Three Dimensional Geometry – JEE Main Maths Formula Sheet & Class 12 Notes | All Formulas for Lines, Planes & Distances

Coordinate Axes, Octants, Distance Formula, Section Formula

Why These Three Dimensional Geometry Fundamentals Are the Entry Point for All JEE Main 3D Questions

Coordinate System in Three Dimensional Geometry (from PDF): Three mutually perpendicular axes OX, OY, OZ intersect at origin O. A point P(x, y, z) is located by: x = perpendicular distance from yz-plane (with sign); y = perpendicular distance from xz-plane (with sign); z = perpendicular distance from xy-plane (with sign). The three coordinate planes (XOY, YOZ, ZOX) divide space into 8 octants with sign patterns for (x,y,z): OXYZ(+,+,+); OX'YZ(–,+,+); OXY'Z(+,–,+); OXYZ'(+,+,–); OX'Y'Z(–,–,+); OX'YZ'(–,+,–); OXY'Z'(+,–,–); OX'Y'Z'(–,–,–).

Special point locations in Three Dimensional Geometry (from PDF):

On x-axis: (λ, 0, 0) — y=0 and z=0. On y-axis: (0, λ, 0). On z-axis: (0, 0, λ). On XY-plane: (λ, μ, 0) — z=0. On XZ-plane: (λ, 0, μ) — y=0. On YZ-plane: (0, λ, μ) — x=0.

Distance Formula in Three Dimensional Geometry (from PDF — JEE Main):

Distance between P(x₁,y₁,z₁) and Q(x₂,y₂,z₂): PQ = √[(x₂–x₁)²+(y₂–y₁)²+(z₂–z₁)²]

In vector form: PQ = (x₂–x₁)î+(y₂–y₁)ĵ+(z₂–z₁)k̂; |PQ| = √[(x₂–x₁)²+(y₂–y₁)²+(z₂–z₁)²]

Distance of P(x₁,y₁,z₁) from origin O: OP = √(x₁²+y₁²+z₁²). Vector: OP = x₁î+y₁ĵ+z₁k̂.

Section Formula in Three Dimensional Geometry (from PDF — JEE Main):

Internal division — R(x,y,z) divides P(x₁,y₁,z₁) and Q(x₂,y₂,z₂) in ratio m:n internally:

x = (mx₂+nx₁)/(m+n); y = (my₂+ny₁)/(m+n); z = (mz₂+nz₁)/(m+n)

Vector form: OR = (m·OQ + n·OP)/(m+n)

External division — ratio m:n externally: x=(mx₂–nx₁)/(m–n); y=(my₂–ny₁)/(m–n); z=(mz₂–nz₁)/(m–n)

Vector form: OR = (m·OQ – n·OP)/(m–n)

Midpoint: ((x₁+x₂)/2, (y₁+y₂)/2, (z₁+z₂)/2)

Centroid of triangle with vertices (x₁,y₁,z₁), (x₂,y₂,z₂), (x₃,y₃,z₃): G=((x₁+x₂+x₃)/3, (y₁+y₂+y₃)/3, (z₁+z₂+z₃)/3)

Centroid of tetrahedron with vertices (x₁,y₁,z₁), (x₂,y₂,z₂), (x₃,y₃,z₃), (x₄,y₄,z₄): G=((x₁+x₂+x₃+x₄)/4, (y₁+y₂+y₃+y₄)/4, (z₁+z₂+z₃+z₄)/4). Download the Free PDF for Three Dimensional Geometry for all fundamental formula examples for JEE Main.

Direction Cosines and Direction Ratios — All Formulas

Why l²+m²+n²=1 and Direction Ratio Conversion Are the Most Fundamental Three Dimensional Geometry Results

Direction Cosines (from PDF — Three Dimensional Geometry): If a line makes angles α, β, γ with the positive directions of x-, y-, z-axes respectively, then l = cosα, m = cosβ, n = cosγ are called the direction cosines (d.c.'s) of the line. Denoted [l, m, n]. Key identity: l² + m² + n² = 1 always.

Direction cosines of coordinate axes (from PDF — Three Dimensional Geometry):

X-axis: angles (0°, 90°, 90°) with X, Y, Z → d.c.'s [1, 0, 0]

Y-axis: angles (90°, 0°, 90°) → d.c.'s [0, 1, 0]

Z-axis: angles (90°, 90°, 0°) → d.c.'s [0, 0, 1]

Position by radius vector and direction cosines (from PDF — Three Dimensional Geometry): Let OP=r (radius vector of P). If l, m, n are d.c.'s of OP, then coordinates of P: x = lr; y = mr; z = nr. So r=√(x²+y²+z²) and l=x/r, m=y/r, n=z/r. The direction cosines of OP = (x/r, y/r, z/r) = (x, y, z)/√(x²+y²+z²).

Direction Ratios (from PDF — Three Dimensional Geometry): Any set of numbers a, b, c proportional to l, m, n (i.e., l/a = m/b = n/c = k) are called direction ratios (d.r.'s). Direction ratios are NOT unique — any scalar multiple (λa, λb, λc) represents the same direction.

Converting direction ratios to direction cosines (from PDF — Three Dimensional Geometry JEE Main):

If a, b, c are direction ratios, then direction cosines:

l = ±a/√(a²+b²+c²); m = ±b/√(a²+b²+c²); n = ±c/√(a²+b²+c²)

(The ± sign corresponds to the two opposite directions of the same line.)

Direction cosines of line joining two points (from PDF — Three Dimensional Geometry JEE Main): D.c.'s of line PQ where P=(x₁,y₁,z₁) and Q=(x₂,y₂,z₂):

l = (x₂–x₁)/|PQ|; m = (y₂–y₁)/|PQ|; n = (z₂–z₁)/|PQ|

where |PQ| = √[(x₂–x₁)²+(y₂–y₁)²+(z₂–z₁)²]

Direction ratios of PQ are proportional to (x₂–x₁, y₂–y₁, z₂–z₁). These three differences always serve as d.r.'s — divide by |PQ| to get d.c.'s.

Relation between d.r.'s and d.c.'s (from PDF — Three Dimensional Geometry):

If (a, b, c) are d.r.'s: l/a = m/b = n/c; and l²+m²+n²=1 → l = a/√(a²+b²+c²).

If l, m, n are d.c.'s of two lines, the angle θ between them: cosθ = l₁l₂+m₁m₂+n₁n₂. Download the Free PDF for Three Dimensional Geometry for all direction cosine and direction ratio examples for JEE Main.

Equation of a Straight Line in 3D — All 6 Forms

Why All 6 Forms of Line Equations Are Essential Three Dimensional Geometry JEE Main Formulas

Form 1 — One-Point Vector Form (from PDF — Three Dimensional Geometry):

Line through point with position vector a, parallel to vector b: r = a + tb (t∈ℝ, parameter)

Here b = lî+mĵ+nk̂ gives the direction of the line.

Form 2 — Cartesian Symmetric Form (from PDF — Three Dimensional Geometry JEE Main):

If a = x₁î+y₁ĵ+z₁k̂ and b = lî+mĵ+nk̂ (direction cosines l,m,n OR direction ratios a,b,c):

(x–x₁)/l = (y–y₁)/m = (z–z₁)/n = t

With direction ratios (a,b,c): (x–x₁)/a = (y–y₁)/b = (z–z₁)/c

This is the standard form for Three Dimensional Geometry JEE Main questions. The line passes through (x₁,y₁,z₁) and has direction ratios a:b:c (or direction cosines l:m:n).

Form 3 — Two-Point Vector Form (from PDF — Three Dimensional Geometry):

Line through points A(position vector a) and B(position vector b): r = a + t(b–a)

Equivalently: r = (1–t)a + tb

Form 4 — Two-Point Cartesian Form (from PDF — Three Dimensional Geometry):

Line through P(x₁,y₁,z₁) and Q(x₂,y₂,z₂):

(x–x₁)/(x₂–x₁) = (y–y₁)/(y₂–y₁) = (z–z₁)/(z₂–z₁) = t

Form 5 — Parametric Form (from PDF — Three Dimensional Geometry):

x = x₁ + t(x₂–x₁); y = y₁ + t(y₂–y₁); z = z₁ + t(z₂–z₁) [for line through two points]

Or: x = x₁+lt; y = y₁+mt; z = z₁+nt [for line through one point with d.c.'s l,m,n]

The parameter t represents the distance along the line from the given point.

Form 6 — Lines parallel/perpendicular to axes (Three Dimensional Geometry):

Line parallel to x-axis through (x₁,y₁,z₁): (x–x₁)/1 = (y–y₁)/0 = (z–z₁)/0 → y=y₁ and z=z₁.

Line through (x₁,y₁,z₁) parallel to y-axis: x=x₁ and z=z₁. Parallel to z-axis: x=x₁ and y=y₁.

General: any line with d.r.'s (a,b,c) passing through (x₁,y₁,z₁) is (x–x₁)/a=(y–y₁)/b=(z–z₁)/c. Download the Free PDF for Three Dimensional Geometry for all line equation forms with examples for JEE Main.

Angle Between Lines, Perpendicularity, Parallelism, and Coplanarity

Why Angle Between Lines and Coplanarity of Lines Are Direct Three Dimensional Geometry JEE Main Questions

Angle Between Two Lines (from PDF — Three Dimensional Geometry JEE Main): The angle θ between two lines with direction cosines [l₁,m₁,n₁] and [l₂,m₂,n₂]:

cosθ = |l₁l₂ + m₁m₂ + n₁n₂|

With direction ratios (a₁,b₁,c₁) and (a₂,b₂,c₂):

cosθ = |a₁a₂+b₁b₂+c₁c₂| / [√(a₁²+b₁²+c₁²)·√(a₂²+b₂²+c₂²)]

Perpendicular lines (from PDF — Three Dimensional Geometry): Two lines are perpendicular iff:

l₁l₂ + m₁m₂ + n₁n₂ = 0 [using d.c.'s]; equivalently: a₁a₂ + b₁b₂ + c₁c₂ = 0 [using d.r.'s]

Parallel lines (from PDF — Three Dimensional Geometry): Two lines are parallel iff their d.c.'s (or d.r.'s) are proportional:

l₁/l₂ = m₁/m₂ = n₁/n₂ [d.c.'s]; equivalently: a₁/a₂ = b₁/b₂ = c₁/c₂ [d.r.'s]

Coplanarity of two lines (from PDF — Three Dimensional Geometry JEE Main): Two lines (x–x₁)/a₁=(y–y₁)/b₁=(z–z₁)/c₁ and (x–x₂)/a₂=(y–y₂)/b₂=(z–z₂)/c₂ are coplanar iff:

det[(x₂–x₁) (y₂–y₁) (z₂–z₁); a₁ b₁ c₁; a₂ b₂ c₂] = 0

In vector form: [a₂–a₁, b₁, b₂] = (a₂–a₁)·(b₁×b₂) = 0. This is also the condition for two lines to intersect or be parallel (both cases are coplanar). If the determinant ≠ 0 → lines are skew.

Summary of line relations in Three Dimensional Geometry (JEE Main):

Intersecting lines: coplanar and determinant = 0, and the system has a solution for t, s.

Parallel lines: direction ratios proportional, no common point.

Skew lines: not coplanar, determinant ≠ 0, don't intersect, not parallel.

Point of intersection of two coplanar lines: use parametric equations of both, set equal, solve for t and s, verify consistency. Download the Free PDF for Three Dimensional Geometry for all angle and coplanarity examples for JEE Main.

Shortest Distance Between Skew Lines and Parallel Lines

Why Shortest Distance Between Skew Lines Is a Guaranteed 4-Mark Three Dimensional Geometry JEE Main Question

Shortest Distance Between Two Skew Lines (from PDF — Three Dimensional Geometry): The shortest distance (SD) between two skew lines r=a₁+tb₁ and r=a₂+sb₂ is:

SD = |(a₂–a₁)·(b₁×b₂)| / |b₁×b₂|

Equivalently using scalar triple product: SD = |[a₂–a₁, b₁, b₂]| / |b₁×b₂|

Condition for coplanarity (SD=0): The two skew lines are actually coplanar (intersecting or parallel) iff [a₂–a₁, b₁, b₂] = 0.

Cartesian form of SD between skew lines (from PDF — Three Dimensional Geometry JEE Main): Lines (x–x₁)/a₁=(y–y₁)/b₁=(z–z₁)/c₁ and (x–x₂)/a₂=(y–y₂)/b₂=(z–z₂)/c₂:

SD = |det[(x₂–x₁) (y₂–y₁) (z₂–z₁); a₁ b₁ c₁; a₂ b₂ c₂]| / √[(b₁c₂–b₂c₁)²+(c₁a₂–c₂a₁)²+(a₁b₂–a₂b₁)²]

The denominator is |b₁×b₂| in cartesian form.

Shortest Distance Between Two Parallel Lines (from PDF — Three Dimensional Geometry): If two parallel lines are r=a₁+tb and r=a₂+sb (same direction vector b):

SD = |(a₂–a₁)×b| / |b|

The shortest distance is the perpendicular distance between the two parallel lines.

Working method for SD between skew lines in Three Dimensional Geometry (JEE Main):

Step 1: Identify a₁, b₁ (point and direction of line 1) and a₂, b₂ (point and direction of line 2).

Step 2: Compute b₁×b₂ using the 3×3 determinant: det[î ĵ k̂; components of b₁; components of b₂].

Step 3: Compute a₂–a₁ (difference of the two given points).

Step 4: Compute (a₂–a₁)·(b₁×b₂) — the scalar triple product (numerator).

Step 5: |b₁×b₂| = magnitude of the cross product from step 2 (denominator).

Step 6: SD = |numerator|/denominator. Download the Free PDF for Three Dimensional Geometry for all SD worked examples for JEE Main.

Equation of a Plane — All 6 Forms

Why All 6 Forms of Plane Equations Are Core Three Dimensional Geometry JEE Main Formulas

Form 1 — General Equation of Plane (from PDF — Three Dimensional Geometry):

Every equation of first degree in x, y, z represents a plane: ax + by + cz + d = 0

The direction ratios of the normal to this plane are (a, b, c). This is the most general Three Dimensional Geometry plane form.

Form 2 — Normal Vector Form (from PDF — Three Dimensional Geometry):

r·n̂ = p where n̂ is the unit normal to the plane and p is the perpendicular distance from origin to the plane. Expanded: if n̂ = lî+mĵ+nk̂ then lx+my+nz = p.

In terms of any normal vector n (not unit): r·n = a·n where a is any point on the plane.

Form 3 — Intercept Form (from PDF — Three Dimensional Geometry JEE Main):

x/a + y/b + z/c = 1 where a, b, c are the x-, y-, z-intercepts of the plane (the plane cuts the axes at A(a,0,0), B(0,b,0), C(0,0,c)).

Form 4 — Normal Cartesian Form (from PDF — Three Dimensional Geometry):

lx + my + nz = p where l,m,n are d.c.'s of the normal and p = perpendicular distance from origin. This is derived by taking P(x,y,z) on plane, N as foot of perpendicular from O with ON=p, and using ON = projection of OP on normal direction ON. Converting from ax+by+cz+d=0: divide by ±√(a²+b²+c²) such that constant term is positive.

Form 5 — One-Point Vector Form (from PDF — Three Dimensional Geometry JEE Main):

Plane through point A(position vector a), perpendicular to vector n: (r – a)·n = 0

Expanded: r·n = a·n. If n=aî+bĵ+ck̂ and point is (x₀,y₀,z₀): a(x–x₀)+b(y–y₀)+c(z–z₀)=0 → ax+by+cz=ax₀+by₀+cz₀.

Form 6 — Three-Point Determinant Form (from PDF — Three Dimensional Geometry JEE Main):

Plane through three non-collinear points (x₁,y₁,z₁), (x₂,y₂,z₂), (x₃,y₃,z₃):

det[(x–x₁) (y–y₁) (z–z₁); (x₂–x₁) (y₂–y₁) (z₂–z₁); (x₃–x₁) (y₃–y₁) (z₃–z₁)] = 0

Alternatively: det[x y z 1; x₁ y₁ z₁ 1; x₂ y₂ z₂ 1; x₃ y₃ z₃ 1] = 0 (4×4 determinant form)

These three points uniquely determine a plane. Download the Free PDF for Three Dimensional Geometry for all 6 plane equation forms with examples for JEE Main.

Angles Between Planes, Line and Plane — All Formulas

Why Angle Formulas Between Planes and Line-Plane Are Direct JEE Main Three Dimensional Geometry Questions

Angle Between Two Planes (from PDF — Three Dimensional Geometry JEE Main): The angle θ between planes with normal vectors n₁ and n₂:

cosθ = |n₁·n₂| / (|n₁||n₂|)

For planes a₁x+b₁y+c₁z+d₁=0 and a₂x+b₂y+c₂z+d₂=0 with normal vectors n₁=(a₁,b₁,c₁), n₂=(a₂,b₂,c₂):

cosθ = |a₁a₂+b₁b₂+c₁c₂| / [√(a₁²+b₁²+c₁²) · √(a₂²+b₂²+c₂²)]

Perpendicular planes (from PDF — Three Dimensional Geometry): θ=90° → n₁⊥n₂ → a₁a₂ + b₁b₂ + c₁c₂ = 0

Parallel planes (from PDF — Three Dimensional Geometry): n₁ || n₂ → a₁/a₂ = b₁/b₂ = c₁/c₂

Angle Between a Line and a Plane (from PDF — Three Dimensional Geometry JEE Main): The angle φ between a line with direction ratios (l,m,n) and the plane with normal having direction ratios (a,b,c):

sinφ = |al + bm + cn| / [√(a²+b²+c²) · √(l²+m²+n²)]

Note: The angle between a line and a plane is the complement of the angle between the line and the normal to the plane. If θ is the angle between line and normal: cosθ = (al+bm+cn)/[√(a²+b²+c²)·√(l²+m²+n²)], and the angle with the plane = 90°–θ → sinφ = cosθ.

Special cases for line and plane (from PDF — Three Dimensional Geometry JEE Main):

(1) Line is parallel to plane: φ=0° → sinφ=0 → al + bm + cn = 0

(2) Line is perpendicular to plane (φ=90°): line is parallel to normal → direction ratios proportional: a/l = b/m = c/n

Line lying in the plane (Three Dimensional Geometry): A line lies in a plane iff: (i) the direction of the line is perpendicular to the normal (al+bm+cn=0), AND (ii) a point on the line lies on the plane. Both conditions must hold simultaneously. Download the Free PDF for Three Dimensional Geometry for all angle examples for JEE Main.

Distances — Point to Plane, Parallel Planes, Plane Through Intersection, Foot and Image

Why Distance Formulas and Foot/Image Results Are the Most Calculation-Heavy Three Dimensional Geometry JEE Main Results

Perpendicular Distance from Point to Plane (from PDF — Three Dimensional Geometry JEE Main):

Distance from P(x₁,y₁,z₁) to plane ax+by+cz+d=0:

d = |ax₁+by₁+cz₁+d| / √(a²+b²+c²)

In vector form: distance from point r₁ to plane r·n=p: d = |r₁·n–p| / |n|.

Derivation from normal form: convert to lx+my+nz=p, then d = |lx₁+my₁+nz₁–p|.

Distance from origin to ax+by+cz+d=0: d = |d|/√(a²+b²+c²) [put x₁=y₁=z₁=0].

Distance Between Two Parallel Planes (from PDF — Three Dimensional Geometry):

Between ax+by+cz+d₁=0 and ax+by+cz+d₂=0 (same normal direction):

d = |d₁–d₂| / √(a²+b²+c²)

In vector form: between r·n=d₁ and r·n=d₂: d = |d₁–d₂|/|n|.

Plane Through the Intersection of Two Planes (from PDF — Three Dimensional Geometry JEE Main):

Family of planes through the intersection of P₁: a₁x+b₁y+c₁z+d₁=0 and P₂: a₂x+b₂y+c₂z+d₂=0:

P₁ + λP₂ = 0 i.e., (a₁x+b₁y+c₁z+d₁) + λ(a₂x+b₂y+c₂z+d₂) = 0 for any λ∈ℝ.

Vector form: (r·n₁–d₁) + λ(r·n₂–d₂) = 0.

Use: if an additional condition is given (e.g., plane passes through a specific point, or is perpendicular to another plane), substitute to find λ. Then write the specific plane from the family.

Foot of Perpendicular from Point to Plane (from PDF — Three Dimensional Geometry JEE Main):

Foot of perpendicular (p,q,r) from point (x₁,y₁,z₁) to plane ax+by+cz+d=0:

(p–x₁)/a = (q–y₁)/b = (r–z₁)/c = –(ax₁+by₁+cz₁+d)/(a²+b²+c²)

So: p = x₁ – a(ax₁+by₁+cz₁+d)/(a²+b²+c²); q = y₁ – b(ax₁+by₁+cz₁+d)/(a²+b²+c²); r = z₁ – c(ax₁+by₁+cz₁+d)/(a²+b²+c²).

Image of Point in a Plane (from PDF — Three Dimensional Geometry JEE Main):

Image (α,β,γ) of point (x₁,y₁,z₁) in plane ax+by+cz+d=0:

(α–x₁)/a = (β–y₁)/b = (γ–z₁)/c = –2(ax₁+by₁+cz₁+d)/(a²+b²+c²)

So: α = x₁ – 2a(ax₁+by₁+cz₁+d)/(a²+b²+c²) [same pattern as foot formula but ×2 instead of ×1].

Key mnemonic: foot formula uses factor k = –(ax₁+by₁+cz₁+d)/(a²+b²+c²); image formula uses 2k. The image is the reflection of the point through the plane — the foot is the midpoint of the original point and its image. Download the Free PDF for Three Dimensional Geometry for all distance, foot, and image examples for JEE Main.