Three Dimensional Geometry

2.1 Core concepts

• A directed line L through the origin making angles α, β, γ with the positive x, y, z-axes has direction cosines l = cos α, m = cos β, n = cos γ (NCERT §11.2, p. 377). These three numbers completely encode the direction of L in space.
• Reversing the direction of L replaces the angles by their supplements, so signs of all three DCs flip; a line in space has two sets of DCs unless taken as directed (NCERT §11.2, p. 377–378). The line itself is undirected; the DCs come with ± sign ambiguity.
• For a line not through the origin, draw a parallel line through O and read off its DCs — parallel lines share the same set of DCs (NCERT §11.2 Remark, p. 378). Direction is a translation-invariant property.
• Any three numbers proportional to l, m, n are called direction ratios (DRs); if a, b, c are DRs then a = λl, b = λm, c = λn for some nonzero λ (NCERT §11.2, p. 378). DRs are convenient because they often come out as integers, while DCs typically involve square roots.
• From l² + m² + n² = 1, the DCs in terms of DRs are l = ±a/√(a²+b²+c²), m = ±b/√(a²+b²+c²), n = ±c/√(a²+b²+c²) (NCERT §11.2, p. 379). The ± reflects the direction-reversal ambiguity.
• The DCs of the line joining P(x1, y1, z1) and Q(x2, y2, z2) are (x2−x1)/PQ, (y2−y1)/PQ, (z2−z1)/PQ, where PQ = √((x2−x1)² + (y2−y1)² + (z2−z1)²); the DRs may be taken as x2−x1, y2−y1, z2−z1 (NCERT §11.2.1, p. 379–380).
• A line is uniquely determined by either (i) a point and a direction, or (ii) two points (NCERT §11.3, p. 381). These two specifications underpin the two standard line equations.
• Vector equation of a line through point A (position vector a) parallel to b: r = a + λb, λ ∈ R (NCERT §11.3.1, p. 381–382). The parameter λ "slides" along the line; varying λ over R sweeps out the whole line.
• Writing b = a î + b ĵ + c k̂, the parametric form is x = x1+λa, y = y1+λb, z = z1+λc, giving the Cartesian symmetric form (x−x1)/a = (y−y1)/b = (z−z1)/c (NCERT §11.3.1, p. 382). Each fraction equals the common parameter λ.
• If l, m, n are DCs of the line, the same equation reads (x−x1)/l = (y−y1)/m = (z−z1)/n (NCERT §11.3.1 Note, p. 382). DC form is occasionally more natural when angles are given directly.
• The angle θ between two lines with DRs a1, b1, c1 and a2, b2, c2 satisfies cos θ = (a1a2 + b1b2 + c1c2) / [√(a1² + b1² + c1²) · √(a2² + b2² + c2²)] (NCERT §11.4 eqn (1), p. 383).
• In DC form (since l² + m² + n² = 1): cos θ = |l1l2 + m1m2 + n1n2| (NCERT §11.4 eqn (3), p. 384). The absolute value enforces the acute-angle convention.
• Two lines are perpendicular iff a1a2 + b1b2 + c1c2 = 0; parallel iff a1/a2 = b1/b2 = c1/c2 (NCERT §11.4, p. 384). These are vector statements: perpendicular ⇔ dot product 0; parallel ⇔ direction vectors collinear.
• In vector form, for r = a1 + λb1 and r = a2 + μb2, the acute angle satisfies cos θ = |b1·b2| / (|b1||b2|) (NCERT §11.4, p. 384).
• Skew lines are non-coplanar lines in space — neither intersecting nor parallel; the shortest distance segment between them is perpendicular to both (NCERT §11.5, p. 385–386). The room-corner example (Fig 11.5) makes this geometrically intuitive.
• Shortest distance between skew lines r = a1 + λb1 and r = a2 + μb2 is d = |(b1 × b2) · (a2 − a1)| / |b1 × b2| (NCERT §11.5.1, p. 386–387). The numerator is the volume of the parallelepiped spanned by a2−a1, b1, b2; the denominator is the area of the base.
• Cartesian form of the shortest distance uses the scalar triple product (a determinant with rows x2−x1, y2−y1, z2−z1; a1, b1, c1; a2, b2, c2) divided by √[(b1c2−b2c1)² + (c1a2−c2a1)² + (a1b2−a2b1)²] (NCERT §11.5.1, p. 387). Two skew lines intersect iff this shortest distance is zero — equivalent to the determinant vanishing (the coplanarity condition).
• Distance between two parallel lines r = a1 + λb and r = a2 + μb is d = |b × (a2 − a1)| / |b| (NCERT §11.5.2, p. 387–388). The cross product picks up the component of (a2 − a1) perpendicular to b.
• Two parallel lines are coplanar; the distance equals the length of the perpendicular from any point of one to the other (NCERT §11.5.2, p. 387).
• Planes are out of scope in the current reprint; equations of planes, angle between planes, distance from a point to a plane, and intersection of a line with a plane are deferred to advanced study.
• A practical guide to picking the right formula: if the problem gives angles → DCs; if it gives a position and direction → vector form r = a + λb; if it gives two points → two-point form; if it asks the angle → cos θ formula; if it asks distance and lines are parallel → parallel distance formula; if skew → skew distance formula.
• The vector approach is computationally lighter than the Cartesian: dot/cross products handle direction-vector arithmetic uniformly. Convert to vector form whenever a problem looks messy in Cartesian.
• A practical tip for shortest distance: compute b₁ × b₂ first; if it equals 0, lines are parallel; else they are either skew or intersecting. They intersect iff the scalar triple product (a₂−a₁)·(b₁×b₂) = 0; otherwise they are skew with positive shortest distance.
• Standard 3D hint: changing coordinates (e.g. translating the origin to one of the line's points) often simplifies a problem; work with vectors based at the origin where possible.

2.2 Definitions to memorise

2.3 Diagrams / processes to remember

• Fig 11.1 (p. 378): a directed line L making angles α, β, γ with the coordinate axes — the picture that justifies l = cos α etc.
• Fig 11.2 (p. 379): the line PQ between P(x1, y1, z1) and Q(x2, y2, z2) with perpendiculars to the XY-plane; gives the formula cos γ = (z2 − z1)/PQ.
• Fig 11.3 (p. 381–382): point A with position vector a, line through A parallel to b, arbitrary point P with position vector r — the geometric picture of r = a + λb.
• Fig 11.4 (p. 383): two lines through the origin with DRs a1, b1, c1 and a2, b2, c2 — used to derive the cos θ formula.
• Fig 11.5 (p. 385): the room of dimensions 1 × 3 × 2 with the diagonal GE on the ceiling and DB along the wall — the canonical example of skew lines.
• Fig 11.6 (p. 386): skew lines l1, l2 with the perpendicular segment PQ and the foot of perpendicular construction used to derive d = |(b1 × b2)·(a2 − a1)| / |b1 × b2|.
• Fig 11.7 (p. 387): parallel lines l1, l2 with foot of perpendicular from T to l1, justifying d = |b × (a2 − a1)| / |b|.
• Process — find DCs: identify DRs from problem; compute √(a²+b²+c²); divide each DR by the magnitude.
• Process — angle between two lines: identify direction vectors b₁, b₂ (or DRs); compute dot product and magnitudes; apply cos θ = |b₁·b₂|/(|b₁||b₂|).
• Process — line equation through two points: compute b = a₂ − a₁; use r = a₁ + λb or Cartesian symmetric form.
• Process — shortest distance (skew): compute b₁ × b₂; compute a₂ − a₁; dot the cross with the difference; divide by the magnitude of the cross.

2.4 Common confusions / NTA trap points

• Mixing up DCs and DRs: only DCs satisfy l² + m² + n² = 1. If a question gives "direction ratios 2, −1, −2", do NOT just plug them in — first normalise to (2/3, −1/3, −2/3).
• Forgetting that DCs come with a sign ambiguity (line vs directed line). Numerically equal sets like (1/√3, 1/√3, 1/√3) and (−1/√3, −1/√3, −1/√3) refer to the same line in opposite directions.
• Using cos θ = b1·b2 / (|b1||b2|) without absolute value — θ is taken to be the acute angle, so |b1·b2| is required.
• Confusing the shortest-distance formula for skew lines (scalar triple over |b1 × b2|) with the parallel-line formula (|b × (a2 − a1)|/|b|). Apply the parallel formula only after checking that b1, b2 are proportional.
• Reading the symmetric Cartesian form: the numerators give the point on the line (x1, y1, z1), the denominators give the DRs (a, b, c). Students often swap these.
• Treating l² + m² + n² = 1 as a "check" rather than a constraint: when DCs are asked, you must divide DRs by their magnitude; you cannot just report the DRs as DCs.
• Forgetting that two parallel direction vectors are proportional, not equal — the proportionality test a₁/a₂ = b₁/b₂ = c₁/c₂ is the correct one.
• Using the skew-line formula when lines are parallel — b₁ × b₂ = 0, so the formula blows up. Always test for parallelism first.
• Sign-error in cross product: b₁ × b₂ = −(b₂ × b₁). The formula for distance uses absolute value, so the sign does not change the answer, but it does matter for directed-distance calculations.
• Mis-reading the parametric form: (x − x₁)/a = (y − y₁)/b = (z − z₁)/c, NOT (x − a)/x₁. The constants on the side of x are the point coordinates; the denominators are direction components.
• Confusing parallel and coincident lines. Coincident means same line; parallel means non-intersecting with same direction.
• Treating the shortest-distance formula as a magnitude of a vector. It is a scalar.

2.5 Key formulas & theorems

2.6 Solved examples (NCERT-grounded)

Example A (NCERT Example 1, p. 380). DCs of a line making angles 90°, 60°, 30° with axes.

Step 1 — write cosines: l = cos 90° = 0; m = cos 60° = 1/2; n = cos 30° = √3/2. Step 2 — verify identity: 0 + 1/4 + 3/4 = 1 ✓. Step 3 — answer: (0, 1/2, √3/2).

Example B (NCERT Example 2, p. 380). DCs from DRs 2, −1, −2.

Step 1 — magnitude: √(4 + 1 + 4) = 3. Step 2 — divide: (2/3, −1/3, −2/3). Step 3 — check identity: 4/9 + 1/9 + 4/9 = 1 ✓. Answer: (2/3, −1/3, −2/3).

Example C (NCERT Example 6, p. 382). Equation of line through (5, 2, −4) parallel to 3î + 2ĵ − 8k̂.

Step 1 — vector form: r = 5î + 2ĵ − 4k̂ + λ(3î + 2ĵ − 8k̂). Step 2 — Cartesian form: (x − 5)/3 = (y − 2)/2 = (z + 4)/−8. Step 3 — state: both forms above.

Example D (NCERT Example 7, p. 384). Angle between r = 3î + 2ĵ − 4k̂ + λ(î + 2ĵ + 2k̂) and r = 5î − 2ĵ + μ(3î + 2ĵ + 6k̂).

Step 1 — b₁·b₂: 1·3 + 2·2 + 2·6 = 19. Step 2 — magnitudes: |b₁| = 3, |b₂| = 7; product = 21. Step 3 — cosine: cos θ = 19/21; θ = cos⁻¹(19/21).

Example E (NCERT Example 9, p. 388). Shortest distance between r = î + ĵ + λ(2î − ĵ + k̂) and r = 2î + ĵ − k̂ + μ(3î − 5ĵ + 2k̂).

Step 1 — cross product: b₁ × b₂ = 3î − ĵ − 7k̂; magnitude = √(9 + 1 + 49) = √59. Step 2 — a₂ − a₁: î + 0ĵ − k̂; dot with cross = 3 + 0 + 7 = 10. Step 3 — divide: d = 10/√59. Answer: 10/√59.