like quadrants in 2D, but now we have up and down too). The coordinates of a point P are (x, y, z) — the distances along the three axes. The distance between two points P(x₁, y₁, z₁) and Q(x₂, y₂, z₂) is √((x₂−x₁)² + (y₂−y₁)² + (z₂−z₁)²) — it's just the 2D distance formula with z added. The section formula extends naturally too: if R divides PQ internally in ratio m:n, then R = ((mx₂ + nx₁)/(m+n), (my₂ + ny₁)/(m+n), (mz₂ + nz₁)/(m+n)). The midpoint is the special case with m = n. Next come direction cosines and direction ratios — these describe which way a line points. Direction cosines (l, m, n) are the cosines of the angles that the line makes with the X, Y, and Z axes. They satisfy the identity l² + m² + n² = 1. Direction ratios (a, b, c) are any three numbers proportional to (l, m, n). They're easier to find — you just take the differences in coordinates between two points on the line. To convert direction ratios to direction cosines: divide each by √(a² + b² + c²). Then we learn how to write the equation of a line in 3D space. In vector form, r = a + λb, where a is the position vector of a point on the line and b is the direction vector. In Cartesian (symmetric) form, (x − x₁)/a = (y − y₁)/b = (z − z₁)/c = λ. Each value of λ gives a different point on the line. The parametric form writes x = x₁ + aλ, y = y₁ + bλ, z = z₁ + cλ — useful for finding specific points. You'll learn to find the angle between two lines using their direction cosines (cos θ = l₁l₂ + m₁m₂ + n₁n₂) or direction ratios (cos θ = |a₁a₂ + b₁b₂ + c₁c₂| / √((a₁²+b₁²+c₁²)(a₂²+b₂²+c₂²))). If the numerator is zero, the lines are perpendicular. If the direction ratios are proportional, they're parallel. We also cover conditions for a point to lie on a line, and how to find the distance of a point from a line using cross product. Finally, we tackle skew lines — lines that are neither parallel nor intersecting. They exist in different planes. The shortest distance between two skew lines is the length of the common perpendicular. The formula uses the scalar triple product: SD = |(a₂ − a₁)·(b₁ × b₂)| / |b₁ × b₂|. This is the distance between the two lines at their closest approach. By the end of this chapter, you'll be comfortable navigating 3D space and describing lines in multiple equivalent forms. The skills here feed directly into the next chapter on line and plane, where we add planes to the mix.

The 3D Coordinate System

Three mutually perpendicular axes — X (horizontal, left-right), Y (horizontal, forward-backward), Z (vertical, up-down) — intersect at the origin O(0,0,0). These axes divide space into 8 octants. A point P(x, y, z) has x-coordinate (distance from YZ-plane), y-coordinate (distance from XZ-plane), and z-coordinate (distance from XY-plane). The sign of each coordinate determines which octant the point lies in.

Distance Formula in 3D

The distance between two points P(x₁, y₁, z₁) and Q(x₂, y₂, z₂) is PQ = √((x₂−x₁)² + (y₂−y₁)² + (z₂−z₁)²). This is derived by applying the 2D distance formula twice — first to find the distance in the XY-plane, then incorporating the Z difference. It's the same formula you know, just with a third term added under the square root.

Section Formula in 3D

If point R divides PQ internally in ratio m:n, the coordinates of R are ((mx₂ + nx₁)/(m+n), (my₂ + ny₁)/(m+n), (mz₂ + nz₁)/(m+n)). For external division: ((mx₂ − nx₁)/(m−n), (my₂ − ny₁)/(m−n), (mz₂ − nz₁)/(m−n)). For the midpoint (m = n): ((x₁ + x₂)/2, (y₁ + y₂)/2, (z₁ + z₂)/2). This is a direct extension of the 2D section formula.

Direction Cosines of a Line

A line making angles α, β, γ with the positive X, Y, Z axes has direction cosines l = cos α, m = cos β, n = cos γ. Fundamental identity: l² + m² + n² = 1. Direction cosines uniquely determine a line's direction (not its sense — opposite direction gives cos(π−α) = −l, etc.). If a line passes through P and Q, then l = (x₂−x₁)/PQ, m = (y₂−y₁)/PQ, n = (z₂−z₁)/PQ.

Direction Ratios of a Line

Direction ratios (a, b, c) are any three numbers proportional to (l, m, n). If a line passes through P(x₁, y₁, z₁) and Q(x₂, y₂, z₂), then a = x₂−x₁, b = y₂−y₁, c = z₂−z₁ (or any multiples). To convert ratios to cosines: l = a/√(a²+b²+c²), m = b/√(a²+b²+c²), n = c/√(a²+b²+c²). Direction ratios are easier to work with because they don't require the actual distance.

Equation of a Line — Vector Form

A line is determined by a point and a direction. Vector form: r = a + λb, where a is the position vector of a known point on the line, b is the direction vector (parallel to the line), and λ is a parameter (real number). Each value of λ gives a different point on the line. For λ = 0, r = a (the given point). This is the most compact form of a line equation.

Equation of a Line — Cartesian (Symmetric) Form

Starting from r = a + λb with a = (x₁, y₁, z₁) and b = (a, b, c), we get x = x₁ + aλ, y = y₁ + bλ, z = z₁ + cλ. Eliminating λ gives (x − x₁)/a = (y − y₁)/b = (z − z₁)/c = λ. This is the symmetric form. If any direction ratio is zero, say b = 0, the equation becomes (x − x₁)/a = (z − z₁)/c, y = y₁. This form is used when you need to find the equation of a line through two given points.

Equation of a Line Through Two Points

If a line passes through P(x₁, y₁, z₁) and Q(x₂, y₂, z₂), the direction ratios are (x₂−x₁, y₂−y₁, z₂−z₁). The symmetric form is (x−x₁)/(x₂−x₁) = (y−y₁)/(y₂−y₁) = (z−z₁)/(z₂−z₁). The vector form is r = a + λ(b − a), where a is P and b is Q. You can also write r = (1−λ)a + λb, which is the two-point form.

Angle Between Two Lines

The angle θ between two lines with direction cosines (l₁, m₁, n₁) and (l₂, m₂, n₂) is given by cos θ = l₁l₂ + m₁m₂ + n₁n₂. For direction ratios (a₁, b₁, c₁) and (a₂, b₂, c₂): cos θ = (a₁a₂ + b₁b₂ + c₁c₂) / √((a₁²+b₁²+c₁²)(a₂²+b₂²+c₂²)). Lines are perpendicular if cos θ = 0 (numerator = 0). Lines are parallel if direction ratios are proportional.

Collinearity of Three Points in 3D

Three points P, Q, R are collinear if they lie on the same line. This happens if the direction ratios of PQ and PR are proportional. That is, (x₂−x₁)/(x₃−x₁) = (y₂−y₁)/(y₃−y₁) = (z₂−z₁)/(z₃−z₁). Alternatively, the area of triangle PQR should be zero (cross product of PQ and PR is zero). Collinearity also means one point's position vector can be expressed as a linear combination of the other two.

Distance of a Point from a Line

To find the perpendicular distance from a point R to a line through P with direction b, use cross product: distance = |(PR) × b| / |b|. This is the area of the parallelogram formed by PR and b divided by the length of b. Alternatively, find the foot of the perpendicular by projecting R onto the line and then compute the distance between R and its foot.

Skew Lines — Definition and Condition

Skew lines are lines that are NOT parallel and do NOT intersect. They lie in different planes. Two lines with direction vectors b₁ and b₂ are skew if: (1) b₁ is not parallel to b₂ (ratios not proportional), and (2) the lines don't intersect (no common point). The condition for intersecting lines: there exist λ, μ such that a₁ + λb₁ = a₂ + μb₂. If no solution exists, the lines are skew.

Shortest Distance Between Two Skew Lines

The shortest distance (SD) between two skew lines r = a₁ + λb₁ and r = a₂ + μb₂ is the length of the common perpendicular. Formula: SD = |(a₂ − a₁)·(b₁ × b₂)| / |b₁ × b₂|. This is the projection of the vector joining a point on each line onto the direction perpendicular to both lines. The scalar triple product in the numerator gives the volume, dividing by |b₁×b₂| gives the height.

Distance Between Parallel Lines

For two parallel lines with direction b, one through point A and the other through point B, the distance between them is: d = |AB × b| / |b|. This is the same as the distance of point B from the first line. Since the lines are parallel, the perpendicular distance is constant everywhere — so just pick any point on one line and find its distance from the other line.

Key Points

  • 3D coordinates: P(x, y, z) — distances from YZ, XZ, XY planes respectively
  • Distance between P and Q: √((x₂−x₁)² + (y₂−y₁)² + (z₂−z₁)²)
  • Section formula (internal): R = ((mx₂+nx₁)/(m+n), (my₂+ny₁)/(m+n), (mz₂+nz₁)/(m+n))
  • Direction cosines l, m, n satisfy l² + m² + n² = 1
  • Direction ratios (a, b, c) are proportional to (l, m, n)
  • Line (vector): r = a + λb; Line (Cartesian): (x−x₁)/a = (y−y₁)/b = (z−z₁)/c
  • Line through two points P, Q: (x−x₁)/(x₂−x₁) = (y−y₁)/(y₂−y₁) = (z−z₁)/(z₂−z₁)
  • Angle between lines: cos θ = l₁l₂ + m₁m₂ + n₁n₂
  • Lines perpendicular iff l₁l₂ + m₁m₂ + n₁n₂ = 0
  • Lines parallel iff direction ratios are proportional
  • Collinearity: PR is proportional to PQ for three collinear points
  • Distance from point to line: d = |PR × b| / |b|
  • Skew lines: not parallel, not intersecting — no common solution
  • Shortest distance between skew lines: SD = |(a₂−a₁)·(b₁×b₂)| / |b₁×b₂|
  • Distance between parallel lines: d = |AB × b| / |b|
  • In symmetric form, each ratio equals the same parameter λ

Practice Questions

  • Find the direction cosines of the line joining the points (1, 2, 3) and (4, 5, 6).
  • Find the equation of the line through (2, 3, −1) with direction ratios (3, 2, 4) in both vector and Cartesian form.
  • Show that the lines (x−1)/2 = (y+1)/3 = (z−2)/4 and (x−3)/4 = (y−2)/6 = (z−1)/8 are parallel.
  • Find the angle between the lines whose direction ratios are (1, 2, 2) and (2, 3, 6).
  • Find the shortest distance between the lines r = i + j + λ(2i − j + k) and r = 2i + j − k + μ(3i − 5j + 2k).
  • Find the distance of the point (2, 3, 4) from the line (x−1)/2 = (y−2)/3 = (z−3)/4.
  • Show that the points A(1, 2, 3), B(3, 5, 7), and C(5, 8, 11) are collinear.
  • Find the foot of the perpendicular from (1, 2, 3) to the line (x−2)/3 = (y−3)/4 = (z−4)/5.